Structural Analysis Example Problems
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Nguyễn Sơn LâmStructural Analysis Example Problems
Uploaded by
Nguyễn Sơn LâmStructural Analysis
Prof. Victor Saouma
Prepared by Christopher Segura
Structural Analysis Example Problems
Victor E. Saouma
Prepared by Christopher Segura
Summer 2011
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Structural Analysis
Prof. Victor Saouma
Prepared by Christopher Segura
List of Examples
2.1 Stress Vectors
2.2 Stress Vectors 2
2.3 Principal Stresses
2.4 Stress Tensor Operations
2.5 Stress Transformation
2.6 Stress Transformation 2
2.7 Strain Invariants & Principal Strains
2.8 Equilibrium Equation
2.9 Stress-Strain
2.10 Stress-Strain 2
3.1 Displacement by Double Integration
4.1 Live Load Reduction
4.2 Wind Load
4.3 Earthquake Load on a Frame
4.4 Earthquake Load on a Tall Building
4.5 Hysdrostatic Load
4.6 Thermal Expansion/Stress
5.1 Statically Indeterminate Cable Structure
5.2 Simply Supported Beam
5.3 Parabolic Load
5.4 Three Span Beam
5.5 Three Hinged Gable Frame
5.6 Inclined Supports
7.1 Truss, Method of Joints
7.2 3D Truss
7.3 Truss I, Matrix Method
7.4 Truss II, Matrix Method
7.5 Truss, Method of Sections
8.1 Funicular Cable Structure
8.2 Design of Suspension Bridge
10.1 Simple Shear and Moment Diagram
10.2 Frame Shear and Moment Diagram
10.3 Frame Shear and Moment Diagram: Hydrostatic Load
10.4 Shear and Moment Diagrams for Frame
10.5 Shear and Moment Diagrams for Inclined Frame
10.6 3D Frame
11.1 Three Hinged Arch, Point Loads
11.2 Semi-Circular Arch
11.3 Statically Indeterminate Arch
11.4 Semi-circular Box Girder
11.5 Internal Forces in an Helicoidal Cantilevered Girder, Point Load
11.6 Internal Forces in an Helicoidal Cantilevered Girder, Uniform Load
12.1 Moment Area, Cantilevered Beam
12.2 Moment Area, Simply Supported Beam
12.3 Maximum Deflection
12.4 Frame Deflection
12.5 Frame Subjected to Temperature Loading
12.6 Conjugate Beam
12.7 Conjugate Beam 2
13.1 Deflection of a Cantilever Beam
13.2 Beam Deflection
13.3 Deflection of a Frame
13.4 Rotation of a Frame
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13.5 Truss Deflection
13.6 Torsional and Flexural Deformation
13.7 Flexural and Shear Deformation in a Beam
13.8 Thermal Effects in a Beam
13.9 Deflection of a Truss
13.10 Thermal Deflection of a Truss: I
13.11 Thermal Deflection of a Truss: II
13.12 Truss with Init ial Camber
13.13 Prestressed Concrete Beam with Continuously Variable I
14.1 Steel Building Frame Analysis
14.2 Analysis of Irregular Building Frame
14.3 Redundant Truss Analysis
14.4 Truss with Two Redundants
14.5 Analysis of Nonprismatic Members
14.6 Fixed End Moments for Nonprismatic Beams
14.7 Rectangular Frame, External Load
14.8 Frame with Temperature and Support Displacements
14.9 Braced Bent with Loads and Temperature Change
15.1 Approximate Analysis of a Frame Subjected to Vertical and Horizontal Loads
16.1 Propped Cantilever Beam
16.2 Two-Span Beam, Slope Deflection
16.3 Two-Span Beam, Slope Deflection, Initial Deflection
16.4 Frames, Slope Deflection
16.5 Box Culvert, Slope Deflection
16.6 Continuous Beam
16.7 Continuous Beam, Simplified Method
16.8 Continuous Beam, Inintial Settlement
16.9 Frame
16.10 Frame with Side Load
17.1 Beam
17.2 Frame
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2.1 Stress Vectors
For the following stress tensor at point P
7 -5 0
σ := -5 3 1
0 1 2
t1
7 -5 0
t := σ = -5 3 1
2
t3 0 1 2
Determine the traction (or stress vector) t passing through P and parallel to the plane ABC where A(4,0,0),
B(0,2,0) and C(0,0,6)
Solution:
The vector normal to the plane can be found by taking the cross products of vectors AB and AC
e 1 e2 e 3
N = AB AC = -4 2 0 = 12 e + 24 e + 8 e
1 2 3
-4 0 6
The unit normal of N is given by
3 6 2
n= e + e + e
7 1 7 2 7 3
n := 3 6 2
7 7 7
Hence the stress vector traction will be
t := n σ = -
9 5 10
7 7 7
9 5 10
and thus t = - e + e + e
7 1 7 2 7 3
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2.2 Stress Vectors 2
The state of stress through a continuum is given with respect to the cartesian axes by
3 x x 5 x 2 0
1 2 ( 2)
2
( )
σ( x) := 5 x 2 0 2 x
3
MPa
0 2x 0
3
Determine the stress vector at P( 1 , 1 , 3) of the plane that is normal to the tangent to the cylindrical surface
(x2) + (x3)
2 2
= 4 at P
Solution:
Vector Equation
( 2) + (x3)
2 2
v( x) := x -4
Gradient
0
2 x
Vect1( x) := ∇ v( x) 2
x
2 x
3
For x := 1
2
x =0
1
x =1
2
( 2)
2
x := 4- x = 1.732
3
Normalized Vector
0
= 0.5
x
n :=
x
0.866
2.5
σ( x) n = 3 MPa
1.732
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2.3 Principal Stresses
The stress tensor is given at a point by point P
3 1 1
σ := 1 0 2
1 2 0
Determine the principal stress values and the corresponding directions
Solution:
3 - λ 1 1
1 0 -λ 2 =0
1 2 0 - λ
Simplifying the determinant
3 - λ 1 1
1 0 - λ 2 = 0 simplify -( λ - 1 ) ( λ + 2) ( λ - 4) = 0
1 2 0 - λ
thus the roots are
1
λ := -( λ - 1 ) ( λ + 2 ) ( λ - 4 ) = 0 solve -2
4
1
λ = -2
4
Note that those are the three eigenvalues of the stress vector. If we let the x1 axis be the one corresponding to the
direction of λ2 and ni2 be the direction cosines of this axis, then we have
( 3 - λ 2) n 1
2 2 2 2 2 2
+ n 2 + n 3 = 0 5 n1 + n2 + n3 = 0
2 2 1 2 1
n3 = -
( )
2 2 2 2 2 2 n1 = 0 n2 =
n 1 + 0 - λ n 2 + 2n 3 = 0 n 1 + 2 n 2 + 2 n 3 = 0 2 2
2
( )
2 2 2 2 2 2
n1 + 2 n 2 + 0 - λ n 3 = 0 n1 + 2 n 2 + 2 n 3 = 0
2
Similarly, if we let x2 be the one corresponding to the direction of λ1 and ni1 be the direction cosines of this axis,
( 3 - λ 1) n 1
2 2 2 2 2 2
+ n 2 + n 3 = 0 2 n1 + n2 + n3 = 0
1 1 1 1 1 1
n2 = - n3 = -
( )
2 2 2 2 2 2 n1 =
n 1 + 0 - λ n 2 + 2n 3 = 0 n 1 - n 2 + 2 n 3 = 0 3 3 3
1
( )
2 2 2 2 2 2
n1 + 2 n 2 + 0 - λ n 3 = 0 n1 + 2 n 2 - n 3 = 0
1
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Finally, if we let x3 be the one corresponding to the direction of λ3 and ni3 be the direction cosines of this axis,
( 3 - λ 3) n 1
2 2 2 2 2 2
+ n 2 + n 3 = 0 -n 1 + n 2 + n 3 = 0
3 2 3 1 3 1
n1 = - n2 = - n3 = -
( )
2 2 2 2 2 2
n 1 + 0 - λ n 2 + 2n 3 = 0 n 1 - 4 n 2 + 2 n 3 = 0 6 6 6
3
( )
2 2 2 2 2 2
n1 + 2 n 2 + 0 - λ n 3 = 0 n1 + 2 n 2 - 4 n 3 = 0
3
0 1
-
1
2 2
1 1 1
n pd := -
3 3 3
2 1 1
- - -
6 6 6
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2.4 Stress Tensor Operations
For the following stress tensor
6 -3 0
σ := -3 6 0
0 0 8
a) Determine directly the three Invariants Iσ, IIσ and IIIσ
b) Determine the principal stresses and principal stress directions
c) Show that the transformation tensor of direction cosines transforms the original stress tensor into the diaganol
axes stress tensor
d) Recompute the three invariants from the principal stresses
e) Split the stress tensor into its spherical and deviator parts
f) Show that the first invariant of the deviator is zero
Solution:
a) Iσ := σ +σ +σ = 20
1, 1 2, 2 3, 3
IIσ := σ σ +σ σ +σ σ -σ σ = 123
1, 1 2, 2 2, 2 3, 3 3, 3 1, 1 1, 2 2, 1
IIIσ := σ = 216
3
b) σp := eigenvals( σ) = 8
9
-0.707 0 -0.707
n p := eigenvecs( σ) = -0.707 0 0.707
0 1 0
3 0 0
n p σ n p = 0 8 0
c) T
0 0 9
d) Iσpd := σp + σp + σp = 20
1 2 3
IIσpd := σp σp + σp σp + σp σp = 123
1 2 2 3 3 1
IIIσpd := σp σp σp = 216
1 2 3
Iσpd
e) σmean := = 6.667
3
σmean 0 0
6.667 0 0
σsperical := 0 σmean 0 = 0 6.667 0
0 0 σmean
0 0 6.667
-0.667 -3 0
f)
σdeviator := σ - σsperical = -3 -0.667 0
0 0 1.333
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2.5 Stress Transformation
Show that the transformation tensor of direction cosines for the stress tensor
3 1 1
σ := 1 0 2
1 2 0
transforms the original stress tensor into the diaganol principal axes stress tensor
Solution:
0 1
-
1
2 2
1 1 1
n pd := - -
3 3 3
2 1 1
- - -
6 6 6
T
σ := n pd σ n pd =
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2.6 Stress Transformation 2
The octahedral plane is the plane which makes equal angles with the principal stress directions. Show that the
shear stress on this plane, the so-called octahedral shear stress, is given by
1
( ) ( σI - σII) ( ) ( )
2 2 2
σoct σI , σII , σIII := + σII - σIII + σIII - σI
3
Solution:
1
v := 1
1
T
n := v = ( 0.577 0.577 0.577 )
v
σI 0 0
σ( σI , σII , σIII) := 0 σII 0
0 0 σ
III
( ) ( ) (
t σI , σII , σIII := n σ σI , σII , σIII 0.577 σI 0.577 σII 0.577 σIII )
( ) (
tn σI , σII , σIII := n t σI , σII , σIII )T 0.333 σI + 0.333 σII + 0.333 σIII
(
t shear σI , σII , σIII := ) ( )(
t σI , σII , σIII t σI , σII , σIII )T - tn(σI, σII , σIII) tn(σI , σII , σIII)
( )
2 2 2
t shear σI , σII , σIII simplify 0.222 σI - 0.222 σI σII - 0.222 σI σIII + 0.222 σII - 0.222 σII σIII + 0.222 σIII
( ) (
t shear σI , σII , σIII - σoct σI , σII , σIII = 0 )
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2.7 Strain Invariants & Principal Strains
Determine the planes of principal strains for the following strain tensor
1 3 0
E := 3 0 0
0 0 1
Solution:
The strain invariants are given by
IE = Eii = 2
1
IIE =
2
( )
Eij Eij - Eii Ejj = -1 + 3 = 2
IIIE = Eij = -3
The principal strains by
1 - λ 3 0
1 + 13 1 - 13
Eij - λδij = 3 0 - λ 0 = ( 1 - λ) λ - λ -
0 2 2
0 1 - λ
1+ 13
E1 = λ 1 = = 2.3
2
E2 = λ 2 = 1
1 - 13
E3 = λ 3 = = -1.3
2
1+ 13 1
The eigenvectors for E1 = give the principal directions n
2
1 + 13 13 1
1 - 3 0 -n 1 , 1 - + 3 n 2 , 1
2 2 2
n1 , 1
1 + 13 1
3 0- 0
n , 1 -n 2 , 1
13
+ + 3 n1 , 1
2 2 2 2
n3 , 1
1 + 13
13 1
0 0 1- -n 3 , 1 -
2 2 2
13 1
-n 1 , 1 2 - 2 + 3 n 2 , 1
0
-n , 1 13 + 1 + 3 n , 1 = 0
2 2 2 1
0
13 1
-n 3 , 1 -
2 2
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1+ 13
n1 , 1 = n2 , 1
2 3
n3 , 1 = 0
1 1
n n = 1
1 + 2 13 + 13
+ 1 ( n 2 , 1 ) = 1
2
12
n 2 , 1 = 0.6
1+ 13 1+ 13
n1 , 1 = n2 , 1 = 0.6 = 0.8
2 3 2 3
1
n = ( 0.8 0.6 0 )
For the second eigenvector λ2=1
n ,2
1 - 1 3 0 1 3 n 2 , 2
3 0 - 1 0 n 2 , 2 -( n 2 , 2 ) + 3 n 1 , 2
0 1 - 1 n 3 , 2
0
0
3 n2 , 2 0
-( n 2 , 2 ) + 3 n 1 , 2 = 0
0
0
2
n = (0 0 1 )
Finally, the third eigenvector can be obtained by the same manner, but more easily from
e1 e2 e3
3 1 2
n = n n = det 0.8 0.6 0 = 0.6e1 - 0.8e2
0
0 1
Therefore
n1
0.8 0.6 0
j 2
ai = n a := 0 0 1
3 0.6 -0.8 0
n
0.8 0.6 0 1 3 0 0.8 0 0.6 2.3 0 0
a E a = 0 0 1 3 0 0 0.6 0 -0.8 = 0 1 0
T
0.6 -0.8 0 0 0 1 0 1 0 0 0 -1.3
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2.8 Equilibrium Equations
In the absence of body forces, does the following stress distribution
x 2 + ν x 2 - x 2
( 2) ( 1) ( 2)
-2 ν x x
1 2
0
x 2 - x 2
( 1) ( 2) ( 1)
2
T( x , ν) = -2 ν x x x + ν 0
1 2
0 0
ν (x ) + (x )
2 2
1 2
where ν is a constant, satisfy equilibrium?
Solution:
T1j = T11 + T12 + T13 = 2 ν x - 2 ν x = 0
x x x x 1 1
j 1 2 3
T2j = T21 + T22 + T23 = -2 ν x + 2 ν x = 0
x x x x 2 2
j 1 2 3
T3j = T31 + T32 + T33 = 0
x x x x
j 1 2 3
Therefore, equilibrium is satisfied.
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2.9 Stress-Strain
Determine the stress tensor at a point where the Lagrangian strain tensor is given by
30 50 20
E := 50 40 0 10
-6
20 0 30
and the material is steel with λ := 119.2GPa and μ := 79.2GPa
Solution:
λ + 2 μ λ λ 0 0 0 277.6 GPa 119.2 GPa 119.2 GPa 0 0 0
λ λ + 2 μ λ 0 0 0 119.2 GPa 277.6 GPa 119.2 GPa 0 0 0
λ λ λ + 2 μ 0 0 0 119.2 GPa 119.2 GPa 277.6 GPa 0 0 0
C := simplify
0 0 0 2 μ 0 0 0 0 0 158.4 GPa 0 0
0 0 0 0 2 μ 0
0 0 0 0 158.4 GPa 0
0 0 0 0 0 2 μ 0 0 0 0 0 158.4 GPa
E1 , 1 3 10- 5
E2 , 2 4 10- 5
E3 , 3 3 10- 5
γ := =
E1 , 2 5 10- 5
E -5
1, 3 2 10
E 0
2, 3 0 10
0.01667
0.01826
0.01667
σ := C γ = GPa
0.00792
0.00317
0
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2.10 Stress-Strain
Determine the stress tensor at a point where the Cauchy stress tensor is given by
100 42 6
σ := 42 -2 0 MPa
6 0 15
with E := 207GPa , μ := 79.2GPa, and ν := 0.3
Solution:
1 -ν -ν 0 0 0 1 -
0.001
-
0.001
0
E E E
207 GPa GPa GPa
0 0
-ν 1 -ν 0 0 0 - 0.001 1
-
0.001
0 0 0
E E E GPa 207 GPa GPa
ν ν 1 0.001 0.001 1
- - 0 0 0 - - 0 0 0
E E E simplify GPa GPa 207 GPa
S :=
1 0.013
0 0 0 μ 0 0 0 0 0
GPa
0 0
0 0 0 0 1 0 0 0 0 0
0.013
0
μ GPa
1 0.013
0 0 0 0 0 0 0 0 0 0
μ GPa
σ1 , 1
σ2 , 2 100
-2
σ3 , 3 15
s := = MPa
42
σ1 , 2
6
σ
1, 3 0
σ
2, 3
4.643 10- 4
-1.763 10- 4
-6.957 10- 5
γ := S s =
5.303 10- 4
-5
7.576 10
0
0 10
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γ
4
γ 5
γ1
2 2 4.643 10- 4 2.652 10- 4 3.788 10- 5
γ γ
E := 4 6 - 4 - 4 0
γ = 2.652 10 -1.763 10 0 10
2
2 2
-5 0 - 5
γ γ 3.788 10 0 10 -6.957 10
5 6
γ
3
2 2
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3.1 Displacement by Double Integration
Determine the deflection at B for the following cantilevered beam.
B
A
2L/3 L/3
Solution:
2L
At 0 x
3
1. Moment Equation
2 w L 5
2
E I y = Mx = x - w L
2 3 18
x
2. Integrate Once
w L 2 5 2
E I y = x - w L x + C 1
x 6 18
However we have at x=0, dy/dx=0, C1=0
3. Integrate Twice
2
w L 3 5 w L 2
E I y = x - x + C2
18 36
Again we have at x=0, y=0, C2=0
2L
At xL
3
1. Moment Equation
x - 2L
2 3
w L - w x -
w L 2 5 2L
E I
2
y = Mx = x -
x 3 18 3 2
2. Integrate Once
3
w L 2
w L x - x -
5 2 w 2L
E I y = x - + C3
x 6 18 6 3
2L dy
Applying the boundary condition at x = , we must have equal to the value coming frmo the left C =0
3 dx 3
3. Integrate Twice
2 4
w L 3 5 w L 2 w 2L
E I y = x - x - x - + C4
18 36 24 3
Again following the same argument as above, C4=0
Substituting for x=L we obtain
4
163 wL
y =
1944 E I
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4.1 Live Load Reduction
A four storey office building has interior columns spaced 30ft apart in the two directions. If the flat roof loading is
50 lb/ft2, determine the reduced live load supported by a typical interior column located on the ground level
Solution:
L0 := 50psf
2
AT := 30ft 30ft = 900 ft
KLL := 4
Lfloor L0 , AT , KLL := L0 0.25 + psf
15
( ) KLL AT
L0 AT
Lfloor := Lfloor , , KLL = 25 psf
psf ft2
Lfloor
Reduction := = 50 %
L0
Lroof := 0.6 L0 = 30 psf
( )
F1 := 3 Lfloor AT + Lroof AT = 94.5 kip
Note that without reduction the total load woul dhave been
F2 := 4 L0 AT = 180 kip
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4.2 Wind Load
Wind blows on the side of the fully enclosed agricultural building located on open flat terrain in Oklahoma. Determine the external
pressure acting on the roof. Also, what is the internal pressure in the building which acts on the roof? Use linear interpolation to
determine qh and Cp.
100 15 ft
ft 10°
50 ft
Solution:
( ) 2
q z Kz , KZt , Kd , V , I := 0.00256 Kz KZt Kd V I psf
KZt := 1
Kd := 1
V := 90mph
I := 0.87
q z Kz , KZt , Kd , , I 18.04 Kz psf
V
mph
Kz15 := 0.85
q z15 := q z Kz15 , KZt , Kd ,
V
, I = 15.334 psf
mph
Kz20 := 0.9
q z20 := q z Kz20 , KZt , Kd ,
V
, I = 16.236 psf
mph
1
h := 15ft + 25ft tan( 10deg) = 17.204 ft
2
q h - q z15 q z20 - q z15
=
h - 15ft 20ft - 15ft
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q z20 - q z15
q h := ( h - 15ft) + q z15
20ft - 15ft
q h = 15.732 psf
External pressure on windward side of roof
( )
p q h , G , C p := q h G Cp psf
G := 0.85
L := 50ft
h
= 0.344
L
-0.9 - -0.7 -0.9 - Cp
=
0.5 - 0.25 h
0.5 -
L
-0.9 - -0.7
0.5 -
h
Cp := -0.9 -
0.5 - 0.25 L
Cp = -0.775
qh
p Windward := p , G , Cp = -10.367 psf
psf
External pressure on leeward side of roof
-0.5 - -0.3 -0.5 - Cp
=
0.5 - 0.25 h
0.5 -
L
-0.5 - -0.3
0.5 -
h
Cp := -0.5 -
0.5 - 0.25 L
Cp = -0.375
qh
p Leeward := p , G , Cp = -5.018 psf
psf
Internal pressure
( )
p int q h , G , Cpi := -q h G Cpi
Cpi1 := 0.18
Cpi2 := -0.18
( )
p int1 := p int q h , G , C pi1 = -2.407 psf
p int2 := p int( q h , G , C pi2) = 2.407 psf
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4.3 Earthquake Load on a Frame
Determine the approximate earthquake forces for the ductile hospital frame structure shown below. The dead load for each floor is
lbf lbf
DL := 200 and the live load is LL := 400 . The structure is built on soft clay. Use DL+0.5LL as the weight of each floor. The
ft ft
building is in seismic zone 3.
400 lb/ft
12 ft
400 lb/ft
12 ft
20 ft
Solution:
1. The fundamental period of vibration is
3
( )
4
T Ct , h n := Ct h n sec
Ct := 0.03
h := 24ft
T := T Ct , = 0.325s
h
ft
2. The C coefficient is
1.25 S
C( S , T) :=
2
3
T
S := 2.0
C := C S , = 5.286 > 2.75
T
sec
Use C := 2.75
3. The other coefficients are Z := 0.3, I := 1.25, R W := 12
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4. Check
C 2.75
= = 0.23 > 0.075 OK!
RW 12
5. The total vertical load is
L := 20ft frame width
W := 2 ( DL + 0.5 LL) L = 16000 lbf
6. The total seismic base shear is
Z I C
V := W = 1375 lbf
RW
7. Since T < 0.7sec, there is no whiplash
8. The toal load on each floor is given by
h 1 := 12ft
h 2 := 24ft
V h 2
F2 := = 916.7 lbf
h1 + h2
V h 1
F1 := = 458.3 lbf
h1 + h2
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4.4 Earthquake Load on a Tall Building
Determine the approximate critical lateral loading for a 25 storey, ductile, rigid space frame concrete structure in the short direction. The rigid
frames are spaced 25 ft apart in the cross section and 20 ft in the longitudinal direction. The plan dimension of the building is 175x100 ft, and
the structure is 25(12ft)=300 ft high. This office building is located in an urban environment with a wind velocity of V := 70mph and in seismic
zone 4. For this investigation, an average building total dead load of DL := 192psf is used. Soil conditions are unknown.
Solution:
1. The total building weight is
L := 100ft
B := 175ft
W := DL L B 25 = 84000 kip
2. The fundamental period of vibration for a rigid frame is
3
( )
4
T Ct , h n := Ct h n sec
Ct := 0.03
h := 25 12ft = 300 ft
T := T Ct , = 2.16 s > 0.7s OK!
h
ft
3. The C coefficient is
1.25 S
C( S , T) :=
2
3
T
S := 1.5
C := C S , = 1.12 < 2.75
T
sec
4. The other coefficients are Z := 0.4, I := 1, RW := 12
5. Check
C 1.12
= = 0.093 > 0.075 OK!
