SAINT LOUIS COLLEGE

San Fernando City, La Union

Review
Material
in

MECHANICS OF
DEFORMABLE
BODIES

MARIA VICTORIA B. MUNAR

Facilitator
2nd Semester, SY 2021-2022
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SIMPLE STRESS P
 max 
Normal Stress A
P P
Stress is defined as the strength of a material per unit Sheering Stress:   sin cos  sin 2
area or unit strength. It is the force on a member divided by A A
the area which carries the force, expressed as psi or in N/m2 *the sheering stress is maximum when θ = 45° or
or MPa. P
 max 
P 2A

A SIMPLE STRAIN
where P is the applied normal load
and A is the area. The maximum
tension or compression occurs over
a section normal to the load.
Also known as a unit deformation, strain is the ratio of
the change in length caused by an applied force, to the
original length.
Shearing Stress

Forces parallel to the area resisting the force cause 
L
shearing stress. It differs from tensile and compressive where δ is the elongation and L is the original length, thus
stresses which are caused by forces perpendicular to the area
 is dimensionless.
on which they act. Shearing stress is also known as
tangential stress.
V Shear Strain – describes distortion (changes in angles)
 s
A Ps
s
where V is the resultant shearing


force which passes through the L
L
centroid of the area A being sheared. where: γ = shear strain
δs = shearing deformation
Ps
L = length
Stress-Strain Diagram

Bearing Stress
Bearing stress is the
contact pressure between
separate bodies. It differs from
compressive stress, as it is an
internal stress caused by
compressive forces. Modulus of Elasticity, E or Young’s modulus
Pb – the slope of the stress-strain diagram in the linearly
b  elastic region, and its value depends upon the particular
Ab
material being used. It is equal to the ratio of stress to
strain.
Proportional Limit
– the greatest stress a material is capable of developing
Stress on an Oblique Plane Under Axial Loading without deviation from the straight-line proportionality
between stress and strain.

Modulus of Resilience
– the work done on a unit volume of material as the force
P is gradually increased from O to A, in N-m/m3. This may
Normal Stress:  cos2 
A be calculated as the area under stress-strain curve from
*the normal stress is maximum when θ = 0 or the origin O up to the elastic limit (right after point A).

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The resilience of a material is its ability to absorb energy Factor of Safety – the ratio of the ultimate or yield strength to
without creating a permanent distortion. the allowable strength
Elastic Limit ultimate strength
Factor of safety 
– the greatest stress a material is capable of developing allowable strength
without permanent or residual deformation when the load
is entirely removed. Hooke’s Law
Yield Stress  = E
– the stress at which considerable elongation occurs, with
no noticeable increase in the stress. Axial Deformation
PL L
Strain Hardening  
AE E
– the material undergoes changes in its atomic and
crystalline structure, resulting in increased resistance of This equation is subject to the following restrictions:
the material to further deformation. 1. The load must be axial.
2. The bar must have a constant cross section and be
Ultimate Stress homogeneous.
– the maximum stress a material is capable of developing.
3. The stress must not exceed the proportional limit.
Rupture Strength
– the stress at which fracture occurs. If, however, the cross-sectional area is not uniform, the
axial deformation can be determined by considering a
Elasticity differential length and applying integration.
– the property of a material which makes it return to its
P L dx

original dimension when the load is removed.
 
Ductility E 0 A
– the ability of a material to deform in the plastic range where A = ty and y and t, if variable, must be expressed in
without breaking. terms of x.
Plasticity For a rod of unit mass ρ suspended vertically from one
– a property of a material where, if the specimen be end, the total elongation of due to its own weight is
unloaded, it will not return to its original length; rather gL2 MgL
it will retain a permanent elongation called a permanent  
2E 2 AE
set.
where ρ is in kg/m3, L is the length of the rod in m, M is the
Stiffness total mass of the rod in kg, A is the cross-sectional area of
– the property of a material to withstand high stress the rod in m2, and g = 9.81 m/s2.
without great strain.
Brittleness Stiffness, k
– implies the absence of any plastic deformation prior to – the ratio of the steady force acting on an elastic body to
failure. the resulting displacement. It has a unit of N/mm.
Malleability P AE
k 
– the property of a material enabling it to undergo  L
considerable plastic deformation under compressive Flexibility
 L
load before actual rupture. 
P AE
Toughness
– the property of a material enabling it to endure high Shear Modulus or modulus of rigidity
impact loads or shock loads. – the ratio of the shear stress τ and the shear strain γ.
Resilience 
G
– the property of a material enabling high impact loads 
without inducing a stress in excess of the elastic limit. The relationship between the shearing deformation and the
Modulus of Toughness applied shearing force is
VL L
– the work done on a unit volume of material as the force s  
is gradually increased from O to E, in N-m/m3. This may As G G
be calculated as the area under entire stress-strain curve where: V is the shearing force acting over an area As.
(from O to E). The toughness of a material is its ability Strain Energy, u
to absorb energy without causing it to break.
P2L
Working Stress u
– the actual stress of a material under a given loading. 2 AE
Allowable Stress
– the maximum safe stress that a material can carry.