RW 12
6. The total seismic base shear along the critical short direction is
Z I C
V := W = 3139 kip
RW
7. Since T > 0.7sec, the whiplash effect must be considered
T
Ft := 0.07 V = 475 kip
sec
Hence the total triangular load is
Ftot := V - Ft = 2664 kip
8. Let us check if wind load governs. From table xx we conservatively assume a uniform wind pressure of 29 psf resulting in a total lateral
force of
p := 29psf
Pw := p h B = 1522.5 kip < 3108 kip
The magnitude of the total seismic load is clearly larger than the totoal wind force
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4.5 Hydrostatic Load
The basement of a building is 12 ft below grade. Ground water is located 9 ft below grade. What thickness concrete slab is
required to exactly balance the hydrostatic uplift?
Solution:
The hydrostatic pressure must be countered by the pressure caused by the weight of the concrete. Since p=γh we equate the two
pressures and solve for h, the height of the concrete slab.
h := 12ft
h water := 9ft
d := h - hwater = 3 ft
lbf
γwater := 62.4
2
ft
lbf
γconcrete := 150
2
ft
γwater d = γconcrete h
γwater d
h := = 15 in
γconcrete
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4.6 Thermal Expansion/Stress
-6 in
A low-rise building is enclosed along one side by a 100 ft-long clay masonry ( α := 3.6 10 , E := 2400000psi)bearing wall. The
in Δ°F
structure was built at a temperature of T := 60Δ°F and is located in the northern part of the United States where the temperature range is
between Tlow := -20 Δ°F and Thigh := 120 Δ°F.
Solution:
1. Assume that the wall can move freely with no restraint from cross-walls and foundations. The wall expansion and contraction
(summer and winter) are given by
ΔL = α ΔT L
( )
ΔLSummer := α Thigh - T 100ft = 0.26 in
( )
ΔLWinter := α Tlow - T 100ft = -0.35 in
2. We now assume (conservatively) that the free movement cannot occur (ΔL=0) hence the resulting stress would be equal to
ΔL α ΔT L
σ = E ε = E = E = E α ΔT
L L
σ = E α ΔT
( )
σSummer := E α Thigh - T = 518 psi
( )
σWinter := E α Tlow - T = -691 psi
Note the tensile stress being beyond the masonry capacity. Cracking will occur.
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5.1 Statically Indeterminate Cable Structures
A rigid plate is supported by two aluminum cables and a steel one. Determine the force in each cable.
Al S teel Al h
P
If the rigid plate supports a load P, determine the stress in each of the three cables.
Solution:
1. We have three unknowns and only two independent equations of equilibrium. Hence the problem is statically indeterminate to
the first degree.
Mz = 0 PAl = PAl
left right
Fy = 0 2PAl + PSt = P
Thus we effectively have two unknowns and one equation
2. We need to have a third equation to solve for the three unknowns. This will be derived from the compatability of the
displacements in all three cables i.e. all three displacements must be equal:
P
σ=
A
ΔL
ε =
L
σ
ε =
E
Combine to obtain
PL
ΔL =
AE
PAl L PSt L PAl ( EA )
Al
= =
EAl AAl ESt ASt PSt ( EA )
St
or -( EA ) PAl + ( EA ) PSt = 0
St Al
3. Solution of this system of two equations with two unknowns yield:
2 1 PAl P
=
-( EA ) ( EA )
St Al St
P
0
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-1
PAl 2 1 P ( EA ) Al -1 P
=
1
=
PSt -( EA ) St ( EA ) Al
0 2 ( EA ) + ( EA ) ( EA ) 2 0
Al St St
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5.2 Simply Supported Beam
Determine the reactions of the simply supported beam shown below.
60 k
4 k/ft
36 k
a b c d
6 ft 6 ft 6 ft
The beam has 3 reactions. We have 3 equations of static equilibrium. Hence it is statically determinate.
Solution:
Fx = 0 Rax - 36kip = 0
kip
Fy = 0 Ray + Rdy - 60kip - 4 ft 12ft = 0
Mz = 0 12 Ray - 6Rdy - 60kip 6ft = 0
c
or
1 0 0 ax
R
36kip
0 1 1 R = 108kip
ay
0 12 -6 Rdy
360kip
ax 36kip
R
Ray = 56kip
R 52kip
dy
Alternatively we could have used another set of equations:
kip
a
M z = 0 60kip 6 ft + 4 12ft 12ft - Rdy 18ft = 0 Rdy = 52kip
ft
kip
Mz
d
=0 Ray 18ft - 60kip 12ft - 4 12ft 12ft = 0 Ray = 56kip
ft
Check:
Fy = 0 56kip + 52kip - 60kip - 48kip = -0 kip
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5.3 Parabolic Load
2
Determine the reactions of the simply supported beam of length L subje cted to a parabolic load w = w0
x
L
dW
x
w = w 0 ( )2
L
A B
dx
L
Solution:
Since there are no axial forces, there are two unknowns and two equations of equilibrium. We have two equations of equilibrium (
Fy
and
M), we judiciously start with the second one, as it would directly give us the reaction at B. Considering an infinitesimal element of
length dx, weight dW, an moment
x=L dM:
2
w0 dx x - RB L = 0
x
A
Mz = 0
L
x=0
L4 1
w0 = w L
1
RB =
L 2 4 0
4L
With RB determined, we solve for RA from
Fy = 0
x=L
2
w0 dx = 0
1 x
RA + w0 L -
4
L
x=0
w0 L3 1 1
RA = - w0 L = w L
2 3 4 12 0
L
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5.4 Three Span Beam
Determine the reactions of the following three span beam
40 k 50 k 30 k
4 k/ft
4
3
a c d
b
4 ft 5 ft 3 ft 3 ft 8 ft 2 ft 2 ft
40 k 40 k 30 k
4 k/ft
30 k
R ax F F
a c d
b
R ay R cy R dy
S S
4 ft 5 ft 3 ft 3 ft 8 ft 2 ft 2 ft
Solution:
We have 4 unknowns (R ,R ,R , and R ), three equations of equilibrium and one equation of condition (
ax ay cy dy Mb = 0), thus the structure is
statically determinate. Though there are many approaches to solve for those four unknowns (all of them correct), a few are simpler to pursue.
In this case, it is easiest to "break" the structure into substructures and examine the free body diagram of each one of them separately.
1. Isolating ab:
40kip 5ft
Mz
b
=0 9ft R ay - 40kip 5 ft = 0 Ray := = 22.2 kip
9ft
40kip 4 ft
Mz = 0
a
40kip 4 ft - S 9 ft = 0 S := = 17.8 kip
9ft
Fx = 0 Rax := 30kip
2. Isolating bd:
kip
Mz
d
=0 -S 18ft - 40kip 15ft - 4 12ft 6 ft - 30kip 2 ft + Rcy 12ft = 0
ft
kip
S 18ft + 40kip 15ft + 4 12ft 6 ft + 30kip 2 ft
ft
Rcy := = 105.7 kip
12ft
kip
Mz
c
=0 -S 6 ft - 40kip 3 ft + 4 12ft 6 ft + 30kip 10ft - Rdy 12ft = 0
ft
kip
-S 6 ft - 40kip 3 ft + 4 12ft 6 ft + 30kip 10ft
ft
Rdy := = 30.1 kip
12ft
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3. Check
kip
Fy = 0 Ray - 40kip - 40kip + R cy - 4
ft
12ft - 30kip + R dy = -0 kip OK!
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5.5 Three Hinged Gable Frame
The three-hinged gable frames are spaced at 30 ft on center. Determine the reactions components on the frame due to: 1) Roof
dead load of 20 psf of roof area; 2) Snow load of 30 psf of horizontal projection; 3) Wind load of 15 psf of vertical projection.
Determine the critical design values for the horizontal and vertical reactions.
15 ft
ft
30
20 ft
ft
30
60 ft
V
B
15 ft
17.5 ft
20 ft
17.5 ft
A C Ahv
15 ft 15 ft
Ahv Chv
30 ft 30 ft Avv Avv Cvv
Solution: a) b) c)
1. Due to smmetry, there is no vertical force transmitted by the hinge for snow and dead load, and thus we can consider only the left (or right)
side of the frame.
2. Point equivalent loads:
a) Roof dead load per one side of frame is
2 2
DL := 20psf 30ft ( 30ft) + ( 15ft) = 20.12 kip
b) Snow load per one side of frame is
SL := 30psf 30ft 30ft = 27 kip
c) Wind load per per frame (ignoring the suction) is
WL := 15psf 30ft 35ft = 15.75 kip
3. There are 4 reactions, 3 equations of equilibrium and one equation of condition; therefore, statically determinate. Alternatively, by
symmetry there is no shear at the hinge, and we would have for the substructure two reactions at the support and one (horizontal) at the
hinge.
4. The relationship between the horizontal and vertical reactions at A due to a centered vertical load, Ahv and Avv respectively is determined
by taking the moment with respect to the hinge (b):
Mz = 0
B
15 V - 30 Avv + 35 Ahv = 0
Fy = 0 Avv - V = 0
15Avv
Ahv =
35
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Substituting for roof dead and snow load we obtain:
AvvDL := DL = 20.12 kip BvvDL := AvvDL = 20.12 kip
15AvvDL
AhvDL := = 8.62 kip BhvDL := AhvDL = 8.62 kip
35
AvvSL := SL = 27 kip BvvSL := AvvSL = 27 kip
15AvvSL
AhvSL := = 11.57 kip BhvSL := AhvSL = 11.57 kip
35
5. The reactions due to wind load (blowing from left) are determined as follows:
a) Vertical reaction at A is determined by considering the entire structure and taking the moment with respect to C (c)
20ft + 15ft
15.75kip
C
Mz = 0 - 60ft Avh = 0
2
WL 20ft + 15ft
Avh :=
2 = 4.59 kip
60ft
Avh is the vertical reaction at A due to the horizontal load, and from equilibrium of forces in the y-direction, we have
Cvh := -Avh = -4.59 kip
(note that wind load does not have any vertical component)
b) The horizontal reaction at B is determined by considering the right substructure and taking the moment with respect to
the internal hinge at B
Mz
B
=0 35ft Chh - Cvh 30ft = 0
C vh 30ft
Chh := = -3.94 kip
35ft
c) Horizontal reactino at A is taken by considering the entire structure and summing forces in the x-direction
Fx = 0 WL + Chh - Ahh = 0
Ahh := WL + Chh = 11.81 kip
and note that A carries most of the horizontal load.
6. Finally, the supports should be designed for the most critical (plausible) combination of reactions
H := AhvDL + AhvSL + Ahh = 32.01 kip
V := AvvDL + AvvSL + Avh = 51.72 kip
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5.6 Inclined Supports
Determine the reactions of the following two span beam resting on inclined supports.
40 k 30 k 50 k
2
1
3 4
4 a b c 3
8 ft 6 ft 6 ft 6 ft 6 ft
Solution:
A priori we would identify 5 reactions; however, we do have 2 equations of condition (one at each inclined support), thus with three equations
of equilibrium, we have a statically determinate system.
Ray 20ft - 40kip 12ft - 30kip 6 ft + 50kip sin atan 6 ft - Rcy 12ft = 0
2
b
Mz = 0
1
Ray 20ft - Rcy 12ft = 40kip 12ft + 30kip 6 ft - 50kip sin atan 6 ft
2
1
3 2 4
Fx = 0 R - 50kip cos atan - Rcy = 0
4 ay 1 3
Rcy = Ray - 50kip cos atan =
3 3 2 9
Ray - 16.77kip
4 4 1 16
9
R - Rcy = 16.77kip
16 ay
Solving for those two equations:
20 -12 Ray
9 = 391.672 kip
-1 R cy 16.77kip
16
ay 14.37kip
R
:=
Rcy -8.69kip
The horizontal components of the reactions at a and c are
3
Rax := R ay = 10.78 kip
4
4
Rcx := - Rcy = 11.59 kip
3
Finally, we solve for Rby
40kip 8 ft + 30kip 14ft - R by 20ft + 50kip sin atan 26ft - Rcy 32ft = 0
2
Mz
a
=0
1
40kip 8 ft + 30kip 14ft + 50kip sin atan 26ft - Rcy 32ft
2
Rby :=
1 = 109.04 kip
20ft
We check our results
Ray - 40kip - 30kip - 50kip sin atan
2
Fy = 0
+ Rby + Rcy = 0 kip
1
OK!
Rax - 50kip cos atan
2
Fx = 0
+ Rcx = 0 kip OK!
1
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7.1 Truss Method of Joints
Using the method of joints, analyze the following truss
8
12
G
10' H F
32'
E
B C D
A
20 k 40 k 40 k
24' 24' 24' 24'
Solution:
1. R=3, m=13, 2j=16, and m+R=2j
2. We compute the reactions
Mz
E
=0 ( 20kip + 12kip) 72ft + ( 40kip + 8kip) 48ft + 40kip 24ft - RAy 96ft = 0
( 20kip + 12kip) 72ft + ( 40kip + 8kip) 48ft + 40kip 24ft
RAy :=
96ft
RAy = 58 kip
Fy = 0 20kip + 12kip + 40kip + 8kip + 40kip - R Ay - REy = 0
REy := 20kip + 12kip + 40kip + 8kip + 40kip - RAy
REy = 62 kip
3. Consider each joint separately:
Node A: Clearly AH is under compression and AB is under tension
FAH
A FAB
58 k
Fy = 0 -FAHy + R Ay = 0
FAHy := -RAy = -58 kip
2 2
Ly := 32ft Lx := 24ft L := Ly + Lx = 40 ft
L
FAH := F = -72.5 kip Compression
Ly AHy
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Fx = 0 -FAHx + FAB = 0
Lx
FAB := R = 43.5 kip Tension
Ly Ay
Node B:
FBH
B
43.5 k FBC
20 k
Fx = 0 FBC := 43.5kip Tension
Fy = 0 FBH := 20kip Tension
Node H:
12 k
FHG
H FHGy
FAHx
FHCy FHGx
FAHy
FHCx
72 k FHC
20 k
Fx = 0 FAHx - FHCx - FHGx = 0
24ft 24ft
43.5kip - FHC - FHG = 0
2 2 2 2
( 24ft) + ( 32ft) ( 24ft) + ( 10ft)
Fy = 0 FAHy + FHCy - 12kip - FHGy - 20kip = 0:
32ft 10ft
58kip + FHC - FHG - 12kip - 20kip = 0
2 2 2 2
( 24ft) + ( 32ft) ( 24ft) + ( 10ft)
This can be most conveniently written as
24ft 24ft
( 24ft) 2 + ( 32ft) 2 2
( 24ft) + ( 10ft)
2
FHC 43.5kip
=
- 32ft 10ft
FHG 26kip
( 24ft) 2 + ( 32ft) 2 2 2
( 24ft) + ( 10ft)
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Solving we obtain
-1
24ft 24ft
FHC ( 24ft) 2 + ( 32ft) 2 ( 24ft) + ( 10ft)
2 2
43.5kip -7.5
:= = kip
FHG 32ft 10ft 26kip 52
-
( 24ft) 2 + ( 32ft) 2 2 2
( 24ft) + ( 10ft)
FHC = -7.5 kip Compression
FHG = 52 kip Tension
Node E:
FEF
FED E
62 k
Fy = 0 FEFy := 62kip
2 2
( 24ft) + ( 32ft)
FEF := 62kip = 77.5 kip Compression
32ft
Fx = 0 FED = FEFx:
ft
FED := 24 62kip = 46.5 kip Tension
32ft
The results of the analysis are summarized below
8
12
52 52
43.5 46.5
58 62
32
2.5
7.5
4
0
77
20
.5
.5
72
43.5 43.5 46.5 46.5
58 20 40 40 62
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4. We would check our calculations by verifying equilibrium of forces at a node not previously used, such as D.
7.2 3D Truss
Z
Pz=600
8'
C
A
4' 2'
B
2' 3'
Y
X
Solution:
1. Consider the free body diagram of the entire truss
600 3ft
MAB = 0 Cz 5 ft - 600 3 ft = 0 Cz :=
5ft
Cz = 360
600 2ft
MCB = 0 600 2ft - Az 6 ft = 0 Az :=
6ft
Az = 200
Fz = 0 Bz + 200 + 360 - 600 = 0 Bz := 600 - 200 - 360
Bz = 40
Fx = 0 Bx := 0
Fy = 0 Ay - Cy = 0
Mz = 0 Ay 3 ft + Cy 2 ft = 0 Ay = Cy = 0
Ay := 0
Cy := 0
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2. Consider the free body diagram of joint B
z FBD
8 ft
2 ft
t FBC
FBA 3f
B y
x 40.0 lb
Joint B
2 2 2
LBD := ( 2ft) + ( 3ft) + ( 8ft) = 8.775 ft
-8 ft
Fz = 0 LBD
FBD + 40 = 0
-LBD
FBD := 40 = -43.875 Compression
8ft
Fx = 0 FBDx - FBC = 0
-3 ft
FBDx := F = 15 Tension
LBD BD
Fy = 0 FBDy - FBA = 0
-2 ft
FBDy := F = 10 Tension
LBD BD
3. FBD of joint A
FAD
8 ft
4 ft FAC
A
ft 39.8°
3
y
10.0 lb
x
200 lb
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2 2 2
LAD := ( 8ft) + ( 3ft) + ( 4ft) = 9.434 ft
-8 ft
Fz = 0 LAD
FAD + 200 = 0
-LAD
FAD := 200 = -235.85 Compression
8ft
Fx = 0 FADx - FACx = 0
α := atan
5ft 5
tan( α) = = 39.8 deg
6ft 6
3ft
FAD - FAC sin( α) = 0
LAD
3ft
-FAD
LAD
FAC := = 117.2 Tension
sin( 39.8deg)
Fy = 0 FBDy - FBA = 0
α := atan
5ft 5
tan( α) = = 39.8 deg
6ft 6
-2 ft
FBA := F = 10 Tension
LBD BD
5. Joint C
FCD
8 ft
39.8° ft
2
2 ft c y
FAC =117 lb
x 360 lb
2 2 2
LCD := ( 8ft) + ( 2ft) + ( 2ft) = 8.485 ft
-8 ft
Fz = 0 FCDz + 360 = 0
LCD
FCD + 360 = 0
-LCD
FCD := -360 = -381.8 Compression
-8 ft
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7.3 Truss I, Matrix Method
Determine all member forces for the following truss
y
10
3 3
2
5
1 x
1 2
10
Solution:
1. We first determine the direction cosines
Member 1 (Nodes 1-2)
Node1: α11 := 1 β11 := 0 Node2: α21 := -1 β21 := 0
Member 2 (Nodes 2-3)
- 2 2 2 - 2
Node2: α22 := = -0.707 β22 := = 0.707 Node3: α32 := = 0.707 β32 := = -0.707
2 2 2 2
Member 3 (Nodes 3-1)
- 2 - 2 2 2
Node2: α33 := = -0.707 β33 := = -0.707 Node3: α31 := = 0.707 β31 := = 0.707
2 2 2 2
2. Next we write the equations of equilibrium
α11 0 α31 1 0 0 F1
0
β11 0 β31 0 1 0 F2 0
α α22 0 0 0 0 F3 0
21 + =0
β21 β22 0 0 0 1 R1x 0
0 α32 α33 0 0 0 R1y 0
0 β32 β33 0 0 0
R -10
2y
-1
F1 α11 0 α31 1 0 0
0 5
F2 β11 0 β31 0 1 0 0 -7.071
F α α22 0 0 0 0
3 := 21 0 -7.071
- =
R1x β21 β22 0 0 0 1 0 0
0 5
R1y 0 α32 α33 0 0 0
R 0 -10 5
β32 β33 0 0 0
2y
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7.4 Truss II, Matrix Method
Set up the statics matrix for the truss shown below
8
12
G
10' H F
32'
E
B C D
A
20 k 40 k 40 k
24' 24' 24' 24'
Solution:
7
12 13
10' 8
6
5 6 7 8 9 10 11
32'
1 5
1 2 2 3 3 4 4
24' 24' 24' 24'
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1. We first determine the direction cosines
24ft
= 0.6
2 2
( 24ft) + ( 32ft)
32ft
= 0.8
2 2
( 24ft) + ( 32ft)
10ft
= 0.385
2 2
( 24ft) + ( 10ft)
24ft
= 0.923
2 2
( 24ft) + ( 10ft)
2. Next we write the equations of equilibrium
2. Next we write the equations of equilibrium
F1
F2
1 0 0 0 0.6 0 0 0 0 0 0 0 0 1 0 0
F 0
0 0 0 0 0.8 0 0 0 0 0 0 0 0 0 1
0 3
0
-1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 F4 0
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 F -20
5
-1 1 -0.6 0 0
0 0 0 0 0.6 0 0 0 0 0 0
F6 0
0 0 0 0 0 0 0.8 1 0.8 0 0 0 0 0 0 0 -40
F
0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 7 0
0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 F8 -40
+ =0
0 0 0 -1 0 0 0 0 0 0 -0.6 0 0 0 0 0 F9 0
0 0 0 0 0 0 0 0 0 0 0.8 0 0 0 0
1 F
0
10
0 0 0 0 0 0 0 0 -0.6 0 0.6 0 -0.923 0 0 0
0
F11
0 0 0 0 0 0 0 0 -0.8 -1 -0.8 0 0.385 0 0 0 0
F12
-0.923 0.923 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0
0 0 0 0 0 0 0 -1 0 0 0 -0.385 -0.385 0 0 0 F13 -8
0 0 0 0 -0.6 0 0.6 0 0 0 0 0.923 0 0 0 0 R 0
1x
0 0 0 0 -0.8 -1 -0.8 0 0 0 0 0.385 0 0 0 0 -12
R
1y
R
5y
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F1
F2 -1
F 1 0 0 0 0.6 0 0 0 0 0 0 0 0 1 0 0 0 1
0 0
3 0 0 0 0.8 0 0 0 0 0 0 0 0 0 1 0 1 43.5
F4
-1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 43.5
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 -20 3 46.5
F5
4
-1 1 -0.6 0 46.5
F 0 0 0 0 0.6 0 0 0 0 0 0 0
0
6
0 0 0 0 0 0 0.8 1 0.8 0 0 0 0 0 0 0 -40 5 -72.5
F
7 0 -1 1 0 0 6 20
0 0 0 0 0 0 0 0 0 0 0 0
7
F8 -40 7.479
0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
-
:= =
0 -1 -0.6 0 0 8 32.033
F9 0 0 0 0 0 0 0 0 0 0 0 0
0 0 9 2.479
F 0 0 0 0 0 0 0 0 0 0.8 0 0 0 0 1
10
10 -0.6 0 -0.923 0 0
40
0 0 0 0 0 0 0 0 0.6 0 0 0
F11
0 0 0 0 0 0 0 0 -0.8 -1 -0.8 0 0.385 0 0 0 0 11 -77.5
12 -51.991
F12 -0.923 0.923 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 13 -51.991
F
0 0 0 0 0 0 0 -1 0 0 0 -0.385 -0.385 0 0 0 -8
13 14 0
0 -0.6 0 0 0
R1x 0 0 0 0.6 0 0 0 0 0.923 0 0 0
15 58
0 0 0 0 -0.8 -1 -0.8 0 0 0 0 0.385 0 0 0 0 -12 16
R1y 62
R
5y
7.5 Truss, Method of Sections
Determine FBC and FHG in the truss below
8
12
G
10' H F
32'
E
B C D
A
20 k 40 k 40 k
24' 24' 24' 24'
Solution:
Cutting through members HG, HC, and BC, we first take the summation of forces with respect to H:
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8
12
G FHG
10' H F H
32'
FHC
E
A FBC
B C D B
A
20 k 40 k 40 k
58 k 20 k
24' 24' 24' 24'
Mz
H
=0 RAy 24ft - FBC 32ft = 0
RAy := 58kip (From previous example)
24ft
FBC := R = 43.5 kip Tension
32ft Ay
Mz
C
=0 58kip 48ft - ( 20kip + 12kip ) 24ft - FHGx 32ft - FHGy 24ft = 0
24ft 10ft
58kip 48ft - ( 20kip + 12kip ) 24ft - 32ft FHG - 24ft FHG =0
2 2 2 2
( 24ft) + ( 10ft) ( 24ft) + ( 10ft)
FHG := 52kip Compression
8.1 Funicular Cable Structures
Determine the reactions and the tensions for the cable structure shown below
Ax A D
Dx
θA C
HC
h = 6'
B θC Dy
Ay θB
12 k 6k not to scale
30' 40' 30'
Solution:
We have 4 external reactions, however the horizontal ones are equal and we can use any one of a number of equations of conditions in
additiona to the three equations of equilibrium. First, we solve for the vertical reactions Ay, Dy and then for the horizontal ones (which are
equal and opposite (|H| = Ax = -Dx). For this problem we could use the following 3 equations of static equilibrium ΣFx = ΣFy = ΣM = 0, however
since we do not have any force in the x direction, the first equation is of no avail. Instead, we will consider the following set ΣFy = ΣMA = ΣMB =
0. Alternatively, we can consider the problem as one with 8 unknowns (Ax, Ay, Dx, Dy, θA, θB, θC, and hC), to be solved through the 2
equations of equilibrium expressed at each of the four points of interesect.
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equations of equilibrium expressed at each of the four points of interesect.
1. Solve for Dy
Mz
A
=0 12kip 30ft + 6kip 70ft - Dy 100 ft = 0
12kip 30ft + 6kip 70ft
Dy := = 7.8 kip
100ft
2. Solve for Ay
Fy = 0 Ay - 12kip - 6kip + Dy = 0
Ay := 12kip + 6kip - Dy = 10.2 kip
3. Solve for the horizontal force by isolating the free body diagram AB
Mz
B
=0 Ay 30ft - H 6 ft = 0
Ay 30ft
H := = 51 kip
6ft
4. Solve for the sag at point C by isolating the free body diagram CD
Mz
C
=0 -Dy 30ft + H hc = 0
30ft Dy
h c := = 4.6 ft
H
5. Solve for the cable internal forces or tractions in this case
θA := atan
6ft 6ft
( )
tan θA =
30ft 30ft
= 11.31 deg
H
TAB := = 52.01 kip
cos θA( )
6ft - h c 6ft - hc
( )
tan θB =
40ft
θB := atan
40ft
= 2.02 deg
H
TBC := = 51.03 kip
( )
cos θB
hc hc
( )
tan θC =
30ft
θC := atan = 8.7 deg
30ft
H
TCD := = 51.59 kip
cos θC ( )
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8.2 Design of Suspension Bridge
Design the following 4 lane suspension bridge by selecting the cable diameters assuming an allowable cable strength of σall := 190ksi.. The
bases of the tower are hinged in order to avoid large bending moments. The total dead load is estimated at 200 psf. Assume a sag to span
ratio of 1/5.
B C
B
HAB HBC
120' 50'
A E F D TAB TBC
A
E
FBE
100' 300' 100'
Solution:
1 kip
1. The dead load is carried by each cable with one half the total deal load or p 1 := 200 psf 50ft = 5
2 ft
2. Using the HS 20 truck (or its distributed equivalent load of 0.64kip/ft per lane), the uniform additional load per cable is
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kip kip
p 2 := 2 0.64 = 1.28
ft ft
Thus the total design load is
kip
p := p 1 + p 2 = 6.28
ft
3. The thrust H is determined by
L := 300ft
h := 60ft
2
p L
H := = 1177.5 kip
8 h
4. The maximum tension is
1
r := sag to span ratio
5
2
Tmax := H 1 + 16r = 1508 kip
5. Note that if we used the approximate formula we woul dhave obtained
(
TmaxApp := H 1 + 8 r
2
) = 1554.3kip
6. The required cross sectional area of the cable along the main span should be equal to
Tmax 2
A := = 7.94 in
σall
which corresponds to a diameter
4 A
d := = 3.18 in
π
7. We seek to determine the cable force in AB. Since the pylon cannot take any horizontal force, we should have the horizontal component of
Tmax equal and opposite to the horizontal component of TAB or
2 2
( 100ft) + ( 120ft)
TAB := H = 1839 kip
100ft
The cable area should be
TAB 2
A := = 9.68 in
σall
which corresponds to a diameter
4 A
d := = 3.51 in
π
8. To determine the vertical load acting on the pylon, we must add the vertical components of Tmax and TAB (VBC and VAB respectively). We can
determine VBC from H and Tmax, thus
120ft 2 2
P := H + Tmax - H = 2355 kip
100ft
Using A36 steel with an allowable stress of σA36 := 21ksii, the cross sectional area of the tower should be
P 2
A := = 112 in
σA36
Note that buckling of such a tower might govern the final dimensions.
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Note that buckling of such a tower might govern the final dimensions.
9. If the cables were to be anchored to a concrete block, the volume of the block shoul dbe at least equal to
120ft
H
100ft 3
V := = 9420 ft
lbf
150
3
ft
10.1 Simple Shear and Moment Diagram
Draw the shear and moment diagram for the beam shown below.
11 k 10 k
3
4 2 k/ft
B C D
A 4ft 6ft 4ft E
4ft
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Solution:
The free body diagram is drawn below
Free Body Diagram
11 k
10 k 8k 2 k/ft
6k A 6k
11 k
13 k D 14 k
E
13 k 2k 2k C 8k
B
13 k
Shear Diagram
2k
A=(13)(4)=52 A=(6)(2)=12
A=(-6)(4)=-24 A=-4(6+14)/2=-40
-6 k
Slope= dV/dx=w=-2
-14 k
Momen Diagram
x =+ 2 dM / d
dM / d x= - 6 -1 4
<d
52+12=64
M/
0+52=52
13 dx
64-24=40
= + < -6
/d x
dM
A B C D E
Reactions are determined from the equilibrium equations
Fx = 0 -Ax + 6kip = 0 Ax := 6kip
3 kip
MA = 0 11kip 4 ft + 10kip 10ft + 2
5 ft
4 ft 16ft - Ey 18ft = 0
4 kip
11kip 4 ft + 10kip 10ft + 2 4 ft 16ft
5 ft
Ey := = 14 kip
18ft
4 kip
Fy = 0 Ay - 11kip - 10kip - 2
5 ft
4 ft + Ey = 0
kip
Ay := 11kip + 8kip + 2 4 ft - Ey = 13 kip
ft
Shear are determined next
1. At A the shear is equal to the reaction and is positive.
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1. At A the shear is equal to the reaction and is positive.
2. At B the shear drops (negative load) by 11 k to 2 k
3. At C it drops again by 8 k to -6 k
4. It stays constant up to D and then it decreases (constant negative slope since the load is uniform and negative) by 2 k
per linear foot up to -14 k
5. As a check, -14 k is also the reaction previously determined at E.
Moment is determined last
1. The moment at A is zero (hinge support)
2. The change in moment between A and B is equal to the area under the corresponding shear diagram, or
ΔMBA := 13kip 4 ft = 52 kip ft
3. Changes between other points are determined the same by taking the area under the shear diagram
10.2 Frame Shear and Moment Diagram
Draw the shear and moment diagram of the following frame
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3k/ft
B C
3k
/ft
4
12'
36.0 k 3
A
D
52.96 k
64.06 k
30' 9'
Solution:
3k/ft
36 36 3(30) 2
-432- +64.06(30)=139.8
2
432 64.06 C
64.06 64.06-3(30)=25.94
B
36
25.96
139.8
36 432 36 C
432
13.22 42.37
64.06
139.8
42.37
(36)(12)=432 139.8
64.06
B 36
3(15)-31.78=13.22
36
A D
3 4
64.06 (52.96)=31.78 (52.96)=42.3
5 5
52.96
64.06
42.3
36
64.06
+
13.22
- +
25.94
-
36 Shear Diagram
-
252 31.78
139.8
+
432 139.8
-
- +
432
Moment Diagram
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Reactions are determined first
4 kip 4 kip
Fx = 0 RAx -
5
3
ft
15ft = 0 RAx :=
5 ft
3
15ft = 36 kip
kip 3 kip 4 kip
MA = 0 3
ft
30ft 15ft + 3
5 ft
15ft 34.5ft - 3
5 ft
15ft 6 ft - RDy 39ft = 0
kip 3 kip 4 kip
3 30ft 15ft + 3 15ft 34.5ft - 3 15ft 6 ft
ft 5 ft 5 ft
RDy := = 52.96 kip
39ft
kip 3 kip
Fy = 0 RAy - 3
ft
30ft -
ft
3
15ft + R Dy = 0
5
kip kip 3
RAy := 3 30ft + 3 15ft - R Dy = 64.04 kip
ft 5 ft
We isolate each member and draw its free body diagram for each force component.
Shear
1. For A-B the shear is constant, equal to the horizontal reaction at A and negative according to our previously defined
sign convention, VA := -36kip
2. For member B-C at B, the shear must be equal to the vertical force which was transmitted along A-B, and which is
equal to the vertical reaction at A, VB := R Ay = 64.04 kip
3. Since B-C is subjected to a uniform load, the shear along B-C will have a slope equal to -3 and in terms of x (measured
from B to C) is equal to
VBC( x) := RAy - 3x
4. The shear along C-D is obtained by decomposing the vertical reaction at D into axial and shear components. Thus, at
3
D the shear is equal to RDy = 31.78 kip and is negative. Based on our sign convention for the load, the slope of the
5
5 kip 3
shear must be equal to -3 along C-D. Thus the shear at point C is such that Vc - 9 ft 3 = - RDy or
3 ft 5
5 kip 3
Vc := 9 ft 3 - RDy = 13.22 kip. The equation for shear is given by (for x going from C to D)
3 ft 5
VCD( x) := Vc - 3x
Moment
1. Along A-B, the moment is zero at A (since we have a hinge) and its slope is equal to the shear, thus at B the moment
is equal to -36kip 12ft = -432 kip ft
2. Along B-C, the moment is equal todx
x x 2
3x
M BC = M B + VBC( x) dx = -432 kip ft +
(RAy - 3x) dx = -432kip ft + RAy x - 2
0 0
which is a parabola. Substituting for x=30, we obtain at node C:
kip 2
3 ( 30ft)
ft
M C := -432 kip ft + RAy 30ft - = 139.2 kip ft
2
dMBC
3. If we need to determine the maximum moment along B-C, we know that = 0 at the point where where V =0,
dx BC
RAy
that is VBC( x) = R Ay - 3x = 0, xmax := = 21.3 ft. In other words, the maximum moment occurs where the shear is
kip
3
ft
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zero. Thus
kip
( )
2
3 xmax
ft
M BCmax := -432 kip ft + R Ay xmax - = 251.5 kip ft
2
4. Finally, along C-D the moment varies quadratically (since we had a linear shear), the moment first increases (positive
shear), and then decreases (negative shear). The moment along C-D is given by
x x 2
3x
M CD = M C + VCD( x) dx = M C +
(Vc - 3x) dx = MC + Vc x - 2
which is a parabola. Substituting for
0 0
x =15ft, we obtain at node D
kip 2
( 15ft)
ft
M D := M C + Vc 15ft - 3 = -0 kip ft OK!
2
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10.3 Frame Shear and Moment Diagram: Hydrostatic Load
The frame shown below is the structural support of a flume. Assuming that the frames are spaced 2 ft apart along the length of the
flume,
1. Determine all internal member end actions
2. Draw the shear and moment diagrams
3. Locate and compute maximum internal bending moments
4. If this is a reinforced concrete frame, show the location of reinforcement
A F
H 2O Density of water=62.4 lb/ft
6 ft.
Spacing of frames=2 ft.
E
B
C D
3 ft. 10 ft. 3 ft.
Solution:
The hydrostatic pressure causes lateral forces on the vertical members which can be treated as cantilevers fixed at the lower end. The
pressure is linear and is given by p ( γ , h ) := γ h. Since each frame supports a 2 ft wide section of the flume, the equation for w (pounds/ft) is
lbf 124.8 h lbf
w( h ) := 2ft 62.4 h
3 2
ft ft
kip
At the base wbase := w( 6ft) = 0.749 . Note that this is both the lateral pressure on the end walls as well as the uniform load on the
ft
horizontal members.
y y y y
Shear
ratic
x x x x
Qu
x 3.744 k
Quad
adr
2.246 k
a
tic
- + -
Shear
B C m D + E
2.246 k 2.246 k
-2.246 k - - -2.246 k
-2.246 k
x
2 ft -3.744 k
2 ft
2.246 k 2.246 k y
y
.749 k/ft .749 k/ft .749 k/ft
4.493 k-ft 4.493 k-ft
1.5 k-ft
4.493 k-ft
4.493 k-ft
x x
B C + D E
Cubic
Cubic
2.246 k m
2.246 k - -
4.493 k-ft 5.99 k 5.99 k 4.493 k-ft
4.493 k-ft -7.864 k-ft -7.864 k-ft 4.493 k-ft
End Actions
1
1. Base force at B is FBx := wbase 6 ft = 2.246 kip
2
1
2. Base moment at B is M B := FBx 6 ft = 4.493 kip ft
3
3. End forces at B for member B-E are equal and opposite
1
4. Reaction at C is RCy := wbase 16ft = 5.99 kip
2
Shear forces
1. Base at B the shear force was determined eariler and was equal to FBx = 2.246 kip. Based on the orientation of the x-y
axis, this is a negative shear.
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2. The vertical shear at B is zero (neglecting the weight of A-B)
3. The shear to the left of C is VCleft := 0 - wbase 3 ft = -2.246 kip
4. The shear to the right of C is VCright := VCleft + RCy = 3.744 kip
Moment diagrams
1. At the base: B M B = 4.493 kip ft
3ft
2. At the support C, M C := -M B - wbase 3 ft = -7.862 kip ft
2
5ft
3. The maximum moment is equal to M max := M C + wbase 5 ft = 1.498 kip ft
2
Design: Reinforcement should be placed along the fibers which are under tension, that is on the side of the negative moment. The figure
below schematically illustrates the location of the flexural reinforcement.
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10.4 Shear Moment Diagrams for Frame
8' 12' 10' Vba 2k/ft 10k
Vbc
5k/ft 30k 2k/ft 10k
H bc B C
B C
A H bd
E M ba M bc M bc
5' M bd V bc
VA 20k G Vbd (10)+(2)(10)
30k
10k
V bc
52.5k 30k
15'
0 0 M bc
0 -200'k
650'k 200'k (10)(10)+(2)(10)(10)/2
D HD
4k/ft 450'k
VD 82.5k
CHECK
30k
5k/ft bd V
M bd
B H ba 450'k H bd
A
(50)-(4)(15)/2
M ba
17.5k V ba
20k 450'k 20k
(50)(15)-[(4)(5)/2][(2)(15)/3)]
17.5k
-22.5k 17.5-(5)(8)
3.5'
17.5-5x=0 50k
-22.5+(-30)=-52.5 V ba
4k/ft 50k
30.6'k (17.5)(3.5)/2
82.5k
-20'k (17.5)(3.5)/2+(-22.5)(8-3.5)/2
(-52.5)(12)+(-20)-650'k M ba
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10.5 Shear Moment Diagrams for Inclined Frame
26k 10' 10'
26k 13'
13' 20k
13' C
13 5
5 3 15'
12 4
B D
2k/ft
20'
A Ha E
36' 20' Ve
Va
(60)(20)-(2)(20)(20)/2
19.2k 48.8k
(20)(20)+(60-20)(20)/2
Fx
800k'
20k 800'k
F z
60-(2)(20)
20k
0k' y
Fy
x
2k/ft 60k
60k F/Fy=z/x
AB ED F/Fx=z/y
19.2k 48.8k Fx/Fy=y/x
1 2 3 4 5 6
28.8k
26k 778k' 17.2
1k
10k 8k
23.
26k 0k 12k
10k (20)(5)/13=7.7
24k
20k
24k
20-10-10
16k
778k'
800k' 26.6k 11.1k (20)(12)/(13)=18.46
7.69k 0k'
28.8k 0k (19.2)(5)/(13)=7.38
20k
18.46k
19.2k 29.3k (19.2)(12)/(13)=17.72
17.72k 7.38k 39.1k
BC CD 48.9k
19.2k 7 10 (26)(12)/(13)=24
-23 (26.6)(13)/(12)=28.8
7.7 .1k
17.7+.4k
+25 -3 (26.6)(5)/(12)=11.1
k -23 9.1k
k -26.6 .1-
-0.58 -26 -0.6-26 16 (28.8)(4)/(5)=23.1
25.42 8 B-C 11 C-D
1,122-(26.6)(13)
(28.8)(3)/(5)=17.28
488+(23.1)(12.5)
'
1,130-(.58)(13) 777k
'
(20)(4)/(5)=16
7k
k' (39.1)(12.5)
800+(25.4)(13) 1122
77
0 'k 488'k (20)(3)/(5)=12
113
800'k (39.1)(5)/(4)=48.9
9 B-C 12 C-D
(39.1)(3)/(4)=29.3
13 14 777k'
2k'
112
0'k 48
k -23 113 8'k
+25.4 .1k
8k 6k
-0.5 - 26 . 800'k
+20k
-39
.1k
+60k
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10.6 3D Frame
y 20 2m
kN
/m
4m
C kN
60
D
4m
1. The frame has a total of 6 reactions (3 forces and 3 moments) at the support, and we have a total of 6 equations of equilibrium,
thus it is staticall determinate.
2. Each member has the following internal forces (defined in terms of the local coordinate system of each member x' - y'- z' such
that x is along the member)
Member Internal Forces
Axial Shear Moment Torsion
Member
Nx' V y' V z' My' Mz' Tx'
C-D
B-C
A-B
3. The numerical calculations for the analysis of the three dimensional frame are quite simple, however the main complexity
stems from the difficulty in visualizing the inter-relations between internal forces of adjacent members.
4. In this particular problem, rather than starting by determining the reactions, it is easier to determined the internal forces at the
end of each member starting with member C-D. Note that temporarily we adopt a sign convention which is compatible with the
local coordinate systems.
C-D
kN
Fy' = 0 Vy'C := 20
m
2 m = 40 kN
Fz' = 0 Vz'C := 60kN
My' = 0 M y'C := -60kN 2 m = -120 kN m
kN 2m
Mz' = 0 M z'C := 20
m
2 m
2
= 40 kN m
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B-C
Fx' = 0 Nx'B := Vz'C = 60 kN
Fy' = 0 Vy'B := Vy'C = 40 kN
My' = 0 M y'B := M y'C = -120 kN m
Mz' = 0 M z'B := Vy'C 4 m = 160 kN m
Tx' = 0 Tx'B := -M z'C = -40 kN m
A-B
Fx' = 0 Nx'A := Vy'B = 40 kN
Fy' = 0 Vy'A := Nx'B = 60 kN
My' = 0 M y'A := -Tx'B = 40 kN m
Mz' = 0 M z'A := M z'B + Nx'B 4 m = 400 kN m
Tx' = 0 Tx'A := M y'B = -120 kN m
The interaction between axial forces N and shear V as well as between moments M and torsion T is clearly highlighted by this example.
z' 120 kN-m kN 120 kN-m m y'
60 -
m kN
C N- 40
4 0k z' x' 20 kN/m
y' kN
x' k N 60
B kN N
40 kN 60 60 60
k
-m 40 kN
160 kN -m
40 kN 40 kN
kN 40
-m 40 kN - m 120 kN-m 40 kN
kN 0 kN 120 kN-m
40 6 120 kN-m
120 kN-m 40 kN
kN
60
m
N-
4 0k
16
0k
-m N
kN 160 -m
40 kN
-m
kN 40 kN
60
120 kN-m
120 kN-m 40 kN kN
60
m
k N-
40
160
x' kN
-m
y' z'
120 kN-m
40
0k
-m N-
40 kN
kN m
40
kN
60
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y' y' y'
x' x'
C C C
B B 60
z'
z' 160 z'
D
60 120
M
40 x'
V
40
y'
M
V C
T
40
z' 120
D
x'
x'
x'
x'
60
B
B B
160
40 120
A A
A 40
0
z'
y' z'
y'
V M T
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11.1 Three Hinged Arch, Point Loads
Determine the reactions of the three-hinged arch shown below
20 k B
30 k 30 k 20 k
HB HB
B
20' 20' VB VB
33.75'
80 k 80 k HC
C C
30' VC
26.25' HA A
A
80' 60' VA
Solution:
Four unknowns, three equations of equilibrium, one equation of condition - staticall determinate
Mz
C
=0
RAy 140 ft + 80kip ( 30ft - 26.25ft) - 30kip ( 60ft + 20ft) - 20kip ( 60ft - 20ft) + RAx 26.25ft = 0
RAy 140 ft + RAx 26.25ft = 30kip 80ft + 20kip 40ft - 80kip 3.75ft
RAy 140 ft + RAx 26.25ft = 2900kip ft
Fx = 0 80kip - RAx - RCx = 0
RAx + RCx = 80kip
Fx = 0 RAy + RCy - 30kip - 20kip = 0
RAy + RCy = 50kip
Mz
B
=0 RAx 60ft - 80kip 30ft - 30kip 20ft + RAy 80ft = 0
RAy 80ft + RAx 60ft = 80kip 30ft + 30kip 20ft
RAy 80ft + RAx 60ft = 3000kip ft
Solving those four equations simultaneously we have:
RAy
140 26.25 0 0 2900
R
Ax 80
0 1 0 1
=
1 0 1 0 R Cy 50
80 60 0 0
R 3000
Cx
-1 RAy
Ay 140 26.25 0 0 2900 15.1 Ay 15.1
RAx
Ax := 0 1 0 1 80 29.8 Ax kip = 29.8 kip
= :=
Cy 1 0 1 0 50 34.9 Cy 34.9
RCy
Cx 80 60 0 0 3000 50.2 Cx 50.2
R
Cx
We can check our results by considering the summation with respect to B from the right:
Mz
B
=0 -20kip 20ft - RCx 33.75ft + RCy 60ft = -0 kip ft
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11.2 Semi-Circular Arch
Determine the reactions of the three-hinged staticall determinate semi-circular arch under its own dead weight w (per unit arc
length s, where ds=rdθ
dP=wRd ?
B ? r B
R R
A
? ? C
A
R cos ?
Solution:
Reactions The reactions can be determined by integrating the load over the entire structure
1. Vertical Reaction is determined first
θ =π
MA = 0 -C y 2 R +
θ=0
wR dθ R( 1 + cos( θ) ) = 0
θ =π
wR π wR
Cy = ( 1 + cos( θ) ) dθ =
2 θ =0
2
2. Horizontal Reactions are determined next
π
θ=
2
MB = 0 -C x R + C y R -
θ=0
wR dθ Rcos( θ) = 0
π
θ=
2
π wR wR
cos( θ) dθ = - 1 wR
π
Cz = -
2 2 θ=0 2
Internal Forces can now be determined
dP=wRda
? M r N
V
V
?
N
Cx
?
Cy
V
a
d
?
a π
C x = -1 wR
2
Rcosθ
R(1-cosθ)
Rcos α
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1. Shear Forces: Considering the free body diagram of the arch, and summing the forces in the radial direction (ΣFR=0)
θ
- - 1 wRcos( θ) + wRsin( θ) -
π π
wR dα sin( θ) + V = 0
2 2
α =0
V = wR - 1 cos( θ) + θ -
π π
sin( θ)
2 2
2. Axial Forces: Similarly, if we consider the summation of forces in the axial direction (ΣFθ=0)
θ
π - 1 wRsin( θ) + π wRcos( θ) -
wR dα cos( θ) + N = 0
2 2 α =0
N = wR θ -
ππ
cos( θ) - - 1 sin( θ)
2 2
2. Moment: Now we can consider the third equation of equilibrium (ΣMz=0)
θ
π - 1 wR Rsin( θ) + π wR R ( 1 - cos( θ) ) +
wR dα R( cos( α) - cos( θ) ) + M = 0
2 2
α =0
M = wR ( 1 - sin( θ) ) + θ - cos( θ)
2 π π
2 2
Deflection are determined last
1. The real curvature ϕ is obtained by dividing the moment by EI
2
( 1 - sin( θ) ) + θ - cos( θ)
M wR π π
ϕ= =
EI EI 2 2
2. The virtual force δ P will be a unit vertical point load in the direction of the desired deflection, causing a virtual internal
moment
R π
δ M = ( 1 - cos( θ) - sin( θ) ) 0θ
2 2
3. Hence, application of the virtual work equation yields:
π
2
( )
2 4
( 1 - sin( θ) ) + θ - cos( θ) ( 1 - cos( θ) - sin( θ) ) R dθ =
wR π π R wR 2
1 Δ = 2 7 π - 18 π - 12
EI 2 2 2 16EI
θ =0
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11.3 Statically Indeterminate Arch
Determine the value of the horizontal reaction component of the indicated two-hinged solid rib arch, as caused by a concentrated
vertical load of 10 k at the center line of the span. Consider shearing, axial, and flexural strains. Assume the rib is a W24x130 with
a total area of 38.21 in2, that it has a web area of 13.70 in2, a moment of inertia equal to 4,000 in4, E of 30,000 k/in2, and a
shearinf modulus of 13,000 k/in2.
P=10 k
B
35'
A
C
.3 6 '
0
R=16 0.898 rad
Solution: 200'
1. Consider that end C is placed on rollers, as shown below.
P=10 k
B N=
sQ
35' C sin
Q
V=co
A
1k Virtual
0.36'
N=P/2 cosQ
R=16
125.36' Q
P/2
0.898 rad
Q V=P/2 sin Q
200'
A unit ficticious horizontal force is applied at C. The axial and shearinf components of this ficticious force and of the
vertical reaction at C, acting on any section θ in the right half of the rib, are shown at the right end of the rib in the figure
above.
2. The expression for the horizontal displacement of C is
B B B
M V N
1 ΔCh = 2 δ M ds + 2 δ V ds + 2 δ N ds
EI Aw G AE
C C C
3. From the figure above, for the rib from C to B
P
M = ( 100 - R cosθ)
2
δ M = 1 ( R sinθ - 125.36ft)
P
V = sinθ
2
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δ V = cosθ
P
N = cosθ
2
δ N = -sinθ
ds = Rdθ
4. If the above values are substituted in the equation for the horizontal displacement of C and integrated between the limits of
0.898 and π/2, the result will be
ΔCh = 22.55in + 0.023in - 0.003in
ΔCh := 22.57in
5. The load P is now assumed to be removed fro mthe rib, and a real horizontal force of 1 k is assumed to act toward the right at
C in conjunction with the ficticious force of 1 k acting to the right at the same point. The horizontal displacement of C will be
given by
B B B
M V N
δChCh = 2 δ M ds + 2 δ V ds + 2 δ N ds = 2.309in + 0.002in + 0.002in
EI Aw G AE
C C
C
δChCh := 2.313in
6. The value of the horizontal reaction component will be
ΔCh
HC := kip = 9.76 kip
δChCh
7. If only flexural strains are considered, the result would be
22.55in
HC := kip = 9.77 kip
2.309in
Comments
1. For the given rib and the single concentrated load at the center of the span it is obvious that the effects of shearing and axial
strains are insignificant and can be disregarded.
2. Eroneous conclusions as to the relative importance of shearing and axial strains in the usual solid rib may be drawn, however,
from the values in the equation for ΔCh. These indicated that the effects of the shearing strains are much more significant thatn
those of the axial strains. This is actually the case for the single concentrated load chosen for the demonstration, but
only because the rib does not approximate the funicular polygon for the single load. As a result, the shearing components on
most sections of the rib are more important than would otherwise be the case.
3. The usual arch encountered in practice, however, is subjected to a series of loads, and the axis of the rib will approximate
the funicular polygon for these loads. In other words, the line of pressure is nearly perpindicular to the right section at all points
along the rib. Consequently, the shearing components are so small that the shearing strains are insignificant and are neglected
4. Axial strains, resulting in rib shortening, become increasingly important as the rise-to-span ratio of the arch decreases. It is
advisable to determine the effects of the rib by considering flexural strains only, and then to check for effects or rib shortening.
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11.4 Semi-Circular Box Girder
Determine the reactions of the semi-circular cantilevered box girder subjected to its own weight w.
x x
wRda
? a
d
O
a
O
M
R
B ? r
A
C ? z
y y V
z z
Solution:
Reactions are again determined first
From geometry we have OA=R, OB=Rcosθ, CD = BA = OA - OB = R - Rcosθ, EB = R(1+cosθ) and BC = Rsinθ. The
moment arms for the moments with respect to the x and y axis are BC and EB respectively. Applying three equations
of equilibrium we obtain
θ =π
A A
Fz - wR dθ = 0 Fz = wR π
θ =0
θ =π
A A 2
Mz - wR dθ R sinθ = 0 Mz = 2 w R
θ=0
θ =π
A A 2
My - wR dθ R ( 1 + cosθ) = 0 M y = -w R π
θ=0
Internal Forces are determined next
1. Shear Force:
θ
Fz = 0 V - wR dα = 0
0
V = wRθ
2. Bending Moment:
θ
MR = 0 M - wR dα R sinα = 0
0
2
M = wR ( 1 - cosθ)
3. Torsion:
θ
Mθ = 0 T + wR dα R ( 1 - cosα) = 0
0
2
T = -wR ( θ - sinθ)
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Deflection are determined last
We assume a rectangular cross-section of width b and heigh d=2b and a Poisson's ratio of ν := 0.3
1. Noting that the member will be subjected to both flexural and torsional deformations, we seek to determine the two
stiffnesses.
3 4
b d 2Eb
2. The flexural stiffness EI is given by EI = E =
12 3
3
3. The torsional stiffness of solid rectangular sections J = kb d where b is the shorter side of the section, d is the longer,
E E E 4 4
and k is a factor equal to 0.229 for d/b=2. Hence G = = , and GJ = 0.229 b = 0.176E b .
2 ( 1 + ν) 2.6 2.6
4. Considering both flexural and torsional deformations, and replacing dx by rdθ:
π π
M T
δ M Δ = δ M R dθ + δ T R dθ
EIz GJ
0 0
where the real moments were given above.
5. Assuming a virtual downward force δ P = 1, we have
δ M = R sinθ
δ T = -R ( 1 - cosθ)
6. Substituting these expressions into the equation for displacement (in 4)
2 π 2 π
w R w R
1 Δ = ( R sinθ) ( 1 - cosθ) R dθ + ( θ - sinθ) R ( 1 - cosθ) R dθ
EI 0 GJ 0
4 π
w R ( sinθ - sinθ cosθ) + 1 ( θ - θ cosθ - sinθ + sinθ cosθ) dθ
1 Δ =
EI 0.265
0
4
w R
Δ = 20.56
EI
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11.5 Internal Forces in an Helicoidal Cantilevered Girder, Point Load
Determine the internal forces N, Vs, and Vw and the internal moments T, M, and Mw along the helicoidal cantilevered girder.
P
A
Z Isometric View
R
B
A P
Plan View
B a R
X Y
Solution:
1. We first determine the geometry in terms of the angle θ
H
x( R , θ) := R cos( θ) y( R , θ) := R sin( θ) z( H , θ) := θ
π
2. To determine the unit vector n at any point we need the derivatives
H
dx( R , θ) := x( R , θ) -R sin( θ) dy( R , θ) := y( R , θ) R cos( θ) dz( H , θ) := z( H , θ)
θ θ θ π
and then determine the unit vector
H
-R sin( θ) i + R cos( θ) j + k
-sinθ i + cosθ j + H k
π 1
n= =
1 1 π R
2 2
2 H
2 H
2
R ( sinθ) 2 + R2 ( cosθ) 2 + 1 +
π π R
Since the denominator depends only on the geometry, it will be designated K.
3. The strong bending axis lies in a horizontal plane, and its unit vector can thus be determined
i j
k
1
n k = -sinθ cosθ
H 1
= ( cosθ i + sinθ j )
K π R K
0
0 1
1
and the absolute magnitude of this vector n k = , and thus
K
s = cosθ i + sinθ j
4. The unit vector along the weak axis is determined
i j
k
w =s n = cosθ sinθ 0 1 H
1
= sinθ i -
H
cosθ j + k
K H K π R π R
-sinθ cosθ π R
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5. With the geometry definition completed, we now examine the equilibrium equations.
F = 0 F = -P
Mb = 0 M = -L P
where
L = ( R - R cosθ) i + ( 0 - R sinθ) j + 0 - H k
θ
π
and
i j k
-θ H
M = L P = R 1 - cosθ -sinθ = PR -sinθ i - ( 1 - cosθ) j
π R
0
0 P
and
M = PR sinθ i + ( 1 - cosθ) j
6. Finally, the components of the force F = -Pk and the moment M are obtained by appropriate dot products with the unit vectors
1 H
N = F n = - P
K π R
Vs = F s = 0
1
Vw = F w = - P
K
PR
T = M n = - ( 1 - cosθ)
K
M s = M s = PR sinθ
PH
M w = M w = - ( 1 - cosθ)
π K
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12.1 Moment Area, Cantilevered Beam
Determine the deflection at Point A
4wL2 2wL3
(25 L ) =
EI 25 5
M=
4wL w ρ
25
EI
M=
B ρ wL2
A C
1.5L L 2
tA/C
Solution:
-2w L2 5L 2 5L 1 -w L 9L -29w L
2 4
+
1
EItAC = L =
2 5 2 3 2 3 2 4 24
Thus,
4
-29w L
ΔA =
24 E I
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12.2 Moment Area, Simply Supported Beam
Determine ΔC and θC for the following example
P 3Pa/4
Pa/2
A C B
a a 2a A2
A1 A3
C’
C B ? BC
A
tA/B C
C”
?C
?B
Solution:
t AB
Deflection Δ is determined from ΔC = c' c = c' c'' - c'' c, c'' c = t CB, and c' c'' =
C 2
3
1 3P a a 2a 3P a 3a a + 3a = 5P a
t AB = +
EI 4 2 3 4 2 3 2 E I
This is positive, thus above tangent from B
3
P a 2a 2a P a
1
t CB = =
E I 2 2 3 3 E I
Positive, thus above tangent from B. Finally,
1 5P a
3 3 3
P a 11 P a
ΔC = - =
2 2 E I 3 E I 12 E I
Rotationa θC is
θBC = θB - θC => θC = θB - θBC
θBC = A3
t AB
θB =
L
3 2
5P a P a 2a P a
-
1
θC = =
2 E I 4a 2 2 8E I
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12.3 Maximum Deflection
Determine the deflection at D and the maximum deflection at B
L 4L/5 4Px/5
A B C D 4PL/5
9tCA/5
tDA
P/5 9P/5 ?D
Solution:
Deflection at D:
9
ΔD = t DA - t CA
5
3
-4 P L
L -2 P L
1
EItCA = L =
2 5 3 15
3
-2 P L
t CA =
15E I
3
-4 P L -4 P L -234 P L
17L
1 1 4L 8L
EItDA = L +
=
2 5 15 2 5 5 15 375
3
-234 P L
t DA =
375E I
Substituting we obtain
3
-48 P L
ΔD =
125 E I
Maximum Deflection at B:
3
-2 P L
t CA =
15E I
t CA -2 P L 2
θA = =
L 15E I
2
1 1 4P x -2 P x
θAB = x =
E I 2 5 5 E I
2 2
-2 P L
-2 P x L
θA = θAB => x= =
15E I 5 E I 3
4P x x 2x L
Δmax = at x =
5 E I 2 3 3
3
4 P L
Δmax =
45 3 E I
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12.4 Frame Deflection
Complete the following example problem
x = 4' 72/EI
72/EI
24 k
72/EI
90o ?B
B 6k
C 6? B
D
?B
12' 12 k tC/B
x = 8'
6k
A
6' 6' 6'
12 k tA/B
(a) (b)
120/EI ?B ?B
120/EI B
x = 8'
C B' ? B C C' B'
B
tCB ?B
6' 90o
tangent @ B
?B deflected
10 k position
20 k tangent @ B
6' vertical line ?B
20 k A A D
12' (b) (c)
10 k tAB12? B
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12.5 Frame Subjected to Temperature Loading
Neglecting axial deformation, compute displacement at A for the following frame
?C
? TT = 60
B C
? TB = 200
h2 =10'
?B ?C
?B
h1 =25' D ?B
t = 16"
A A
A” A'
L=20'
Solution:
1. First let us sketch the deformed shape
2. BC flexes => θB = θC 0
3. Rigid hinges at B and C with no load on AB and CD
4. Deflection at A
ΔA = A A'' = A A' + A' A''
A A' = ΔB = ΔC = θC h 2
A' A'' = θB h 1
5. We need to compute θB and θC
t CB
θB =
L
t BC
θC = θCB + θB OR θC =
L
6. In order to apply the curvature area theorem, we need a curvature (or moment diagram)
1 TB - TT M
= α =
ρ h E I
7.
t CB = A => θB = A =
L L 1 A -A
OR θB =
2
2 L 2 2
8.
θCB = A
θC = θCB + θB
A A
θC = A - =
2 2
9. From above
A A A
ΔA = θC h 2 + θB h 1 = h 2 + h 1 =
2 2
h + h2
2 1
( )
TB - TT
A = α L
h
TB - TT 1
ΔA = α
h
2
(
L h 1 + h 2)
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10. Substitute
ΔA := 6.5 10 ( -6 )(200 - 60) 20ft 1 35ft
16in 2
ΔA = 2.867 in
11. Other numerical values
A
θB = θC =
2
θB :=
1
6.5 10 (
- 6 200 - 60
)
20ft = 0.006825 rad
2 16in
θC := θB = 0.006825
M 1 TB - TT
= = α
EI ρ h
1
ρ := = 1465.2 ft
(
6.5 10- 6
) 20016in- 60
12. In order to get M, we need E and I. Note the difference with other statically determinate structures; the stiffer the beam, the higher the
moment; the higher the moment, the higher the stress? NO!!
M y E I y E y
13. σ= = =
I ρ I ρ
2 2 2
M d y d y M cx
14. ρ is constant => BC is on arc of circle M is constant and = => = = c => y = + dx + e
EI 2 2 EI 2
dx dx
15. The slope is a parabola
2
d y
2
1 M dx
= =
ρ EI 3
2
2
1 + dy
dx
Let us get curvature from the parabola slope and compare it with ρ
2
cx
y= + dx + e
2
dy
= cx + d
dx
2
d y
=c
2
dx
at x=0, y=o, thus e=0
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at x=0, y=o, thus e=0
dy
at x=0, = θB = -0.006825rad, thus d = -0.006825rad
dx
dy -4
at x=20 ft, = θC = 0.006825rad, thus c( 20ft) - 0.006825 = 0.006825, thus c = 6.825 10
dx
x2 -4
y = 6.825 10 - 10x
2
dy -4
= 6.825 10 ( x - 10)
dx
2
d y -4
= 6.825 10
2
dx
-4
ϕ := 6.825 10
1
ρ := ft = 1465.2 ft as expected!
ϕ
2
d y
2
dx
If we were to use the exact curvature formula
3
2
2
1 + dy
dx
-4
6.825 10
ϕ := = 0.000682
3
2
1 + ( 0.006825) 2
1
ρ := ft = 1465.3 ft
ϕ
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12.6 Conjugate Beam
Analyze the following beam
P
4PL/5EI
D
C
C D
A A
L 4L/5
Solution:
3 equations of equilibrium and 1 equation of condition = 4 = number of reactions. Deflection at D = Shear at D of the corresponding conjugate
beam (Reaction at D) Take AC and ΣM with respect to C
RA ( L) -
4P L L L
= 0
5E I 2 3
3
2 P L
RA =
15E I
(Slope in real beam at A) As computed before. Let us dram the moment diagram for the conjugate beam
M = L x -
P 2 2
EI 15
2 3
x =
2 P
15 15 E I
2
L x - x
3 ( )
dM
Point of maximum moment (Δ ) occurs when =0
max dx
dM
=
2 P
( 2
L - 3x ) = 0 => x =
2 L
dx 15 E I 3
as previously determined
L
x=
3
2 P 3
L L
2 L
M= -
15 E I
3 3
3
4P L
M= as before
45 3 E I
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12.7 Conjugate Beam
Analyze the following beam
6k 8k 6k
A B
I = I AD = 450in 4
D C E I = I EB = 450in 4
I' = 2I = I DE = 900in4
12 ft 6 ft 6 ft 12 ft
10k 10k
M ( k ft )
144
120
(+)
x ( ft )
12 18 24 36
Solution:
From simple observation, the reactions at A and B are equal to 10 k. The elastic load on the conjugate beam is then shown below.
120 120
EI 72 EI
60 60 720 72 720
EI
EI EI EI EI EI
720
EI
A B A B
C C
12 ft 6 ft 6 ft 12 ft
1116 8 ft 10 ft 10 ft 8 ft 1116
EI EI
720 36
EI EI
360
EI
A C MC
2 ft
3 ft
1116 10 ft
EI 18 ft vC
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We next seek to determine the internal moment at C' in the conjugate beam. It is obtained from equilibrium:
2 2 2 2
1116 kip ft 720 kip ft 360 kip ft 36 kip ft
Mz
B
=0 18ft - 10ft - 3 ft - 2 ft + M C' = 0
E I E I E I E I
3
E := 29 10 ksi
4
I := 450in
ΔC = M C'
2 2 2 2
720 kip ft 360 kip ft 36 kip ft 1116 kip ft
ΔC' := 10ft + 3 ft + 2 ft - 18ft = -1.554 in
E I E I E I E I
13.1 Deflection of a Cantilever Beam
Determine the deflection of the cantilever beam with span L under a point load P at its free end. Assume constant EI.
P x
M=PL
Solution:
1
We = P Δf
2
L
2
M ( x)
U= dx
2 E I
0
M = -P x
2 L 3 2
P 2 L P
U( P , L , EI) := x dx
2 EI 0 6 EI
3 2
1 L P
P Δf =
2 6 EI
3
P L
Δf =
3 EI
Note that the solution of this problem was facilitated by the fact thatΔf is co-alligned with P.
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13.2 Beam Deflection
4
Determine the deflection at point C. E := 29000ksi , I := 100in
2 k/ft 1k
A B
A C C
B dC 0.5 k 1.5 k
15 k 45 k
20' 10'
x x
15x-x2
Virtual Moment
Real Moment -x2 -0.5x -x
Solution:
For the virtual force method, we need to have two expressions for the moment, one due to the real load, and the other to the (unit)
virtual one.
Element x =0 M δM
2
AB A 15x - x - 0.5x
2
BC C -x -x
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For the virtual force method, we need to have two expressions for the moment, one due to the real load, and the other to the (unit) virtual
one.
L
M ( x)
ΔC δ P = δ M dx
EIz
0
20 10
15x - x2 -x2
1 ΔC = ( -0.5x) dx + ( -x) dx
EI
EI
0 0
20 10
15x - x2 -x2 2500
ΔC ( EI) := ( -0.5x) dx + ( -x) dx
EI
EI EI
0 0
3
2500kip ft
ΔC := = 1.49 in
E I
13.3 Deflection of a Frame
6 kN 6 4
Determine both the vertical and horizontal deflection at A for the frame shown. E := 200 10 , I := 200 10 mm
2
m
50 kN
1 kN 1 kN
A B C
2m 2m
5m
D
1 kN
50 kN 5 kN-m
100 kN-m 4 kN-m 1 kN
Solution:
To analyze the frame we must determine analytical expressions for the moments along each member for the real load and the two
virtual ones. One virtual load is a unit horizontal load at A, and the other a unit vertical one at A
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50 kN
1 kN
x x +4 1 kN
+100
+ +
50x
x
-x
-
1 kN
100 kN-m 4 kN-m 5 kN-m
50 kN 1 kN
Element x =0 M δMv δMh
AB A 0 -x 0
BC B - 50x - 2- x 0
CD C 100 4 -x
Note that moments are considered positive when they produce compression on the inside of the frame. Substitution yields:
L
M ( x)
Δν δ P = δ M ( x) dx
EIz
0
2 2 5
1Δν = -x
( 0)
dx + ( -2 - x) ( -50x) dx + 4 100 dx
EI EI EI
0 0 0
2 2 5
Δν( EI) := -x
( 0)
dx + ( -2 - x) ( -50x) dx + 4 100 dx 7000
EI EI EI 3 EI
0 0 0
3
7000kN m
Δν := = 5.833 cm
3 E I
Similarly for the horizontal displacement
L
M ( x)
Δh δ P = δ M ( x) dx
EIz
0
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2 2 5
1Δh = 0
( 0)
dx + ( 0) ( -50x) dx + -x 100 dx
EI EI EI
0 0 0
2 2 5
Δh ( EI) := 0
( 0)
dx + ( 0 ) ( -50x) dx + -x 100 dx - 1250
EI EI EI EI
0 0 0
3
-1250kN m
Δh := = -3.125 cm
E I
13.4 Rotation of a Frame
4
Determine the rotation of joint C for the frame shown. E := 29000ksi , I := 240in
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3 k/ft 1 k-ft
B C
?C
10'
15'
D
A 30 k
30 k 20' 0.05 k
0.05 k
Solution:
In this problem the virtual force is a unit moment applied at joint C, δ M e. It will cause an internal moment δ M i.
Element x=0 M δM
AB A 0 0
2
BC B 30x - 1.5x - 0.05x
CD D 0 0
Note that moments are considered positive when they produce compression on the outside of the frame. Substitution yields:
L
M ( x)
θC δ M e = δ M dx
EIz
0
20
30x - 1.5x2
1 θC = ( -0.05x) dx
EI
0
20
30x - 1.5x2 1000
θC ( EI) := ( -0.05x) dx -
EI EI
0
2
-1000kip ft
θC := = -0.021 rad
E I
13.5 Truss Deflection
Determine the vertical deflection of joint 7 in the truss shown E := 30000ksi
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2 in2 2 in2 2 2 3 3 4
1i
n2
2
2
1 6 7 4
1 in2
in
in
2
1 in2
15'
1 in2
=2
5 8 9
2
A
in
1
1 5
2 in2 2 in2 2 in2 2 in2 10 6 11 7 12 8 13
20k 20k 20k 1k
30k 30k
4 @ 20'
Solution:
Two analyses are required. One with the real load, and the other using a unit vertical load at joint 7. Results for those analyses are
summarized below. Note that advantage was taken of the symmetric load and structure.
A e δPPL/A nδPPL/A
L P δP
Member 2 2 n 2
in ft kip kip k-ft/in k-ft/in
1&4 2 25 -50 -0.083 518.75 2 1037.5
10 & 13 2 20 40 0.67 268 2 536
11 & 12 2 20 40 0.67 268 2 536
5&9 1 15 20 0 0 2 0
6&8 1 25 16.7 0.83 346.5 2 693
2&3 2 20 -53.3 -1.33 708.9 2 1417.8
7 1 15 0 0 0 1 0
Total 4220.3
L
P
Δ δ P = δ P dx
A E
0
Pe L
1 Δ = Σδ Pe
A E
kip ft
4220.3
2
in
Δ := = 1.688 in
E
13.6 Torsional and Flexural Deformation
4 4
Determine the vertical deflection at A in the structure shown. E := 30000ksi, I := 144in , G := 12000ksi, J := 288in
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5'
C
5k
10k B 1k
A
5'
Solution:
1. In this problem we have both flexural and torsional deformation. Hence we should determine the internal moment and torsion
distribution for both the real and the unit virtual load.
2. Then we will use the following relation
M
δ P ΔA = δ M dx + δ T T dx
EI GJ
3. The moment and torsion expressions are given by
Element x=0 M δM T δT
AB A 10x x 0 0
BC B 15x x 50 5
4. Substituting
M
δ P ΔA = δ M dx + δ T T dx
EI GJ
5 5 5
10x
1 ΔA = x dx + x 15x dx + 5 50 dx
EI EI GJ
0 0 0
5 5 5
10x
ΔA( E , I , G , J) := x dx + x 15x dx + 5 50 dx 3125 + 1250
E I E I G J 3 E I G J
0 0 0
3 3
3125kip ft 1250kip ft
ΔA := + = 1.042 in
3 E I G J
13.7 Flexural and Shear Deformations in a Beam
Determine the deflection of a cantilevered beam, of length L, subjected to an end force P due to both flexural and shear
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deformations. Assumed G=0.4E and a square solid beam cross section.
Solution:
1. The virtual work equation is
L L L L
M ( x)
δ P Δ = δ M ( x) dϕ( x) dx + δ V ( x) γxy( x) dx = δ M ( x) dx + δ V( x) λ V( x) dx
EI GA
0 0
0 0
2. The first integral yields for M=Px, and δ M = 1 x
L L
δ M ( x) M ( x) dx = P x2 dx
EI EI 0
0
L 3
P 2 94280057135446.5 N ft
x dx
EI 0 EI
3. The second integral represents the contribution of the shearing action to the total internal virtual work and hence to the total
displacement.
4. Both the real shear V and the virtual shear δ V are constant along the length of the member, hence
L L
δ V( x) λ V( x) dx = λ 1 ( P) dx
GA GA 0
0
L
λ 374606093684.841 GPa N ft
1 ( P) dx
GA 0 GA
5. Since λ := 1.2 for a square beam; hence
4
h 2
I= and A = h
12
then
P L 12L2 3P L 1.33L2
= + 1
9
Δ= +
3 E 4 2 2 2
h h E h h
6. Choosing L := 20ft and h := 1.5ft (L/h=13.3)
3P L 2
1.33 + 1 = 3P L ( 237 + 1 )
20ft
Δ=
E h 1.5ft E h 2
2
Thus the flexural deformation is 237 times the shear displacement. This comparison reveals why we normally neglect
shearing deformation in beams. As the beam get shorter or deeper, or as L/h decreases, the flexural deformation
2
decreases relative to the shear displacement. At L/h=5, the flexural deformation has reduced to 1.33 5 = 33.25 times
the shear displacement.
13.8 Thermal Effects in a Beam
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Consider the cantilever beam shown. If the beam is a steel, wide flange section, 2 m long and 200 mm deep, what is the angle of
rotation, θ1, at the end of the beam caused by the temperature effect? The original uniform temperature of the beam was 40C.
70oC
200 mm e=a? Tdx=11.7x10-6 (80)dx=
936x10-6 dx
230 oC
df
2m -80oC
100 mm
?1
?1 100 mm
+80oC
δM=1
Thermal Gradient dx
Strain
1
Virtual forces
Solution:
1. The external vurtual force conforming to the desired real displacement θ1 is a moment δ M = 1 at the tip of the cantilever,
producing an external work term of moment times rotation. The internal virtual force system for this cantilever beam is a
uniform moment δ M int = 1.
2. The real internal deformation results from (a) the average beam temperature of 150C, which is 110C above that of the original
temperature, and (b) the temperature gradient of 160C across the depth of the beam.
3. The first part of the thermal effect produces only a lengthening of the beam and does not enter into the work equation since the
virtual loading produces no axial force corresponding to an axial chnge in length of the beam.
L
4. The second effect (thermal gradient) produces rotation dϕ, and an internal virtual work term of δ M ϕ dx
0
5. We determine the value of dϕ by considering an extreme fiber thermal strain as shown above. The angular rotation in the length
dx is the extreme fiber thermal strain divided by half the beam depth.
2
M d y dθ ε
= = = =ϕ
EI 2 dx y
dx
L
1 θ1 = δ M ϕ dx
0
L
θ1 =
1
α TB - TT
dx
( )
h
0
-6m
11.7 10 ( 230 - 70) 2 m
m
θ1 := = 0.01872 rad
0.2m
6. This example raises the following points
a) The value of θ1 would be the same for any shape of 200 mm deep steel beam that has its neutal axis of bending at
middepth
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b) Curvature is produced only by thermal gradient and is independent of absolute temperature values.
c) The calculation of rotations by the method of virtual forces is simple and straightforward; the applied virtual force is a
moment acting at a point where rotation is to be calculated.
d) Internal angular deformation dϕ has been calculated for an effect other than load-induced stresses. This extension of
the method of virtual forces to treat inelastic displacements is obvious - all we need to know is a method for
determining the inelastic internal deformations.
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13.9 Deflection of a Truss
Determine the deflection at node 2 for the truss shown
A=5.0 in2 each;
60 k 120 k E=10x103 ksi 60 120
7 -45.0 -0.50 -0.54
4 5
-
.8
0.226
-0.56 -1.124
0 . 56
16.8
6
0.56
6 -117.3 -0.56
- 16
3 4 5 12' -83.8
-0.22
1.574
1 3 37.5 52.5 0.25 0.25 0.45 0.63
1 2
A
12' 12' 0.5 1.0 0.5
75.0 105.0
Solution:
δP P
e
L A E
Member 2 δPPL/AE
kip kip ft in ksi
3 -4
1 0.25 37.5 12 5 10 x 10 22.5 x 10
3 -4
2 0.25 52.5 12 5 10 x 10 31.5 x 10
3 -4
3 -0.56 -83.8 13.42 5 10 x 10 125.9 x 10
3 -4
4 0.56 16.8 13.42 5 10 x 10 25.3 x 10
3 -4
5 0.56 -16.8 13.42 5 10 x 10 -25.3 x 10
3 -4
6 -0.56 -117.3 13.42 5 10 x 10 176.6 x 10
3 -4
7 -0.5 -45 12 5 10 x 10 54 x 10
-4
Total 410.5 x 10
The deflection is thus given by
7
δ PL
δ P Δ =
Pe
AE
n =1
-4
Δ := 410.5 10 ft = 0.493 in
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13.10 Thermal Deflection of a Truss; I
The truss shown is built such that the lower chords are shielded from the rays of the sun. On a summer day the lower chords are
30F cooler than the rest of the truss members. What is the magnitude of the vertical displacement at joint 2 as a result of this
temperature difference?
Solution:
1. The virtual force system remains identical to that of the previous example because the desired displacement component is the
same.
2. The real internal displacements are made up of the shortening of those members of the truss that are shielded from the sun.
3. Both the bottom chord members 1 and 2 thus shorten by
ΔL = α ΔT L
ΔL := 0.0000128
in
( 30 Δ°F) 12ft = 0.0553 in
in Δ°F
4. Then,
1 Δ = δ P ΔL
Δ := 0.25( -ΔL) + 0.25( -ΔL)
Δ = -0.0276 in
5. The negative sign n the displacement indicates that it is in opposite sense of the displacement; the assumed direction is
always identical to the direction of the applied virtual force.
6. Note that the same result would be obtained if we had considered the internal displacements to be made up of lengthening of all
truss members above the bottom chord.
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13.11 Thermal Deflection of a Truss; II
A six panel highway bridge is constructed with sidewalks outside the trusses so that the bottom chords are shaded. What will be the vertical
deflection component of the bottom chord at the center of the bridge when the temperature of the bottom chord is 40F (ΔT) below that of the
1
top chord, endposts, and webs? (coefficient of steel thermal expansion is α := 0.0000065
Δ°F
5 6
28' 4
7 8 9 10 11
3 2 1 1k
6@21'
Solution:
1. The deflection is given by
L
P Pi Li
Δ δ P = δ P
AE
dx =
δPi AE = δPiΔLi = δPiαΔTL
0
where ΔL is the temperature change in length of each member, and δ P are the member virtual internal forces
2. Taking advantage of symmetry (Note that we ignor members 1-3 because we assumed that they had the reference temperature,
and all other members are subjected to a relative temperature increase of 40F)
L δP
Member αΔTL δPΔL
ft kip
4 35 0.0091 0.625 0.00568
5 21 0.00546 0.75 0.00409
6 21 0.00546 1.13 0.00616
7 0 0 0 0
8 35 0.0091 -0.625 -0.00568
9 28 0.00728 0.5 0.00364
10 35 0.0091 -0.625 -0.00568
11 0 0 0 0
Total 0.00821
3. Hence the total deflection is
Δ := 2 0.00821 ft = 0.197 in
4. A more efficient solution would have consisted in considering members 1, 2, and 3 only and applying ΔT=-40F, we would obtain
the same displacement.
5. Note that the forces in members 1, 2, and 3 (-0.75, -0.375, and -0.375 respectively) were not included in the table because the
corresponding ΔT=0.
6. A simple solution would have ΔT := -40 Δ°F in members 1, 2,and 3 thus
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L δP
Member αΔTL δPΔL
ft kip
1 21 -0.00546 -0.75 0.004095
2 21 -0.00546 -0.375 0.0020475
3 21 -0.00546 -0.375 0.0020475
Total 0.00819
Δ := 2 0.00819 ft = 0.197 in
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13.12 Truss with Initial Camber
It is desired to provide 3 in. of camber at the center of the truss shown below by fabricating the endposts and top chord members
additionally long. How much should the length of each endpost and each panel of the top chord be increased?
2 3
1 36'
1k
6 @ 27'
Solution:
1. Assume that eachtop chord is increased 0.1 in.
δPint ΔL
Member δPΔL
kip in
1 0.625 0.1 0.0625
2 0.75 0.1 0.075
3 1.125 0.1 0.1125
Total 0.25
Thus,
2 0.25in = 0.5 in
2. Since the structure is linear and elastic, the required increase of length for each section will be
3in
0.1in = 0.6 in
0.5in
3. If we use the practical value of 0.625in
0.625in 0.5in
= 3.125 in
0.1in
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13.13 Prestressed Conrete Beam with Continuously Variable I
A prestressed concrete beam is made of variable depth for proper location of the straight pretensioning tendon. Determine the midspan
kN
displacement (point c) produced by the dead weight of the girder. The concrete weighs w := 23.6 and has E := 25000MPa. The beam is
3
m
0.25m wide.
1.0 m 2.36 kN/m
0.25 m wide
A 0.6 m 3.54 kN/m
C B
20 m
47.2 kN 47.2 kN
DC
0.5 kN
0.5 kN 0.5 kN
Solution:
1. We seek an expression for the real moment M. This is accomplished by first determining the reactions and then considering
the free body diagram.
2. We have the intermediary resultant forces
3.54 kN x
R1 ( x) := 0.25m x 0.6m w
m
2
1 0.4m 0.118 kN x
R2 ( x) := 0.25m x x w
2 10m 2
m
Hence,
M ( x) := 47.2x - 3.54x
x 2 x 3 2
- 0.118x 47.2 x + -0.039 x + -1.77 x
2 3
3. The moment of inertia of the rectangular beam varies continuously and is given, for the left half of the beam, by
1 3
I( x) = b h
12
1 3 3
I( x) := ( 0.25) ( 0.6 + 0.04x) 0.021 ( 0.04 x + 0.6)
12
4. Thus, the real angle changes produced by the dead load bending are
3 2
M 47.2 x + -0.039 x + -1.77 x
dϕ = dx =
E ( 0.6 + 0.04x) 3
EI 1
48
1
5. The virtual force system corresponding to the desired displacement is shown above with δ M = x for the left half of the span. Since the
2
beam is symmetrical, the virtual work equations can be evaluated for only one half of the beam and the final answer is then obtained by
multiplying the half-beam result by two.
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6. The direct evaluation of the integral δ M dϕ is difficult beause of the expression for dϕ. Hence we shall use a numerical procedure,
M M
replacing the δ M dx with the δ M Δx, where each quantity in the summation is evaluated at the center of the interval Δx
EI EI
and held constant over the interval length. As Δx becomes very short, the solution approaches the exact answer.
7. An interval length of 1 meter, giving 10 elements in the half length of the beam, is chosen to establish an accurate result.
8. The internal virtual work quantity is then
L
2
M ( x) M ( x)
δ M M ( x)
Δx = δ M
M ( x) 48
δ M ( x) EI dx ~ ( x)
EI δ M ( x)
E
0.25 3
Δx =
E
( x) 3 Δx
0 h h
12
9. The summation for the 10 elements in the left half of the beam gives
3 3
Segment x h h M δM MδM/h
1 0.5 0.62 0.238 23.2 0.25 24
2 1.5 0.66 0.288 66.7 0.75 174
3 2.5 0.7 0.343 106.4 1.25 388
4 3.5 0.74 0.405 150 1.75 648
5 4.5 0.78 0.475 173 2.25 820
6 5.5 0.82 0.551 200 2.75 998
7 6.5 0.86 0.636 222 3.25 1134
8 7.5 0.9 0.729 238 3.75 1224
9 8.5 0.94 0.831 250 4.25 1279
10 9.5 0.98 0.941 256 4.75 1292
Total 7981
10. The SI units should be checked for consistency. Letting the virtual force carry the units of kN, the virtual moment has the units of m-kN,
and the units of the equation
1 ( m kN) ( m kN) m
m = = mm
kN MN 4 1000
m
2
m
11. Then
L
δ M M
dx
EI
0
2 7981 1 mm = 30.65 mm
48
25000
and the deflection at midspan is
Δc := 30.65 mm
12. Acceptably accurate results may be obtained with considerably fewer elements (longer intervals Δx). Using four elements with centers at
M
δ M Δx is
2, 5, 8, and 10, the
EI
3 ( 174 ) + 3 ( 820) + 3 ( 1224) + 1 ( 1292) = 7946
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which is only 0.4% lower than the 10 element solution. If we go to two elements, 3 and 8, we obtain a summation of
5 388 + 5 1224 = 8060 which is 1% high. A one element solution, with x=5m and h=0.8m, gives a summation of 9136
which is 14.4% high and much less accurate than the 2 element solution.
13. Finally, it should be noted that the calculations involved in this example are essentially identical to those necessary in the
moment area method.
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14.1 Steel Building Frame Analysis
A small, mass-produced industrial building is to be framed in structural steel with a typical cross as shown below. The engineer is considering
three different designs for the frame: (a) for poor or unknown soil conditions, the foundation for the frame may not be able to develop any
dependable horizontal forces at its bases. In this case the idealized base conditions are a hinge at one of the bases and a roller at the other;
(b) for excellent soil conditions with properly designed foundations, the base of the frame legs will have no tendency to move horizontally, and
the idealized base condition is that of hinges at both points A and D; and (c) a design intermediate to the above cases, with a steel tie member
capable of carrying only tension running between points A and D in the floor of the building. The foundations would not be expected to
provide any horizontal restraint for this latter case, and the hinge-roller details at points A and D would apply. Critical design loads for a frame
of this type are usually the gravity loads (deal load + snow load) and the combination of dead load and wind load. We will restrict our attention
to the first condition, and will use a snow load of S := 30psf and an estimated total dead load of D := 20psf . With frames spaced at
lbf
spacing := 15ftt on centers along the length of the building, the design load is wtotal := spacing ( S + D) = 750 . If the frame is made of
ft
steel beam sections 21 in. deep and weight 62 lb/ft of length (W 21x62), and the member for design (c) is tentatively chosen as a 2 in2 bar,
determine the bending moment digrams for the three designs and discuss the alternate solutions
Roof system
B
C
12' RDh RAh
D 40' A
RDv b) Hinged support
Structure cross section
RDh RDh Tie member
RDv a) Roller support RDv c) Roller with tie RAv
RAv
Solution:
Structure a This frame is statically determinate since it has three possible unknown external forces acting on it. Final bending
moments are shown below.
83.3ft.k 83.3ft.k 81ft.k 81ft.k
66.7ft.k 69ft.k
150 ft.k 6.75 k Tension
6.93k 6.93k
15k 15k 15k 15k 15k 15k
c)
a) b)
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Structure b Hinging both legs of the frame results in another unknown force, making the structure statically indeterminate to the first degree
(one redundant).
1. A lateral release at point A is chosen, with the redundant shearing force R1. The displacement Δ1 in the primary structure, as a result of the
real loading, shown in the figure below, is computed by virtual work.
2. The virtual force system produces virtual bending moment δ M, which is uniform across the top member of the frame.
Q=1
2
M=w/2(Lx-x )
?1
b) Virtual forces
a) Moment caused by actual load on primary
structure
dM=M=12
1
M=1.x
M=1.x
x dM=1.x
f11 dM=1.x
c) Real load for determining f11
d) Moments produced by virtual forces and unit redundant
The virtual moment acting through the real angle changes given the internal work term
40
M
1 Δ1 = δ M dx
EI
0
3. Equating this to the external virtual work of 1 Δ1, we have
40
1
(
0.75 40x - x
2)
2 3
1 Δ1 = 12 dx kip ft
EI
0
40
1
2
(
0.75 40x - x
2
) 3
48000 ft kip
3
Δ1 ( EI) := 12 dx kip ft
EI EI
0
4. The equation of consistent displacement is Δ1 + f11 R1 = 0. The flexibility coefficient f11 is computed by applying a unit horizontal force at
the release and determining the displacement at the same point.
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5. It is seen that the real loading and the virtual loading are identical for the calculations, and
12
20
x 12
1 f11 = 2 x dx + 12 dx
EI
EI
0 0
or
12
20 3
x 12 6912 ft kip
f11( EI) := 2 x dx + dx kip ft
3
12
EI
EI EI
0 0
6. solving for R1
1
EI
(
48000 + 6912R1 = 0 )
or
48000
R1 := kip = 6.944 kip
6912
Structure c The frame with the horizontal tie between the points A and D has three unknown external forces. However, the structure is
statically indeterminate to the first degree since the tie member provides one degree of internal redundancy.
1. The logical release to choose is a longintudinal release in the tie member, with the associated longitudinal displacement and axial force.
2. The primary structure is the frame with the tie member released.
Release in tie member dQ=1
a) Primary structure c) Virtual forces
dM=M=12
M=1.x M=1.x
? 1Q dM=1.x dM=1.x
b) Actual loading on primary structure
e) Moments produced by virtual forces and unit
redundants
1k
d) Real loading for computing f 11 f11
The compatability equation is based on the fact that the displacement at the release must be zero; that is, relative displacement of the two
sections of the tie at the point of release must be zero, or
Δ1 + f11 R 1 = 0
where
Δ1 = displacement at release 1 in the primary structure, produced by the loading
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f11 = relative displacement at release 1 for a unit axial force in the member
R1 = force in the tie member in the original structure
3. Virtual work is used to determine both displacement terms.
4. The value of Δ1 is identical to the displacement for structure (b) because the tie member has no forces (and consequently no deformation)
in the primary structure. Thus,
3
48000kip ft
Δ1 := = 2.083 in
4
30000ksi 1327in
5. The flexibility coefficient f11 is composed of two separate effects: a flexural displacement due to the flexibility of the frame, and the axial
displacement of the stressed tie member. The virtual and real loadins for this calculation are shown in the previous figure.
12
20 PL
x 12
1 f11 = 2 x dx + 12 dx + δ P
EI
EI EA
0 0
6912 1 1 40
1f11 = +
EI EA
6912kip ft3 40 kip ft 1
f11 := + = 0.308
in
4 2 kip kip
30000ksi 1327in 30000ksi 2 in
6. The equation of consistent deformation is
Δ1 + f11 R 1 = 0
or
-Δ1
R1 := = -6.764 kip
f11
7. The two displacement terms in the equation must carry opposite signs to account for their differences in direction.
Comments The bending moment in the frame differes only slightly from that of structure (b). In other words, the tie member has such high
axial stiffness that it provides nearly as much restraint as the foundation of structure (b). Frames with tie members are used widely in industrial
buildings. A lesson to be learned here is that it is easy to provide high stiffness through an axially loaded member.
The maximum moment in frames (b) and (c) is about 55% of the maximum moment in frame (a). This effect of continuity and redundancy is
typical - the positive bending moments in the members are lowered while the joint moments are increased and a more economical design
can be realized. Finally, we should notice that the vertical reactions at the bases of the columns do not change with the degree of horizontal
restraint at the bases.
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14.2 Analysis of Irregular Building Frame
The structural steel frame for the Church of the Holy Spirit, Penfield, New York is shown below. In this example we will discuss the
idealization of the structure and then determine the forces and bending moments acting on the frame.
10'
W4-13
E F
8'
W30-108 B w=0.8k/ft B
A w=0.72k/ft
15' 8'
A G C W14-30 D M=24.6x-0.4x 2 C D
W10-49
E M=7.86x2
d) Moment due to actual loading on structure
61'-5" 21'-8" dQ=1
a) Geometry of frame
M=x/61.42 D
B dM=1 C
A dM=x/21.33
C D e) Moment dM due to dQ=1 at release 1
b) Idealized frame E 47.1 k.ft
A B 23.8 k 354 k.ft
5.5 k
C D
c) Primary frame with angular 35.2 k
f) Reactions and moment for frame
release at B
Solution:
1. The two main horizontal members of the frame are supported at points A and D by masonry walls.
2. The connection used at these points is not intended to transmit axial forces from the frame to the wall; accordingly, the axial forces in the
horizontal members are assumed to be zero and the joints at A and D are idealized as rollers that transmit vertical forces only.
3. The base joint E is designed to resist both horizontal and vertical loads, but not moment, and is assumed to be a hinge.
4. Finally, joints B and C are designed to provide continuity and will be taken as rigid; that is, the angles of intersection of the members at the
joint do not change with applied loading.
5. The frame is simplified for analysis by removing the small 4 in. wide flange members EF and FG and replacing their load effect by applying
the roof load with acts on EF directly to the segment AG.
6. the idealized frame is shown above.
7. The dead load on the higher portion of the frame is p AB := 25psf times the frame spacing of spacing := 13.33ft, or
lbf
wAB := p AB spacing = 333 along the frame.
ft
8. The dead load on CD is less beacuse the weight of the frame member is substantially smaller, and the dead load is about p CD := 19psf ,
lbf
or wCD := p CD spacing = 253 .
ft
lbf
9. Snow load is S := 35psf over both areas, or wSnow := S spacing = 467 .
ft
10. The total loads are then
kip
Member AB: wtotAB := wAB + wSnow = 0.8
ft
kip
Member CD: wtotCd := wCD + wSnow = 0.72
ft
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11. the frame has four unknown reaction components and therefore has one redundant. Although several different releases are possible, we
choose an angular (bending) release at point B.
12. The resulting primary structure is shown as (c) in the figure above, where the redundant quantitity R1 is the bending moment at point B.
13. The equation of compatability is
θ1 + f11 M 1 = 0
where θ1 is the relative angular rotation corresponding to release 1 as produced by the actual loading, and f11 is the
flexibility coefficient for a unit moment acting at the release.
14. From virtual work we have
M
1 θ1 = δ M dx
EI
and
M
1 f11 = δ M dx
EI
where δ M and M are defined in (e) in the figure above.
15. Then
61.42 21.33
θ1 EIAB , EICD :=
( )
1
x
( 24.6x - 0.4x
2
) dx + EI1
x
(
7.68x - 0.36x
2
) dx
EIAB
61.42 CD
21.33
0 0
291.31658667 7763.6329512000000002
(
θ1 EIAB , EICD ) EICD
+
EIAB
4 4
16. With IAB := 4470in and ICD := 290in
θ1 ( E) :=
291 7764 4 118411
+ in
E ICD E IAB 43210 E
17. Similarly
61.42 21.33
8
1 1
2 2
x x dx + 1
(
f11 EIAB , EICD , EIBC := )
EIAB
61.42
dx +
EICD
21.33 EIBC 0
2
( 1 ) dx
0 0
8 20.473 7.11
(
f11 EIAB , EICD , EIBC )
EIBC
+
EIAB
+
EICD
4
18. With IBC := 273in
f11( E) := 8 + 20.473 + 7.11 in4 simplify 0.058
E I
BC E IAB E ICD E
2
kip ft
Note that the numerators of θ and f have units of
1 11 4
in
19. Applying the compatability equation,
118411 0.058
+ M1 = 0
43210 E E
and the bending moment at point B is
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118411
-
43210 E
M 1 := kip ft = -47.248 kip ft
0.058
E
The reactions and moments in the structure are given in (f) in the figure above.
20. Once we have M1, the structure is by now statically determinate and from statics we can complete the shear and moment diagrams.
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14.3 Redundant Truss Analysis
Determine the bar forces in the steel truss shown below using the flexibility method. The truss is part of a supporting tower for a
tank, and the 20 kN horizontal load is produced by wind loading on the tank.
20kN
B 20 0
A
0
3m
-2
0
8 .2
20
Release
8
C D 20
20
20
20 ? 1q
3m
c) Primary structure
b) Primary structure
a) Original Structure subjected to real load
-0.707 -0.707 -10
20 14
+1 1 4. -10
+
-0.707
-0.707
-0.707
-0.707
-14
+10
. 14
1
1
20
-0.707 +10
-0.707
d) Virtual load on 20 20
f
primary structure 11
e) Action of unit force
f) Final results
corresponding to redundant R1
Solution:
1. Applying the criteria for indeterminancy 2 4 = 8 equations, 6 members + 3 reactions => one degree of indeterminancy. A logitudnial
release in any of the six bars may be chosen.
2. Because the truss members carry only axial load, the longitudinal release is identical to actually cutting the member and removing its axial
force capability from the truss.
3. In analyzing the trusses with double diaganols it is both convenient and customary to select the release in one of the diganol members
because the resulting primary structure will be the conventional truss to which we are accustomed.
4. Choosing the diagonal member BC for the release, we cut it and remove its axial stiffness from the structure. The primary structure is
shown in (b) in the figure above.
5. The analysis problem reduces to applying an equation of compatability to the changes in length of the release member. The relative
displacement Δ1 of the two cut ends of member BC, as producued by the real loading, is shown in (d) in the figure above.
6. The displacement is always measured along the length of the redundant member, and since the redundant is unstressed at this stage of
the analysis, the displacement Δ1 is equal to the relative displacement of joint B with respect to joint C.
7. This displacement must be eliminated by the relative displacements of the cut ends of member BC when the redundant force is acting on
the member. The latter displacement is written in terms of the axial flexibility coefficient f11, and the desired equation of consistent deformation
is
Δ1 + f11 R 1 = 0
8. The quantity Δ1 is given by
PL
δ P
1 Δ1 =
AE
where δ P and P are given in (d) and (e) in the figure above, respectively
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9. Similarly,
P L
f11 =
δ P
AE
10. Evaluating these summations in tabular form:
P δP L
Member δPPL δPPL
kN kN m
AB 0 -0.707 3 0 1.5
BD 0 -0.707 3 0 1.5
CD 20 -0.707 3 -42.42 1.5
AC 20 -0.707 3 -42.42 1.5
AD -28.28 1 4.242 -119.96 4.242
BC 0 1 4.242 0 4.242
Total -204.8 14.484
11. Since A = constant for each member
PL -204.8
δ P
14.484
2 2
Δ1 = = kN m and f11 = kN m
AE AE AE
then
1
AE
( )
-204.8 + 14.484 R1 = 0
204.8
12. the solution for the redundant force value is R1 := kN = 14.14 kN
14.484
13. The final values for forces in each of the truss members are given by superimposing the forces due to the redundant and the forces due
to the real loading.
14. The real loading forces are shown in (c) in the figure above, while the redundant force effect is computed by multiplying the member
forces in (d) in the figure above by 2.83, the value of the redundant.
δP R1δP P Ptotal
Member
kN kN kN kN
AB -0.707 -10 0 -10
BD -0.707 -10 0 -10
CD -0.707 -10 20 10
AC -0.707 -10 20 10
AD 1 14.14 -28.28 -14.14
BC 1 14.14 0 14.14
15. It is informative to compare the member forces from this solution to the approximate analysis obtained by assuming that the
double diagonals each carry half the total shear in the panel. The comparison is given in the figure below
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-10 -10
20 20
14
4
.1
4.
4
+1
+1
+10
-10
-10
+10
-1
-1
4.1
4.
4
14
20 20
+10 +10
20 20
20 20
a) Exact analysis b) Approximate analysis
It reveals that the approximate analysis is the same as the exact analysis for this particular truss. The reason for this is that the
stiffness provided by each of the diagonal members (against "shear" deformation of the rectangular panel) is the same, and
therefore they each carry an equal portion of the total shear accross the panel.
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14.4 Truss with Two Redundants
Another panel with a second redundant member is added to the truss of the preceeding example
Solution:
1. The twice redundant truss is converted to a determinate primary structure by releasing two members of the truss; we choose
two diagonals (DB and BF).
2. Releasing both diagonals in a single panel, such as members AE and DB, is inadmissible since it leads to an unstable truss
form.
3. The member forces and required displacements for the real loading and for the two redundant forces in members DB and BF are
given in the figure below.
15k
15k -9.5 -9.5 ?2
4k A B 4 15 -0.707 -f21
C
-7. +1
78
07
43
-0.707
-9.5
-5.5
-15
3.
-0.7
10'
+1 1k
4
D F 4 0 -0.707
E ?1 f11
5.5 9.5
2@10'
b) Primary structure subjected to c) Action to unit force R1=1
real load
a) Original structure; all A=4 in2
-0.707 f22 15
4 -5.82 -3.56
-0.707
-0.707
1k 2.5 -8.
8 38
+5 -5.38
-1.82
-3.56
.20
.05
-5
-0.707
-f12 4
7.68 5.94
d) Action of unit force R2=1
e) Final results
4. Although the rela loading ordinarily stresses all members of the entire truss, we see that the unit forces corresponding to the redundant
stress only those members in the panel that contain the redundant; all other bar forces are zero.
5. Recognizing this fact enables us to solve the double diagonal truss problem more rapidly than a frame with multiple redundants.
6. The virtual work equations for computing the six required displacements (two due to load and four flexibilities) are
PL
δ P1
1 Δ1 =
AE
PL
δ P2
1 Δ2 =
AE
P1 L
1 f11 =
δ P1
AE
P1 L
1 f21 =
δ P2
AE
f12 = f21 by reciporocal theorem
P2 L
1 f22 =
δ P2
AE
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7. If we assume tension in a truss member as positive, use tensile unit loads when computing the flexibility coeeficients
corresponding to the redundants, and let all displacement terms carry their own signs, then in the for the redundants a positive
value of force indicates tension while a negative value menas the member is in compression.
8. The calculation of f22 involves only the six members in the left panel of the truss; f21 invloves only member BE.
9. The simple used for performing the displacement analyses, as summarized in tabular form, leads one quickly to the
compatability equations which state that the cut ends of both redundant members must match (there can be no gaps or overlaps
of members in the actual structure).
Δ1 Δ1 f11 f21 f22
Member P P1 P2
δP1PL δP2PL δP1P1L δP2P1L δP2P2L
AB -9.5 -0.707 0 806 0 60 0 0
BC -9.5 0 -0.707 0 806 0 0 60
CF -9.5 0 -0.707 0 806 0 0 60
EF 0 0 -0.707 0 0 0 0 60
DE 4 -0.707 0 -340 0 60 0 0
AD -5.5 -0.707 0 466 0 60 0 0
AE 7.78 1 0 1322 0 170 0 0
BE -15 -0.707 -0.707 1272 1272 60 60 60
CE 13.43 0 1 0 2280 0 0 170
BD 0 1 0 0 0 170 0 0
BF 0 0 1 0 0 0 0 170
Total 3528 5164 580 60 580
10. The equations are
Δ1 + f11 R 1 + f12 R2 = 0
Δ2 + f21 R 1 + f22 R2 = 0
or
1 580 60 1 -1. 3528
R
=
AE 60 580 R 2 AE 5164
11. The final set of forces in the truss is obtained by adding up, for each member, the three separate effects. In terms of the forces shown in
the figure above, the force in any member is given by F = P + R1 P1 + R2 P2
Member P P1 P2 R1P1 R2P2 Ptotal
AB -9.5 -0.707 0 3.676 0 -5.82
BC -9.5 0 -0.707 0 5.925 -3.56
CF -9.5 0 -0.707 0 5.925 -3.56
EF 0 0 -0.707 0 5.925 5.94
DE 4 -0.707 0 3.676 0 7.68
AD -5.5 -0.707 0 3.676 0 -1.82
AE 7.78 1 0 -5.2 0 2.58
BE -15 -0.707 -0.707 3.676 5.925 -5.38
CE 13.43 0 1 0 -8.38 5.05
BD 0 1 0 -5.2 0 -5.2
BF 0 0 1 0 -8.38 -8.38
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14.5 Analysis of Nonprismatic Members
The nonprismatic beam is loaded with an end moment MA at its hinged end A. Determine the moment induced at the fixed end B
by this loading.
A 2I B
I
R1=1
L
- 3L
L/2 L/2 8EI
16EI
a) Beam with end B fixed e) Rotations produced by a unit value of redundant R 1
MA ?A MB 0.40L
MA
MB=2/3MA
b) Elastic curve
f) Moment diagram for A hinged, B fixed
?B
MB
MA
c) Primary structure
g) Beam with end A fixed
0.286L
MA M L MB
- A
5M A L 8EI
- MA=2/5MB
16EI
d) Rotations produced by MA h) Moment diagram for A fixed, B hinged
Solution:
1. The beam has one redundant force; we select MB as the redundant R1, and obtain the primary structure shown in the figure above. It can
be shown that the flexibility coefficients for unit moments applied at each end are those in (d) and (e) in the figure above, with a sing
convention of counterclockwise as positive.
2. The equation of consistent displacement at B is
-M A L 3 L
+ R = 0
8E I 16 E I 1
and the value of MB is
2
M B = R1 = M A
3
3. The resulting moment diagram is given in (f). We note that the inflection point is 0.4L from the fixed end. If the beam had a uniform value of I
across its span, the inflection point would be L/3 from the fixed end. Thus the inflection point shifts toward the section of reduced stiffness.
4. The end rotation θA is given by
5 M A L 1 2 11 M A L
- M A
L
θA = =
16 E I
8 3 E I 48 E I
MA
5. The ratio of the applied end moment to rotation is called the rotational stiffness and is
θA
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MA 48 E I
=
θA 11 L
6. If we now reverse the boundary conditions, making A fixed and B hinged, and repeat the analysis for an applied moment MB, the resulting
moment diagram will be as given in (h) in the above figure. The moment induced at end A is only 40% of the applied end moment MB. The
inflection point is 0.286L from the fixed end A. The corresponding end rotation θB in (g) in the figure above is
11 M B L
θB =
80 E I
MB
7. The rotational stiffness is
θB
MA 80 E I
=
θA 11 L
8. A careful comparison of the rotational stiffnesses, and of the moment diagrams in (f) and (h) in the figure above, illustrate the fact that
flexural sections of increased stiffness attract more moment, and that inflection points always shift in the direction of decreased stiffness.
9. The approach illustrated here may be used to determine moments and end rotations in any type of non-prismatic member. The end
rotations need in the force analysis may be calculated by either virtual work or moment area (or by other methods). Complex variations in EI
are handles by numerical integration of the virtual work equation or by approximating the resultant M/EI area and their locations in the
moment area method.
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14.6 Fixed End Moments for Nonprismatic Members
The beam of the previous example
Solution:
1. The beam has two redundant forces and we select MA and MB. Releasing these redundants, R1 and R2, we obtain the primary sturcture.
2. The equations of consistent deformations are
Δ1 + f11 R 1 + f12 R2 = 0
Δ2 + f21 R 1 + f22 R2 = 0
where R1 is MA and R2 is MB.
3. The values of Δ1 and Δ2, the end rotations produced by the real loading on the primary structure, can be computed by the virtual work
method.
4. The flexibility coefficients are also separately derived (not yet in these notes)
5. We define counterclockwise end moments and rotations as positive and obtain
5 -1
L 16 8 1 w L -0.352
R 3
=
E I -1 3 R1 E I 0.0273
8 16
from which
2
R1 = M A = 0.0742w L
2
R2 = M B = -0.0961w L
6. The stiffer end of the beam attracts 30% more than the flexible end.
2
w L
7. For a prismatic beam with constant I, the fixed end moments are equal in magnitude ( M A = -M B = ) and intermediate in value
12
between the two end moments determiend above.
8. Fixed end moments are an essential part of indeterminate analysis based on the displacement (stiffness) method and will be used
exclusively in the Moment Distribution method.
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14.7 Rectangular Frame; External Load
QC1
QB EIB QC2
B C
h EIC
A D
L
Solution:
1. The structure is statically indeterminate to the third degree, and the displacements (flexibility terms) are shown in the figure
above.
2. In order to evaluate the 9 flexibility terms, we refer tot he table.
2k 5k
10 k.ft
1k 1k 1 ft.k
Release 1 Release 2 Release 3
5L-10
h 1
L
2h+5L-10
f31 f32
?2 f21
f12 f23
?1 f33
?3 f11
f22
f13
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3. Substituting h := 10ft, L := 20ft, and EIB = EIC = EI, the flexibility matrix then becomes
2667 3000 -300
1
f = 3000 6667 -400
EI
-300 -400 40
and the vector of displacements for the primary structure is
-12833
1
Δ= -31333
EI
1800
where the units are kips and feet.
4. The inverse of the flexibility matrix is
2.4 0 18
EI 00.375 3.75
-1 -3
f = 10
18 3.75 197.5
5. Hence the reactions are determined from
R1 2.4 0 18 -12833 1.601
- 3 1
R2 := 10 EI 0 0.375 3.75 -31333 -5
EI
R 18 3.75 197.5 1800 7.007
3
R1 1.6
R2 = -5
R 7
3
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14.8 Frame with Temperature and Support Displacements
The single bay frame, of the previous example, has a height h := 10ft and span L := 20ft and its two supportsrigidly connected and is
constructed of reinforced concrete. It supports a roof and wall partitions in such a manner that linear temperature variation occurs across the
depth of the frame members when inside and outside temperatures differ. Assume the member depth is constant at 1 ft, and that the
structure was built with fixed bases A and D at a temperature of 85F. The temperature is now 70F inside and 20F outside. We wish to
determine the reactions at D under these conditions. Assume that the coefficient of linear expansion of reinforced concrete is
1
α := 0.0000055 .
Δ°F
Solution:
1. Our analysis proceeds as before, with the [D] vector interpreted appropriately. The three releases shown in the previous example, will be
used.
2. The first stage in the analysis is the computation of the relative displacements Δ1Δ, Δ2Δ, Δ3Δ, of the primary structure caused by temperature
effects. These diaplcements are caused by two effects: axial shortenin g of the members because of the drop in average temperature (a
middepth of the members), and curvature of the members because of the temperature gradient.
3. In the following discussion the contributions to the displacements due to the axial strain are denoted with a single prime (') and those due to
curvature by a double prime (").
4. Consider the axial strain first. A unit length of frame member shortens as a result of the temperature decrease from 85F to 45F at the
midepth of the member. the strain is therefore
αΔT := α 40 Δ°F = 0.00022
5. The effect of axial strain on the relative displacements needs little analysis. The horizontal member shortens by an amount
αΔT 20ft = 0.0044 ft. The shortening of the vertical members results in no relative displacement in the vertical direction. No rotation occurs.
6. We therefore have Δ'1Δ := -αΔT 20ft = -0.0044 ft, Δ'2Δ := 0ft, and Δ'3Δ := 0rad
7. The effect of curvature must also be considered. A frame element of length dx undergoes an angular strain as a result of the temperature
gradient. The change in length at an extreme fiber is
ε ( dx) := α 25 Δ°F dx 0.0001375 dx
8. With the resulting real rotation of the cross section
ε ( dx)
dϕ( dx) := rad 0.000275 dx rad
0.5
9. The relative displacements of the primary structure at D are found by the virtual force method.
10. A virtual force δ Q is applied in the direction of the desired displacement and the resulting moment diagram δ M determined.
11. The virtual work equation
δ Q Δ = δ M dϕ
is used to obtain each of the desired displacements at D.
12. The results, which you should verify, are
Δ''1 := 0.0828ft
Δ''2 := 0.1104ft
Δ''3 := -0.01104 rad
13. Combining the effects of axial and rotational strain, we have
Δ1 := Δ'1Δ + Δ''1 = 0.0784 ft
Δ2 := Δ'2Δ + Δ''2 = 0.1104 ft
Δ3 := Δ'3Δ + Δ''3 = -0.01104 rad
14. We now compute the reduntants caused by temperature effects
-1
R=f ( -Δ)
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R1 -0.0784
0.0106 - 3
- 3
R2 = 10 EI ( 18 3.75 197.5 ) -0.1104 = 10 EI
0.355
R 0.01104
3
where the units are kips and feet.
15. You should construct the moment diagram for the structure using the values of the redundant found in the analysis.
16. Notice the stiffness term EI does not cancel out in this case. Internal forces and reactions in a statically indeterminate subject to effects
other than loads (such as temperature) are dependent on the actual stiffnesses of the structure.
17. The effects of axial strain caused by forces in the members have been neglected in this analsis. This is usual for low frames where
bending strain dominates behavior. To illustrate the significance of this assumption, consider member BC. We have found
- 6 EI - 6 EI
R1 = 10.6 10 . The tension in BC has this same value, resulting in a strain for the member of 10.6 10 .For a rectangular
EA EA
3 2
I b d d -6 -7
member, = . In our case d=1ft, therefore the axial strain is 10.6 10 ( 0.0833) = 8.83 10
= , which is several orders of
A 12 bd 12
magnitude smaller than the temperature strain computed for the same member. We may therefore rest assured that neglecting axial strain
caused by forces does not affect the values of the redundants in a significant manner for this structure.
18. Now considert the effects of foundation movement on the same structure. The intermediate frame behavior depends on a structure that
we did not design: the earth. The earth is an essential part of nearly all structures, and we must understand the effects of foundation behavior
on structural behavior. for the purposes of this example, assume that a foundation has revealed the possibility of a clockwise rotation of the
support at D of 0.001 radians and a downward movement of the support at D of 0.12 ft. We wish to evaluate the redundants R1, R2, and R3
caused by this foundation movement.
19. No analysis is needed to determine the values of Δ1Δ, Δ2Δ, and Δ3Δ for the solution of the redundants. These displacements are found
directly from the support movements, with proper consideration, of the originally chosen sign convention which defined the positive direction
of the relative displacements. From the given support displacements, we find Δ1Δ := 0, Δ2Δ := 0.12ft, and Δ3Δ := 0.001rad. Can you
evaluate these quantities for a casr in which the displacements occured at A instead of D?
20. The values of the redundants is given by
-1
R=f ( -Δ)
R1
-3 -0.12 18 - 3
R2 = 10 EI ( 18 3.75 197.5 ) = 10 EI
R 0.001 -252.5
3
with units in kips and feet.
21. A moment diagram may now be constructed, and other internal force quantities computed from the now known values of the redundants.
The redundants have been evaluated separately for effects of temperature and foundation settlement. These effects may be combin ed with
those due to loading using the principle of superposition.
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14.9 Braced Bent with Loads and Temperature Change
The struss shown below represents an internal braced bent in an enclosed shed, with lateral loads of 20 kN at the panel points. A
temperature drop of 30C may occur on the other members (members 1-2, 2-3, 3-4, 4-5, and 5-6). We wish to analyze the truss for
loading and for temperature effects.
Solution:
1. The first step in the analysis is the definition of the two redundants. The choice of forces in diagonals 2-4 and 1-5 as redundants facilitates
the computations because some of the load effects are easy to analyze.
2. The computations are organized in tabular form in Table ??. The first column gives the force in each bar caused by a unit load (1 kN)
q1
corresponding to release 1. These are denoted p and also represent the bar force caused by a virtual force δ Q1 applied at the same
i δ Q1
location. Column 3 lists the same quantitiy for a unit load and for a virutal force δ Q2 applied at release 2. These three columns constitute a
record of the truss analysis needed for this problem.
Verticals: 500 mm2 ; Webs: 250 mm 2; E=200,000 N/mm2; a=1.0x10-5/ oC
20kN 3 4 4.85
-7.5
R1
.87
-6.
-6
.7
8
4.85
-17
4.85
10
7
L=3m
.6
-7.5
12.5
7.76
20kN
5 -9.0
R2 -2
.11
2 6.2
-4
-4
L=3m
2.91
2.91
.1
39.4
1
-41.4
.3
1 6
30
2.91
-18.5
d) Forces due to
a) Structure b) Redundant forces c) Forces due to lateral load temperature drop
3. Column 4 gives the value of L/EA for each bar in terms of Lc/EAc of the vertical members. This is useful because the term L/EA cancels out
in some of the calculations.
4. The method of virtual work is applied directly to compute the displacements Δ1Q and Δ2Q corresponding to the releases and caused by the
actual loads. Apply a virtual force δ Q1 at release 1. The internal virtual forces q 1 are found in column 2. The internal virtual work q 1 Δl is
Lc
found in column 5 as the product of columns 1, 2 and 4. The summation of column 5 is Δ1q = -122.42 . Similarly, column 6 is the
EAc
Lc
product of columns 1, 3, and 4, giving Δ2q = -273.12 .
EAc
5. The same method is used to compute the flexibilities. In this case, the real loading is a unit load corresponding to release 1 leading to f11
and f21, and then to release 2 leading to f12 and f22. Column 7 shows the computation for f11. It is the product of column 2, representing force
due to the real unit load with column 2 representing force due to a virtual force δ Q1 at the same location (release 1) multiplied by column 4 to
include the Lc/EAc term. Column 8 derives from columns 2, 3, and 4 and leads to f21. Columns 9 and 10 are the computations for the
remaining flexibilities.
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6. We have assumed that a temperature drop of 30C occurs in the outer members. The corresponding length changes are found in column
11. Again using the virtual work method, column 12 tabulates the internal virtual work of th virtual forces through displacements Δl where for
each bar, Δl = αlΔT. Column 12 is therefore the product of columsn 2 and 11. The summation of the elements of column 12 is the
displacement Δl corresponding to release 1. Column 13 repeats this process for Δ2 corresponding to release 2.
7. The tabulated information provides the necessary terms for a matrix solution of the problem. We have
8.66 1 Lc
f =
1 8.66 EAc
-122.42 Lc
Δq =
-273.12 EAc
6.36 - 4
ΔΔ = 10 Lc
4.24
therefore
-1 0.117 -0.0134 EAc
f =
-0.0134 0.117 Lc
8. The redundant forces due to the applied load are
-1
R=f (
-ΔQ )
EAc 122.42 Lc 10.66
R = ( -0.0134 0.117 ) =
Lc 273.12 EAc 30.32
9. Thus R 1 = 10.66kN, R2 = 30.32kN
-1
10. The redundant forces due to the temperature drop are R = f ( )
-ΔΔ
EAc -6.36 - 4 -6.87 - 5
R = ( -0.0134 0.117 ) 10 Lc = 10 EAc
Lc -4.24 -4.11
kN 2
11. Thus with E := 200 , Ac := 500mm
2
mm
-5
R1 := -6.87 10 E Ac = -6.87 kN
-5
R2 := -4.11 10 E Ac = -4.11 kN
12. Using the redundant forces from each of these analyses, the remainder of the bar forces are computed by simple equilibrium. Table ??
contains such computations. The bar forces in any bar is the force in column 1 added to that in column 2 multiplied by R1 plus that is column 3
multiplied by R2. This follows from the fact that columns 2 and 3 are bar forces caused by a force of unitiy corresponding to each of the
redundants. The results of the calcultations are shown for the applied loading and for the temperature drop. The forces caused by the
temperature drop are similar in magnitude to those caused by wind load in this example. Temperature differences, shrinkage, support
settlement, or tolerance errors can cause important effects in statically indeterminate structures.
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15.1 Approximate Analysis of a Frame Subjected to Vertical and Horizontal Loads
Draw the shear and moment diagrams for the following frame.
0.25 k/ft
15 k
12 13 14
0.50 k/ft 5 8
14' 6 7
30 k
9 10 11
4 16'
16' 1 2 3
20' 30' 24' 20' 30' 24'
Solution:
The analysis should be conducted in conjunction with the free body diagram shown below.
Vertical Loads
3.75
3.75
3.0
3.0
2.5
2.5
4.5
6.5
10.1
10.1
4.5
6.5
4.5 5.6 3.6 6.5
0.64 0.8 0.51 0.93
0.64 0.8 0.93
0.51
4.5 5.6 3.6 3.6
7.5
6.0
6.0
5.0
5.0
7.5
9.0
9.0
20.2
13.0
20.2
13.0
4.5 5.6 3.6 6.5
0.56 0.7 0.45 0.81
0.56 0.7 0.45 0.81
4.5 5.6 3.6 6.5
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Gravity Loads
kip kip kip
w12 := 0.25 w13 := w12 = 0.25 w14 := w12 = 0.25
ft ft ft
kip kip kip
w9 := 0.5 w10 := w9 = 0.5 w11 := w9 = 0.5
ft ft ft
Approximate Equations
2
M left( w , L) := -0.045 w L Maximum negative moment at girder end
2
M cent( w , L) := 0.08 w L Maximum positive moment
1. Top Girder Moments
( )
M 12lft := M left w12 , 20ft = -4.5 kip ft
( )
M 12cnt := M cent w12 , 20ft = 8 kip ft
M 12rgt := M 12lft = -4.5 kip ft
( )
M 13lft := M left w13 , 30ft = -10.125 kip ft
( )
M 13cnt := M cent w13 , 30ft = 18 kip ft
M 13rgt := M 13lft = -10.125 kip ft
( )
M 14lft := M left w14 , 24ft = -6.48 kip ft
( )
M 14cnt := M cent w14 , 24ft = 11.52 kip ft
M 14rgt := M 14lft = -6.48 kip ft
2. Bottom Girder Moments
( )
M 9lft := M left w9 , 20ft = -9 kip ft
( )
M 9cnt := M cent w9 , 20ft = 16 kip ft
M 9rgt := M 9lft = -9 kip ft
( )
M 10lft := M left w10 , 30ft = -20.25 kip ft
( )
M 10cnt := M cent w10 , 30ft = 36 kip ft
M 10rgt := M 10lft = -20.25 kip ft
( )
M 11lft := M left w11 , 24ft = -12.96 kip ft
( )
M 11cnt := M cent w11 , 24ft = 23.04 kip ft
M 11rgt := M 11lft = -12.96 kip ft
3. Top Column Moments
M 5top := M 12lft = -4.5 kip ft
M 5bot := -M 5top = 4.5 kip ft
M 6top := -M 12rgt + M 13lft = -5.625 kip ft
M 6bot := -M 6top = 5.625 kip ft
M 7top := -M 13rgt + M 14lft = 3.645 kip ft
M 7bot := -M 7top = -3.645 kip ft
M 8top := -M 14rgt = 6.48 kip ft
M 8bot := -M 8top = -6.48 kip ft
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4. Bottom Column Moments
M 1top := M 5bot + M 9lft = -4.5 kip ft
M 1bot := -M 1top = 4.5 kip ft
M 2top := M 6bot - M 9rgt + M 10lft = -5.625 kip ft
M 2bot := -M 2top = 5.625 kip ft
M 3top := M 7bot - M 10rgt + M 11lft = 3.645 kip ft
M 3bot := -M 3top = -3.645 kip ft
M 4top := M 8bot - M 11rgt = 6.48 kip ft
M 4bot := -M 4top = -6.48 kip ft
0.50 k/
ft +18.0
+8.0 +11.5
-4.5 -5.6 +3.6 +6.5
-4.5 -4.5 -10.1 -10.1 -6.5 -6.5
12 6 0.25 k/ 13 7 14
8 14' +32.0
ft +16.0 +23.0
+4.5 +5.6 -3.6
9 10 11 -9.0 -6.5
4 16 -9.0 -20.2 -20.2 -13.0 -4.5 -5.6 +3.6 +6.5
1 2 3
'
+4.5 +5.6 -3.6 -6.5
20' 30' 24'
5. Top Girder Shear
w12 20ft
V12lft := = 2.5 kip
2
V12rgt := -V12lft = -2.5 kip
w13 30ft
V13lft := = 3.75 kip
2
V13rgt := -V13lft = -3.75 kip
w14 24ft
V14lft := = 3 kip
2
V14rgt := -V14lft = -3 kip
6. Bottom Girder Shear
w9 20ft
V9lft := = 5 kip
2
V9rgt := -V9lft = -5 kip
w10 30ft
V10lft := = 7.5 kip
2
V10rgt := -V10lft = -7.5 kip
w11 24ft
V11lft := = 6 kip
2
V11rgt := -V11lft = -6 kip
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7. Column Shears
h 5 := 14ft h 6 := h 5 = 14 ft h 7 := h 5 = 14 ft h 8 := h 5 = 14 ft
h 1 := 16ft h 2 := h 1 = 16 ft h 3 := h 1 = 16 ft h 4 := h 1 = 16 ft
M 5top
V5 := = -0.643 kip
h5
2
M 6top
V6 := = -0.804 kip
h6
2
M 7top
V7 := = 0.521 kip
h7
2
M 8top
V8 := = 0.926 kip
h8
2
M 1top
V1 := = -0.563 kip
h1
2
M 2top
V2 := = -0.703 kip
h2
2
M 3top
V3 := = 0.456 kip
h3
2
M 4top
V4 := = 0.81 kip
h4
+3.75
+2.5 +3.0
-2.5 -3.0
+0.93
-3.75
+0.51
+7.5
-0.80
-0.64
+5.0 +6.0
-5.0 -6.0
-7.5
-0.70
+0.45
+0.81
-0.56
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8. Top Column Axial Forces
P5 := V12lft = 2.5 kip
P6 := -V12rgt + V13lft = 6.25 kip
P7 := -V13rgt + V14lft = 6.75 kip
P8 := -V14rgt = 3 kip
9. Bottom Chord Axial Forces
P1 := P5 + V9lft = 7.5 kip
P2 := P6 - V10rgt + V9lft = 18.75 kip
P3 := P7 - V11rgt + V10lft = 20.25 kip
P4 := P8 - V11rgt = 9 kip
Horizontal Loads
17.5
17.5
17.5
17.5
17.5
17.5
15
17.5 35 35 17.5
2.5 5 5 2.5
2.5 17.5 5 35 5 35 2.5 17.5
77.5
77.5
77.5
77.5
77.5
77.5
30
60 120 120 60
7.5 15 15 7.5
60 120 60
7.5 15 15 120 7.5
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Horizontal Loads
F5 := 15kip
F1 := 30kip
Approximate Equations
F
Vext( F , n ) :=
2 n
n bays := 3
1. Column Shears
( )
V5 := Vext F5 , n bays = 2.5 kip
V6 := 2 V5 = 5 kip
V7 := 2 V5 = 5 kip
V8 := V5 = 2.5 kip
( )
V1 := Vext F1 + F5 , n bays = 7.5 kip
V2 := 2 V1 = 15 kip
V3 := 2 V1 = 15 kip
V4 := V1 = 7.5 kip
2. Top Column Moments
V5 h 5
M 5top := = 17.5 kip ft
2
M 5bot := -M 5top = -17.5 kip ft
V6 h 6
M 6top := = 35 kip ft
2
M 6bot := -M 6top = -35 kip ft
V7 h 7
M 7top := = 35 kip ft
2
M 7bot := -M 7top = -35 kip ft
V8 h 8
M 8top := = 17.5 kip ft
2
M 8bot := -M 8top = -17.5 kip ft
3. Bottom Column Moments
V1 h 1
M 1top := = 60 kip ft
2
M 1bot := -M 1top = -60 kip ft
V2 h 2
M 2top := = 120 kip ft
2
M 2bot := -M 2top = -120 kip ft
V3 h 3
M 3top := = 120 kip ft
2
M 3bot := -M 3top = -120 kip ft
V4 h 4
M 4top := = 60 kip ft
2
M := -M = -60 kip ft
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M 4bot := -M 4top = -60 kip ft
4. Top Girder Moments
M 12lft := M 5top = 17.5 kip ft
M 12rgt := -M 12lft = -17.5 kip ft
M 13lft := M 12rgt + M 6top = 17.5 kip ft
M 13rgt := -M 13lft = -17.5 kip ft
M 14lft := M 13rgt + M 7top = 17.5 kip ft
M 14rgt := -M 14lft = -17.5 kip ft
5. Bottom Girder Moments
M 9lft := M 1top - M 5bot = 77.5 kip ft
M 9rgt := -M 9lft = -77.5 kip ft
M 10lft := M 9rgt + M 2top - M 6bot = 77.5 kip ft
M 10rgt := -M 10lft = -77.5 kip ft
M 11lft := M 10rgt + M 3top - M 7bot = 77.5 kip ft
M 11rgt := -M 11lft = -77.5 kip ft
14' 14'
16'
16'
20' 30' 24' 20' 30' 24'
17.5 ‘k 17.5 ‘k 17.5 ‘k
17.5 ‘k
‘k
-17.5 ‘k -17.5 ‘k -17.5
35 ‘k 35 ‘k 17.5 ‘k
77.5 ‘k 77.5 ‘k 77.5 ‘k
60 ‘k -17.5 ‘k -35 ‘k -35 ‘k -17.5 ‘k
120 ‘k 120 ‘k 60 ‘k
-77.5 ‘k -77.5 ‘k -77.5 ‘k
-60 ‘k -120 ‘k -120 ‘k -60 ‘k
6. Top Girder Shear
-2 M 12lft
V12lft := = -1.75 kip
20ft
V12rgt := V12lft = -1.75 kip
-2 M 13lft
V13lft := = -1.167 kip
30ft
V13rgt := V13lft = -1.167 kip
-2 M 14lft
V14lft := = -1.458 kip
24ft
V14rgt := V14lft = -1.458 kip
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7. Bottom Girder Shear
-2 M 9lft
V9lft := = -7.75 kip
20ft
V9rgt := V9lft = -7.75 kip
-2 M 10lft
V10lft := = -5.167 kip
30ft
V10rgt := V10lft = -5.167 kip
-2 M 11lft
V11lft := = -6.458 kip
24ft
V11rgt := V11lft = -6.458 kip
8. Top Column Axial Forces (+ve tension, -ve compression)
P5 := -V12lft = 1.75 kip
P6 := V12rgt - V13lft = -0.583 kip
P7 := V13rgt - V14lft = 0.292 kip
P8 := V14rgt = -1.458 kip
8. Bottom Column Axial Forces (+ve tension, -ve compression)
P1 := P5 - V9lft = 9.5 kip
P2 := P6 - V10rgt + V9lft = -3.167 kip
P3 := P7 - V11rgt + V10lft = 1.583 kip
P4 := P8 + V11rgt = -7.917 kip
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16.1 Propped Cantilever Beam
Find the end moments for the beam
20 kN
2 3
1
10 m 5m
Solution:
1. The beam is kinematicall indeterminate to the third degreee (θ2, Δ3, θ3), however by replacing the overhang by a fixed end moment equal
to 100 kN-m at support 2, we reduce the degreee of kinematic indeterminancy to one (θ2).
2. The equilibrium relation is
M 21 - 100kN m = 0
3. The member end moments in terms of the rotations are
2
(
M 12 = 2 E K12 2 θ1 + θ2 =
10
)
E I θ2
4
(
M 21 = 2 E K12 θ1 + 2 θ2 = )
10
E I θ2
4. Substituting into the equilibrium equations
10 250
θ2 = M 21 =
4E I EI
or
2 2 EI 250
M 12 = EI θ2 = = 50kN m
10 10 EI
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16.2 Two-Span Beam, Slope Deflection
Draw the moment diagram for the two span beam
2 kip/ft 5 kips
?2
?3
3
1 2
20' 15' 15'
0
40.97
79.52
Solution:
1. The unknows are θ2 and θ3
2. The equilibrium relations are
M 21 + M 23 = 0
M 32 = 0
3. The fixed end moments are given by
- -2
kip 2
2 ( 20ft)
M 12F = -M 21F =
-w L
M 12F :=
ft
= 66.667 kip ft
12 12
-PL -( -5 kip ) ( 30ft)
M 23F = -M 32F = M 23F := = 18.75 kip ft
8 8
4. The member end moments in terms of the rotations are
2EI EI
M 12 = 2 E K12 θ2 + M 12F = θ2 + M 12F = θ + M 12F
L1 10 2
4EI EI
( )
M 21 = 2 E K12 2θ2 + M 21F =
L1
θ2 + M 21F = θ2 + M 21F
5
2EI EI EI
( )
M 23 = 2 E K23 2θ2 + θ3 + M 23F =
L2
(
2θ2 + θ3 + M 23F = )
7.5 2
θ + θ + M 23F
15 3
2EI EI EI
( )
M 32 = 2 E K23 θ2 + 2θ3 + M 32F =
L2
(
θ2 + 2θ3 + M 32F = )
15 2
θ + θ + M 32F
7.5 3
5. Substituting in the equilibrium equations
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5. Substituting in the equilibrium equations
EI EI EI EI EI EI
θ2 + M 21F + θ2 + θ + M 23F = 0 => θ2 + θ2 + θ = 47.917 kip ft
5 7.5 15 3 5 7.5 15 3
EI EI EI EI
θ + θ + M 32F = 0 => θ + θ = 18.75 kip ft
15 2 7.5 3 15 2 7.5 3
or
5 1 2 718.755
θ
EI =
1 2 θ3 281.25
EIθ2 5 1 - 1 718.755 128.473
:= =
EIθ3 1 2 281.25 76.388
6. Substituting for the moments
EIθ2 kip ft
M 12 := + M 12F = 79.514 kip ft
10
EIθ2 kip ft
M 21 := - M 12F = -40.972 kip ft
5
EIθ2 kip ft EIθ3 kip ft
M 23 := + + M 23F = 40.972 kip ft
7.5 15
EIθ2 kip ft EIθ3 kip ft
M 32 := + - M 23F = 0 kip ft
15 7.5
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16.3 Two-Span Beam, Slope Deflection, Initial Deflection
Determine the end moments for the previous problem if the middle support settles by 6 in.
Solution:
1. Since we are performing a linear elastic analysis, we can separately analyze the beam for support settlement, and then add the moments
to those due to the applied loads.
2. The unknowns are θ2 and θ3
3. The equilibrium relations are
M 21 + M 23 = 0
M 32 = 0
4. The member end moments in terms of the rotations are
M 12 = 2 E K12 θ2 - 3
Δ 2EI 0.5 EI 3EI
= θ + 3 = θ +
L12 20 2 20 10 2 400
M 21 = 2 E K12 2θ2 - 3
Δ 2EI
2θ2 + 3
0.5 EI 3EI
= = θ2 +
L12
20 20 5 400
M 23 = 2 E K23 2θ2 + θ3 - 3
Δ 2EI 0.5 EI EI EI
= 2θ2 + θ3 + 3 = θ2 + θ3 +
L23
30 30 7.5 15 300
M 32 = 2 E K23 θ2 + 2θ3 - 3
Δ 2EI 0.5 EI EI EI
= θ2 + 2θ3 + 3 = θ + θ3 +
30 15 2
L23
30 7.5 300
5. Substituting into the equilibrium equations
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5. Substituting into the equilibrium equations
EI 3EI EI EI EI EI EI -13EI
θ2 + + θ2 + θ3 + =0 θ2 + θ3 =
5 400 7.5 15 300 3 15 1200
EI EI EI EI EI -EI
θ2 + θ3 + =0 θ2 + θ3 =
15 7.5 300 15 7.5 300
or
-13
100 20 2
θ
EI = EI 4
20 40
3
θ
-1
which will give
11
θ2 -1 -13 - 360
100 20
:= 4
θ3 20 40 7
-1 -
720
6. Thus the additional moments due to the settlement are
EI 3EI EI
M 12( EI) := θ2 + simplify
10 400 225
EI 3EI EI
M 21( EI) := θ2 + simplify
5 400 720
EI EI EI
M 23( EI) := θ2 + θ3 +
simplify -0.0013888888888888888889 EI
7.5 15 300
EI EI EI
M 32( EI) := θ + θ + simplify -3.7037037037037037037e-24 EI
15 2 7.5 3 300
16.4 Frame, Slope Deflection
Determine the end moments for the frame
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Solution:
1. The effect of the 35 cantilever can be included by replacing it with its end moment.
L kip ft
M 3 = -w L M 3 := 3 6 ft 6 = 54 kip ft
2 ft 2
2. The unknown displacements and rotations are
Δ2 and θ2 at joint 2
θ3 and θ4 at joints 3 and 4
We observe that due to the lack of symmetry, there will be a lateral displacement in the frame, and neglecting axial
deformations, Δ2 = Δ3
3. The equilibrium relations are
M 23 + M 32 = 0
M 32 + M 34 = -54kip ft
M 43 = 0
V12 + V43 - 10kip = 0
Thus we have four unknown displacements and four equations. However, the last two equations are in terms of of the
shear forces, and we need to have them in terms of the end moments. This can be achieved through the following
equilibrium relations
M 12 + M 21 + 50
V12 =
L12
M 34 + M 43
V43 =
L34
Hence, all four equations are now in terms of the moments.
4. The fixed end moments for member 23 are
4. The fixed end moments for member 23 are
-PL -10kip 10ft
M 21F = M 21F := = -12.5 kip ft
8 8
kip 2
2 -3 ( 20ft)
-w L ft
M 23F = M 21F := = -100 kip ft
12 12
5. The member end moments in terms of the rotations are
Δ2
M 12 = 2 E K12 θ2 - 3
L12
(
+ M12F = 0.2EI θ2 - 0.3Δ2 + 12.5 )
Δ2
M 21 = 2 E K12 2θ2 - 3
L12
(
+ M21F = 0.2EI 2θ2 - 0.3Δ2 - 12.5 )
( ) (
M 23 = 2 E K23 2θ2 + θ3 + M 23F = 0.1EI 2θ2 + θ3 + 100 )
M 32 = 2 E K32 ( θ2 + 2θ3 ) + M 32F = 0.1EI ( θ2 + 2θ3 ) - 100
Δ2
M 34 = 2 E K34 2θ3 + θ4 - 3
L34
(
= 0.2EI 2θ3 + θ4 - 0.3Δ2 + 100 )
Δ 2
M 43 = 2 E K34 θ3 + 2θ4 - 3
L34
(
= 0.2EI θ3 + 2θ4 - 0.3Δ2 + 100 )
6. Substituting into the equilibrium equations and dividing by EI
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6. Substituting into the equilibrium equations and dividing by EI
-875
6θ2 + θ3 - 0.6Δ2 =
EI
460
θ2 + 6θ3 + 2θ4 - 0.6Δ2 =
EI
θ3 + 2θ4 - 0.3Δ2 = 0
and the last equilibrium equation is obtained by substituting V12 and V43 and multiplying by 10/EI
-83.3
θ2 + θ3 + θ4 - 0.4Δ2 =
EI
or
1 0 -0.6
θ2
6 -875
θ
1 6 2 -0.6 3 1 460
EI =
0 1 2 -0.3 θ4 0
EI
1
1 1 -0.4 Δ
-83.3
2
which will give
θ2
-294.8
3 1
θ 68.4
=
θ4 EI -240.6
-1375.7
Δ
2
7. Substituting into the slope deflection equations gives the end moments
M12
36
M21 -47.88
M
23 = 47.88
M32 -115.8
61.78
M34
M 0
43
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16.5 Box Culvert Slope Deflection
Draw the shear and moment diagram for the following box girder
0.2 k/ft 0.2 k/ft
B C B C
18'
18'
A D A D
0.8 k/ft 0.8 k/ft 0.8 k/ft 0.8 k/ft
16' 16'
Solution:
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1. From symmetry θB = -θC, and at the base θA = θD = 0.
2. The fixed end moments are given by
kip 2
2 0.2 ( 16ft)
w L ft
FEMBC = FEMBC := = 4.267 kip ft
12 12
kip 2
2 0.2 ( 16ft)
-w L ft
FEMCB = FEMCB := - = -4.267 kip ft
12 12
kip 2
2 0.8 ( 18ft)
w L ft
FEMAB = FEMAB := = 12.96 kip ft
20 20
kip 2
FEMBA =
(
- w L
2
) FEMBA := -
0.8
ft
( 18ft)
= -8.64 kip ft
30 30
3. The moments are given by
2EI EI
M BC =
16
( 8
)
2θB + θC + FEMBC = θB + 4.267
2EI 2EI
M BA =
18
(
2θB + 0 + FEMBA =
9 B
)
θ - 8.64
2EI EI
M AB =
18
(
2θB + FEM AB =)9 B
θ + 12.96
4. Equilibrium at joint B
M BA + M BC = 0
2EI EI
θB - 8.64 + θ + 4.267 = 0
9 8 B
8.64 - 4.267 12.594
θB ( EI) :=
2EI EI EI
+
9 8
5. Substitute θB to get the moments
M BC :=
EI 12.594
+ 4.267 kip ft 5.841 ft kip
8 EI
M AB := + 12.96 kip ft 14.359 ft kip
EI 12.594
9 EI
M BA := - 8.64 kip ft -5.841 ft kip
2EI 12.594
9 EI
6. Member forces are determined from statics. Careful, the moment diagram is now based on the so-called "design" sign convention.
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0.2 k/ft 1.6
0.2 k/ft
1.93 1.93 VBA=1.93
1.93 5.84
B C 5.84
5.84 1.6 1.6 5.84
9.3'
18'
1.6
V
1.6 6.06
A D
0.8 k/ft 0.56 0.8 14.36
0.8 k/ft
M VAB=5.27 5.27 14.36
16' 5.84 5.84 V M
1.6
16.6 Continuous Beam
Solve for the moments at A and B by moment distribution, using (a) the ordinary method, and (b) the simplified method.
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10
K=5 K=3 k
A B C
12' 10' 10'
10
1 6 From previous step 10 From previous step
k Fixed end
moments due
7.0
25 25 to the load 11.7 23.4 14.1 1.1 2.2 1.3 0.6
2 7 11
Release C,
Release C, Release C,
balance
25 7.0 balance Moment 0.6 balance Moment
Moment
3 8 12
Carry over
Carry over to B Carry over at B
12.5 25 to B 3.5 7.0 0.3 0.6
4
Sum the
moments at B
37.5
0
5 Release B, 9 Release B, 13 Release B,
Distribute Distribute Distribute
Moments Moments Moments
23.4 14.1 0 2.2 1.3 0 0.2 0.1 0
6 10 Carry over from 14 Carry over
Carry over from from B to A
B to A and from
7.0 B to A and and from B to
11.7 23.4 14.1 from B to C 1.1 2.2 1.3 0.6 B to C 0.1 0.2 0.1 0.6
C
MAB=- 11.7 - 1.1 - 0.1= - 12.9 MBC = 25 + 12.5 - 14.1 + 3.5 - 1.3 + 0.3 - 0.1 = 25.8
MBA= - 23.4 - 2.2 - 0.2= - 25.8 MCB = - 25.0 + 25.0 - 7.0 + 7.0 - 0.6 + 0.6 = 0.0
25.8 25.8
12.9
3.22 6.29 37.1 3.71 Free Body Diagram
3.22
12.9 Moment Diagram
25.8
Solution:
For this example the fixed-end moments are computed as follows:
PL 10kip 20ft
M BCF = M BCF := = 25 kip ft
8 8
M CBF := -M BCF = -25 kip ft
2. Since the relative stiffness is given in each span, the distribution factors are
KAB 5
DFAB = = =0
ΣK ∞+ 5
KBA 5
DFBA = = = 0.625
ΣK 5+3
KBC 3
DFBC = = = 0.375
ΣK 5+3
KCB 3
DFCB = = =1
ΣK 3
3. The balancing computations are shown below.
Joint A B C
Member AB BA BC CB 138 of 159
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Joint A B C
Member AB BA BC CB
K 5 5 3 3
DF 0 0.625 0.375 1
25 -25
12.5 25
-11.7 -23.4 -14.1 -7
FEM 3.5 7
-1.1 -2.2 -1.3 -0.6
0.3 0.6
-0.1 -0.2 -0.1
Total -12.9 -25.8 25.8 0
4. The above solution is that referred to as the ordinary method, so named to designate the manner of handling the balancing at
the simple support at C. It is known, of course, that the final moment must be zero at this support because it is simple
5. Consequently, the first step is to balance the fixed-end moment at C to zero. The carry-over is then amde immediately to B.
When B is balanced, however, a carry-over must be made back to C simply because the relative stiffness of BC is based on end C
of this span being fixed. It is apparent, however, that the moment carried back to C (in this case -7.0) cannot exist at this joint.
Accordingly, it is immediately balanced out, and a carry-over is again made to B, this carry-over being considerably smaller than
the first. Now B is again balanced, and the process continues until the numbers involved become too small to have any practical
value.
6. Alternatively, we can use the simplified method. It was previously shown that if the support at C is simple and a moment is
applied at B, then the resistance of the span BC to this moment is reduced to three-fourths the value it would have had wit C fixed.
Consequently, the relative stiffness of span BC is reduced to three-fourths of the value given, it will not be necessary to carry over
to C.
Joint A B C
Member AB BA BC CB
K 5 5 2.25 3
DF 0 0.69 0.31 1
FEM 25 -25
12.5 25
-12.9 -25.8 -11.7
Total -12.9 -25.8 25.8 0
7. From the standpoint of work involved, the advantage of the simplified method is obvious. It should always be used when the external
(terminal) end of a member rests on a simple support, but it does not apply when a structure is continuous at a simple support. Attention is
called to the fact that when the opposite end of the member is simply supported, the reduction factor for the stiffness is always 3/4 for a
prismatic member but a variable quantity for a nonprismatic member.
8. One valuable feature of the tabular arrangement is that of dropping down one line for each balancing operation and making the carry-over
on the same line. The practice clearly indicates the order of balancing the joints, which in turn makes it possible to check back in the event of
an error. Moreover, the placing of the carry-over on the same line with the balancing moments definitely decreases the chances of omitting a
carry-over.
9. The correctness of the answers may ina sense be checked by verifying ΣM = 0 at each joint. However, even though the final answers
satisfy this equation at every join, this is in no way a check on the inital fixed end moments. These fixed end moments, therefore, should be
checked with great care before beginning the balancing operation. Moreover, it occasionally happens that conpensating errors are made in
the balancing, and these errors will not be apparent when cheking ΣM = 0 at each joint.
10. To draw the final shear and moment diagram, we start by drawing the free body diagram of each beam segment with the computed
moments, and then solve from statics for the reactions
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moments, and then solve from statics for the reactions
12.9kip ft + 25.8kip ft
12.9kip ft + 25.8kip ft - 12ft VA = 0 VA := = 3.225 kip
12ft
VA + VBL = 0 VBL := -VA = -3.225 kip
25.8kip ft + 10kip 10ft
25.8kip ft + 10kip 10ft - 20ft VBR = 0 VBR := = 6.29 kip
20ft
6.29kip + VC - 10kip = 0 VC := 10kip - 6.29kip = 3.71 kip
-VBL - VBR + RB = 0 RB := VBL + VBR = 3.065 kip
Check: RA + RB + RC - 10kip = 0 VA + RB + VC - 10kip = 0 kip
M BC := VC 10ft = 37.1 kip ft
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16.7 Continuous Beam
Using the simplified method of moment distribution, find the moments in the following continuous beam. The values of I as
indicated by the various values of K, are different for the various spans. Determine the values of the reactions, draw the shear and
bending moment diagrams, and sketch the deflected structure.
0.5 k 1k 1k 0.2 kips/ft
K = 20 K = 60 K = 30
5' 5' 10'
10' 20' 20' 15'
Solution:
1. Fixed-end moments
M AOF := -0.5kip 10ft = -5 kip ft
For the 1 kip load:
2 2
P a b 1kip 5 ft ( 15ft)
M ABF = M ABF1 := = 2.812 kip ft
2 2
L ( 20ft)
2 2
P a b 1kip ( 5ft) 15ft
M BAF = M BAF1 := = 0.938 kip ft
2 2
L ( 20ft)
For the 4 kip load:
P L 4kip 20ft
M ABF = M ABF4 := = 10 kip ft
8 8
M BAF4 := -M ABF4 = -10 kip ft
For the uniform load:
kip 2
2 0.2 ( 15ft)
w L ft
M CDF = M CDF := = 3.75 kip ft
12 12
M DCF := -M CDF = -3.75 kip ft
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2. The balancing operation is shown below
Joint A B C D
Member AO AB BA BC CB CD DC
K 0 20 15 60 60 40 40
DF 0 1 0.2 0.8 0.6 0.4 0
-5 12.8 -10.9 3.8 -3.8
-7.8 -3.9
2.9 11.9 5.9
FEM -2.9 -5.8 -3.9 -1.9
0.6 2.3 1.1
-0.3 -0.7 -0.4 -0.2
0.1 0.2
Total -5 5 -11.2 11.2 0.5 -0.5 -5.9
3. The only new point in this example is the method of handling the overhanging end. It is obvious that the final internal moment at
A must be 5 kip-ft and, accordingly, the first step is to balance out 7.8 kip-ft of the fixed end moment at AB, leaving the required 5
kip-ft for the internal moment at AB. Since the relative stiffness of BA has been reduced to three-fourths of its original value, to
permit considering the support at A as simple in balancing, no carry-over from B to A is required.
4. The easiest way to determine the reactions is to consider each span as a free body. End shears are first determined as caused
by the loads alone on each span and, following this, the end shears caused by the end moments are computed. These two
shears are added algebraically to obtain the net end shear for each span. An algebraic summation of the end shears at any
support will give the total reaction.
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16.8 Continuous Beam, Initial Settlement
For the following beam find the moments at A, B, and C by moment distribution. The support at C settles by 0.1 in. Use E := 30000ksi
5 kips/ft 10 k
I = 2400 in4 I = 7200 in4
K = 10 K = 20
I = 1200 in4
K = 10
20' 10' 30'
Solution:
1. Fixed-end moments
Uniform load:
kip 2
2 5 ( 20ft)
w L ft
M ABF = M ABF := = 166.667 kip ft
12 12
M BAF := -M ABF = -166.667 kip ft
Concentrated load:
P L 10kip 30ft
M CDF = M CDF := = 37.5 kip ft
8 8
M DCF := -M CDF = -37.5 kip ft
Moments caused by deflection:
4
6EI Δ 6 E 1200in 0.1in
M BCF = M BCF := = 125 kip ft
2 2
L ( 10ft)
M CBF := M BCF = 125 kip ft
4
6EI Δ 6 E 7200in 0.1in
M CDF = - M CDF := - = -83.333 kip ft
2 2
L ( 30ft)
M DCF := M CDF = -83.333 kip ft
2. Moment distribution
Joint A B C D
Member AB BA BC CB CD DC
K 10 10 10 10 15 20
DF 0 0.5 0.5 0.4 0.6 1
167 -167 38 -38
125 125 -83 -83
60 121
FEM -28 -56 -84
17 35 35 17
-3 -7 -10
1 2 1
Total 185 -130 130 79 -79 0
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The fixed-end moments caused by a settlement of supports have the same sign at both ends of each span adjacent to the settling
support. The above computations have been carried to the nearest kip-ft, which for the moments of the magnitudes involved, would
be significantly close for purposes of design.
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16.9 Frame
Find all moments by moment distribution for the following frame. Draw the bending moment diagram and the deflected structure.
18 k
12'
I = 360 in4
26' I = 260 in4 I = 390 in4
K = 10 K = 15
KM = 10/26 KM = 15/26
18'
Solution:
1. The first step is to perform the usual moment distribution. The reader should fully understand that this balancing operation adjusts the
internal moments at the ends of the members by a series of corrections as the joints are considered to rotate, until ΣM = 0 at each joint. The
reader should also realize that during this balancing operation no translation of any joint is permitted.
2. The fixed-end moments are
2
18kip 12ft ( 6ft)
M BCF := = 24 kip ft
2
( 18ft)
2
18kip 6 ft ( 12ft)
M CBF := - = -48 kip ft
2
( 18ft)
3. Moment distribution
Joint A B C D
Member AB BA BC CB CD DC
K 10 10 20 20 15 15
DF 0 0.333 0.667 0.571 0.429 0
24 -48
13.7 27.4 20.6 10.3
-6.3 -12.6 -25.1 -12.5
FEM 3.6 7.1 5.4 2.7
-0.6 -1.2 -2.4 -1.2
0.3 0.7 0.5 0.2
-0.1 -0.2
Total -6.9 -13.9 13.9 -26.5 26.5 13.2
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4. The final moments listed in the table are correct only if there is no translation of any joint. It is therefore necessary to determine whether or
not, with the above moments existing, there is any tendency for side lurch of the top of the frame.
5. If the frame is divided into three free bodies, the result will be as shown below.
Inspection of this sketch indicates that if the moments of the first balance exist in the frame, there is a net force of 1.53kip - 0.8kip = 0.73 kip
tending to sway the frame to the left. In order to prevent side-sway, and thus allow these moments to exist (temporarily for the purpose of the
analysis), it is necessary that an imaginary horizontal force ce considered to act to the right at B or C. This force is designated as the artificial
joint restraint (abbreviated as AJR) and is shown below.
6. This illustration shows the complete load system which would have to act on the structure if the final moments of the first balance are to be
correct. The AJR, however, cannot be permitted to remain, and thus its effect must be cancelled. This may be accomplished by finding the
moments in the frame resulting from a force equal but opposite to the AJR and applied at the top.
7. Although it is not possible to make a direct solution for the moments resulting from this force, they may be determined inddirectly. Assume
some unknown force P acts on the frame, as shown below, and causes it to deflect laterally to the left, without joint rotation, through some
distance Δ. Now, regardless of the value of P and the value of the resulting Δ, the fixed-end moments induced in the ends of the columns
must be proprtional to the respective values of KM.
Δ 1 K
Recalling that the fixed end moment is M F = 6EI = 6EKM Δ, where KM = = we can write
2 2 L
L L
M ABF M DCF
Δ= =
6E KM 6E KM
M ABF KMAB 10
= =
M DCF KMDC 15
These fixed-end moments could, for example, have values of -10 and -15 kip-ft, or -20 and -30, or -30 and -45, or any other combination so
long as the above ratio is maintained. The proper procedure is to choose values for the fixed-end moments of approximately the same order
of magnitude as the original fixed-end moments due to the real loads. This will result in the same accuracy for the results of the balance for
the side-sway correction that was realized in the first balance for the real loads. Accordingly, it will be assumed that P, and the resulting Δ, are
of such magnitude as to result in fixed-end moments shown below.
8. Obviously ΣM = 0 is not satisfied for joints B and C in this deflected frame. Therefore, these joints must rotate until equilibrium is reached.
The effect of this rotation is determined in the distribution below.
Joint A B C D
Member AB BA BC CB CD DC
K 10 10 20 20 15 15
DF 0 0.333 0.667 0.571 0.429 0
-30 -30 -45 -45
12.9 25.8 19.2 9.6
2.8 5.7 11.4 5.7
FEM
-1.6 -3.3 -2.4 -1.2
0.2 0.5 1.1 0.5
-0.3 -0.2 -0.1
Total -27 -23.8 23.8 28.4 -28.4 -36.7
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9. During the rotation of joints B and C, as represented by the above distribution, the value of Δ has remained constant, with P varying in
magnitude as required to maintain Δ.
10. It is now possible to determine the final value of P simply by adding the shears in the columns. The shear in any member, without external
loads applied along its length, is obtained by adding the end moments algebraically and dividing by the length of the member. The final value
of P is the force necessary to maintain the deflection of the frame after the joints have rotated. In other words, it is the force which will be
consistent with the displacement and internal moments of the structure as determined by the second balancing operation. Hence this final
value of P will be called the consistent joint force (abbreviated as CJF).
11. The consistent joint force is given by
27kip ft + 23.8kip ft 28.4kip ft + 36.7kip ft
CJF := + = 4.458 kip
26ft 26ft
and inspection clearly indicates that the CJF must act to the left.
12. Obviously, then, the results of the last balance above are moments which will exist in the frame when a force of 4.45 kip acts to the left at
the top level. It is necessary, however, to determine the moments resulting from a force of 0.73 kip acting to the left at the top level, and some
as yet unknown factor "z" times 4.45 kip will be used to represent this force acting to the left.
13. The free body diagram for the member BC is shown above. ΣH = 0 must be satisfied for this figure, and if forces to the left are
considered as positive, the result is 4.45z - 0.73 = 0, and
0.73
z := = 0.164
4.45
If this factor is applied to the moments obtained from the second balance, the result will be the moments caused by a force of 0.73 kip acting
to the left at the top level. If these moments are then added to the moments obtained from the first balance, the result will be the final moments
for the frame, the effect of the AJR having been cancelled. The combintaion of moments is shown below.
Joint A B C D
Member AB BA BC CB CD DC
M from 1st balance -6.9 -13.9 13.9 -26.5 26.5 13.2
z x M from 2nd balance -4.4 -3.9 3.9 4.7 -4.7 -6
Final Moments -11.3 -17.8 17.8 -21.8 21.8 7.2
14. If the moments are correct, the shears in the two columns of the frame shoul dbe equal and opposite to satisfy ΣH = 0 for the entire
frame. This check is expressed
11.3kip ft + 17.8kip ft -21.8kip ft + -7.2kip ft
+ = 0 kip
26ft 26ft
and
1.12kip - 1.11kip = 0.01 kip (nearly)
The signs of all moments taken from the previous table have been reversed to give the correct signs for the end moments external to the
columns. It will be remembered that the moments considered in moment distribution are always internal for each member. However, the
above check actually considers each column as a free body and so external moments must be used.
15. The moment under the 18 kip load is obtained by treating BC as a free body:
M 18 := 5.77kip 12ft - 17.8kip ft = 51.44 kip ft
16. The direction of side lurch may be determined from the obvious fact that the frame will always lurch in a direction opposite to the AJR. If
required, the magnitude of this side lurch may be found. The procedure which follows will apply.
A force P of sufficient magnitude to result in the indicated column moments and the lurch Δ was applied to the frame. During the second
balance this value of Δ was held constant as the joints B and C rotated, and the value of P was considered to vary as necessary. The final
Δ
value of P was found to be 4.45 kip. Since Δ was held constant, however, its magnitude may be determined from the equation M = 6EI ,
2
L
where M is the fixed-end moment for either column, I is the moment of inertia for that column, and L is the length. This Δ will be the lurch for
4.45 kip acting at the top level. For any other force acting horizontally, Δ would vary proportially and thus the final lurch of the frame would be
the factor z multiplied by Δ determined above.
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16.10 Frame with Side Load
Find by moment distribution the moments in the following frame
18 k
12'
5k
I = 360 in4
26' I = 260 in4 I = 390 in4
K = 10 K = 15
KM = 10/26 KM = 15/26
18'
Solution:
The first balance will give the results shown
AB BA BC CB CD DC
-7.2 -14.6 14.6 -22.5 22.5 0
A check of the member BC as a free body for ΣH = 0 will indicate that an AJR is necessary as follows:
AJR + 0.84kip - 0.87kip - 5kip = 0
from which
AJR := 5kip + 0.87kip - 0.84kip = 5.03 kip in the direction assumed
The values of KM for the two columns are shown, with KM for column CD being K/2L because of the pin at the bottom. The horizontal
displacement Δ of the top of the frame is assumed to cause the fixed-end moments shown there. These moments are proportional to the
values of KM and of approximately the same order of magnitude as the original fixed-end moments due to the real loads. The results of
balancing out these moments are
AB BA BC CB CD DC
-34.4 -28.4 28.4 23.6 -23.6 0
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34.4kip ft + 28.4kip ft + 23.6kip ft
CJF := = 3.323 kip
26ft
and
5.03 - 3.32z = 0
from which
5.03kip
z := = 1.514
CJF
The final results are
AB BA BC CB CD DC
M from 1st balance -7.2 -14.6 14.6 -22.5 22.5 0
z x M 2nd balance -52.1 -43 43 35.8 -35.8 0
Final moments -59.3 -57.6 57.6 13.3 -13.3 0
If these final moments are correct, the sum of the column shears will be 5 kip
59.3 kip ft + 57.6kip ft + 13.3kip ft
ΣV := = 5.01 kip
26ft
The 5 kip horizontal load acting at C enters into the problem only in connection withthe determination of the AJR. If this load had been applied
to the column CD between the ends, it would have resulted in intial fixed-end moments in CD and these would be computed in the usual way.
In addition, such a load would have enetered into the determination of the AJR, since the horizontal reaction of CD against the right end of BC
would have been computed by treating CD as a free body.
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17.1 Beam
Considering the figure, let P1 = 2P, M = PL, P2 = P, and P2 = P. Solve for the displacements.
P1 ?1 P2
?2
B C
A
L/2 L/2 L/2 L/2
FLEXIBILITY STIFFNESS
Statically Determinate Kinematically Determinate
P1 -P2L/8 P2 P2L/8
-P1L/8 P1L/8
A B B C
P1 P2 -P2/2 -P2/2
-P1/2 -P1/2
?1?2 SNEF1 SNEF2
A C
Displacements due to B
Actual Load Nodal Equivalent Forces
K12 K22
1 1 K11 1
f12 f22 K21 1
f11 f21
Applied
Applied Forces
Displacments
[K]{? }={M} +{NEF}
[f]{R}+{? }={0}
Solution:
1. Using the previously defined sign convention (counterclockwise positive)
P1 L P2 L 2P L P L P L
ΣNEF1 = - = - =
8 8 8 8 8
P L
ΣNEF2 =
8
2. If it takes 4EI/L ( k44AB) to rotate AB and 4EI/L ( k22BC) to rotate BC, it will take a total force of 8EI/L to simultaneously rotate AB and BC
(Note that a rigid joint is assumed)
8EI 2EI
3. Hence, K11 which is the sum of the rotational stiffnesses at global d.o.f. 1, will be equal to K11 = ; similarly, K21 = ( k42BC)
L L
4EI 2EI
4. If we rotate dof 2 by a unit angle, then we will have K22 = and K12 =
L L
5. The equilibrium relation can be written as
8EI 2EI PL
L L θ1 PL 8
= +
2EI 4EI θ2 0
PL
L L 8
or
PL + PL 8EI 2EI
8 L L θ1
=
PL 2EI 4EI θ2
8 L L
We note that the matrix corresponds to the structure's stiffness matrix, and not the augmented one.
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6. The two by two matrix is next inverted
-1 17 P L2
8EI 2EI PL + PL
θ1 L L 8 112 EI
=
=
θ2 2
2EI 4EI PL
L L 8 -5 P L
112 EI
7. Next we need to determine both the reactions and the internal forces.
8. Recall that for each element p = k δ, and in this case p = P and δ = Δ for element AB. The element stiffness matrix has been previously
derived, and in the case of the global and local d.o.f. are the same.
9. hence, the equilibrium equation for element AB, at the element level, can be written as
12EI 6EI - 12EI 6EI
2P
L3 L
2
L
3 2
L
2
1
p 6EI 4EI 6EI 2EI
0
- 0 2PL
p 2 L2 L 2 L 8
L
= 0 +
p3 12EI 6EI 12EI 6EI 2P
- - - 2
p L3 2 3 2 17 P L 2
4 L L L
112 EI -2 PL
6EI 2EI 6EI 4EI 8
-
L2 L 2 L
L
Solving
p 1 p 2 p 3 p 4 = PL
107 31 5 5
( ) 56
P
56
PL
56
P
14
10. Similarly, for element BC:
12EI 6EI 12EI 6EI
-
P
L3 L
2
L
3
L
2
0 2
p1 6EI 4EI 6EI 2EI
2
- 17 P L PL
L L
p2 L2 L
2
112 EI 8
= +
p3 - 12EI -
6EI 12EI
-
6EI
0
P
p L3 L
2
L
3
L -5 P L2
2 2
4 -PL
6EI 2EI 6EI 4EI 112 EI
- 8
L2 L 2 L
L
or
-1
p 2 p 3 p 4 = P P 0
7 9
( p1 )
8 14
PL
7
11. This simple example calls for the following observations:
a) Node A has contributions from element AB only, while node B has contributions from both AB and BC
b) We observe that p 3AB p 1BC even though they both correspond to a shear force at node B, the difference between
them is equal to the reaction at B. Similarly, p 4AB p 2BC due to the externally applied moment at B
c) From this analysis, we can draw the complete free body diagram and then the shear and moment diagrams
which is what the engineer is most interested in for design purposes.
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Structural Analysis
Prof. Victor Saouma
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17.2 Frame
Whereas in the first example all local coordinate systems were identical to the global one, in this example we consider the
orthogonal frame shown below.
P
y ?2 ?3 P
x
M
A P/2 A A
B B x B
EI ?1
w H
w Y
C C
X C
L/2 L/2 y
K21 K31
K22 K32
1 K11 A 1 rad.
A B 1 K12 A
B K23 K33
B K13
C
C C
Solution:
1. Assuming axial deformations, we do have three global degress of freedom Δ1, Δ2, and θ3.
2. Constrain all degrees of freedom, and thus make the structure kineamatically determinate.
3. Determine the nodal equivalent forces for each element in its own local coordinate system (the first three values are associated with the first
node, and the last three with the second node):
-P -PL -P PL
p 1 v1 m1 p 2 v2 m2 = 0
( ) 0 Member AB
2 8 2 8
-wH -w H2
-wH w H
2
( 1 1 1 2 2 2) 2
p v m p v m = 0
12
0
2 12
Member BC
4. Summing the nodal equivalent forces at node B in global coordinates we have
-wH -P PL w H 2
( P1 P2 P3 =
2
) 2 8
-
12
5. Next, we apply a unit displacement in each of the three global degrees of freedom, and we seek to determine the structure global stiffness
matrix. Each entry Kij of the global stiffness matrix will correspond to the internal force in degreee of freedome i, due to a unit displacement in
degree of freedom j.
6. Recalling the force displacement relations derived earlier, we can assemble the global stiffness matrix in terms of contributions from both AB
and BC.
152 of 159
Structural Analysis
Prof. Victor Saouma
Prepared by Christopher Segura
Ki1 Ki2 Ki3
Δ1 Δ2 Δ3
AB EA/L 0 0
K1j 3 2
BC 12EI/H 0 6EI/H
3 2
AB 0 12EI/L -6EI/L
K2j
BC 0 EA/H 0
2
AB 0 -6EI/L 4EI/L
K3j 2
BC 6EI/H 0 4EI/H
7. Summing up, the structure global stiffness matrix [K] is
EA + 12EI 0
6EI
L H
3
H
2
12EI EA -6 EI
K= 0 +
L
3 H
L
2
6EI -6 EI 4EI 4EI
+
2 2 L H
H L
8. The global equation of equilibrium can now be written
EA + 12EI 6EI
0 wH
L H
3
H
2
2
-P
Δ1
12EI EA -6 EI 2 P
0 + Δ =
2 -
3 H 2 0 2
L L
θ M
6EI -6 EI 4EI 4EI 3 -PL w H2
+ 8 + 12
2 2 L H
H L
9. Solve for the displacements
-1
EA + 12EI 6EI
0 -P - wH
L 3 2
H H 2 2
Δ1
12EI EA -6 EI -P
0 +
Δ2 =
3 H 2 2
θ L L
3 2
6EI -6 EI 4EI 4EI PL w H
+ M + 8 - 12
2 2 L H
H L
10. To obtain the element internal forces, we will multiply each element stiffness matrix by the local displacements. For element AB, the local
and global coordinates match, thus
153 of 159
Structural Analysis
Prof. Victor Saouma
Prepared by Christopher Segura
EA -EA
L 0 0
L
0 0
0 12EI 6EI 12EI 6EI 0
0 - P
p1 3 2 3 2
L L L L 0
2
p2 6EI 4EI 6EI 2EI 0
0 0 - PL
p L L 0
8
2 2
3 =
L L
Δ1 +
p4 -EA EA 0
0 0 0 0
L L Δ2 P
p5
p 0 - 12EI - 6EI 0 12EI 6EI θ 2
- 3
6 2
-PL
3 2 3
L L L L
8
0 6EI 2EI
0 -
6EI 4EI
2 L 2 L
L L
11. For element BC, the local and global coordinates do not match, hence we will need to transform the dispalcements from their global to
their local components. But since vector (displacement and load) and matrix transformation have not yet been covered, we not by inspection
that the relationship between global and local coordinates for element BC is
Local δ1 δ2 θ3 δ4 δ5 θ6
Global 0 0 0 Δ2 -Δ1 θ3
and we observe that there are no local or global displacements associated with dof 1-3; Hence, the internal forces for element BC
are given by:
EA -EA
L 0 0
L
0 0
0
0 12EI 6EI
0 -
12EI 6EI wH
p1 3 2 3 2 -
L L L L 0
2
0
p2 6EI 4EI 6EI 2EI w H2
0 0 -
p 2 L 2 L 0 -
3 =
L L 12
+
p4 -EA EA Δ2 0
0 0 0 0
L wH
L -Δ1
p5 -
p 0 - 12EI - 6EI 0 12EI 6EI θ 2
6
- 3
L
3
L
2
L
3
L
2 2
w H
0 6EI 2EI
0 -
6EI 4EI 12
2 L 2 L
L L
Note that the element is defined as going from C to B hence x,y,z correspond to Y, -X, Z.
154 of 159
Structural Analysis
Prof. Victor Saouma
Prepared by Christopher Segura
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Structural Analysis
Prof. Victor Saouma
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Structural Analysis
Prof. Victor Saouma
Prepared by Christopher Segura
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Structural Analysis
Prof. Victor Saouma
Prepared by Christopher Segura
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Structural Analysis
Prof. Victor Saouma
Prepared by Christopher Segura
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