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Statically Indeterminate Members Solution:

Indeterminate members are structures where the reactive PS PB
forces or the internal resisting forces over a cross section exceed
the number of independent equations for equilibrium. Such
structures are called statically indeterminate and their analysis
require the use of additional relations that depend on the elastic δS
δB
deformations of the members; i.e., additional geometric
relations between the elastic deformations produced by the


loads, called equations of conditions or equations of
compatibility, and force-displacement relations that would
relate the forces with the displacements.
M A 0 0.6 PS  1.6 PB  50(2.4)  0
0.6 PS  1.6 PB  120 103
PROBLEMS 0.6 S (600 10 6 )  1.6 B (300 10 6 )  120 103
360 S  480 B  120 10 9
1. An 80-m-long wire of 5-mm diameter is made of a steel with
E = 200 GPa and an ultimate tensile strength of 400 MPa. If 30 S  4 B  1109 Eq. 1
a factor of safety of 3.2 is desired, determine
From the similarity of triangles:
a. the largest allowable tension in the wire,
S B
b. the corresponding elongation of the wire.  ; 8 S  3 B
0.6 1.6
Solution:
 S (1)  B (2)
 ultimate 400 8 3 Eq. 2
FS = ; 3.2  ; σ = 125 MPa 200 83
 allowable 
Solve Eq. 1 and Eq. 2 simultaneously:
P P
a.   ; 125  10 6  ; P = 2454.37 N σs = 31.043 MPa ; σs = 17.177 MPa
A  (5) 2  10 6
4

PL L 125  10 6 (80) 4. A short post constructed from a hollow circular tube of

b.     ; δ = 50×10–3 m = 50 mm aluminum supports a compressive load of 26 kips. The inner
AE E 200  10 9 and outer diameters of the tube are di = 4.0 in. and do = 4.5
in., respectively, and its length is 16 in. The shortening of
2. An aluminum pipe must not stretch more than 0.05 in. when the post due to the load is measured as 0.012 in. Disregard
it is subjected to a tensile load. Knowing that E = 10.1 × 106 the weight of the post itself, and assume that the post does
psi and that the maximum allowable normal stress is 14 ksi, not buckle under the load. Use E = 10600 ksi.
determine a. Determine the compressive stress in the post.
a. the maximum allowable length of the pipe, b. Determine the strain in the post.
b. the required area of the pipe if the tensile load is 127.5
kips. Solution:
Solution: Given:
L 14L P = 26 kips L = 16 in
a.   ; 0.05  ; L = 36.071 in. = 3.006 ft di = 4.0 in δ = 0.012 in
E 10.1103
do = 4.5 in
P 127.5
b.   ; 14  ; A = 9.107 in2 P 26
a.   ;    7.789 ksi
A A
A 
4

4.5 2  4 2 
3. The figure shows a rigid bar that is supported by a pin at A b.   E  ; 7.789 10600; ϵ = 7.384
and two rods; one made of steel and the other of bronze.
Neglecting the weight of the bar, compute the stress in each 5. The lap joint shown is fastened by four ¾-in diameter rivets.
rod caused by the 50-kN load. For steel, A = 600 mm2 and Calculate the maximum safe load P that can be applied if the
E = 200 GPa. For bronze, A = 300 mm2 and E = 83 GPa. shearing stress in the rivets is limited to 14 ksi and the
bearing stress in the plates is limited to 18 ksi. Assume the
applied load is uniformly distributed among the four rivets.

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Solution: a. the largest load P that can be safely supported,

b. the corresponding shearing stress in the splice.
From sheering of a rivet:
P Solution:
P P
τ  V ; 14  4
2
; P = 24.74 kips a.   ; 75 
A  3 A 3(5)
 
4 4 P = 1125 lb
P sin 2 1125 sin 60
From bearing of rivet on the plate: θ b.   
2A 2(3)(5)
P
F τ = 32.476 psi
σ b  ; 18  4 ; P = 47.25 kips
A 37
 
48
Hence, the max. safe load P = 24.74 k
8. A steel bar 8.0 ft long has a circular cross section of diameter
6. The composite shaft, consisting of aluminum, copper, and d1 = 0.75 in. over one-half of its length and diameter
steel sections, is subjected to the loading shown. Neglect the d2 = 0.5 in. over the other half. The modulus of elasticity
size of the collars at B and C. E = 30 × 106 psi. How much will the bar elongate under a
a. Determine the normal stress in each section. tensile load P = 5000 lb?
b. Determine the displacement of end A with respect to end
D.
A B C

Solution:
PL 5000 (4)(12)
 ;  AB   0.018 in.
AE 
(0.75 )(30  10 )
2 6
4
5000 (4)(12)
 BC   0.041 in.

(0.5 )(30  10 )
2 6
Solution: 4
Axial forces: PAB = 2.00 kips (T)  total  0.018 0.041 0.059 in.
PBC = 5.00 kips (C)
PCD = 1.50 kips (C) 9. The rigid bar ABC shown is hinged at A and supported by
P 2 a steel rod at B. Determine the largest load P that can be
a.   ;  AB   22.222 ksi applied at C if the stress in the steel rod is limited to 30 ksi
A 0.09 and the vertical movement of end C must not exceed
5
 BC   41.667 ksi 0.10 in.
0.12 PS
1.5
 CD   25 ksi
0.06 Steel AH A C
L = 4 ft
PL 2(18) A=
b.  AB    0.04 in. (elongation) AV
AE 0.09(10 103 ) 
P
A B C
PL 5(12) 2 ft 3 ft
 BC    0.028in. (contraction) 
AE 0.12(18103 ) P 
Solution: s
C
PL 1.5(16) MA = 0] 5P – 2Ps = 0;
 AB    0.014 in. (contraction) Ps = 2.5P
AE 0.06(29 103 )
Δ C δs ;
 δs  2 (0.1) = 0.04 in.
 A / D   AB   BC   CD = 0.04 – 0.028 – 0.014 5 2 5

 A / D  0.002 in From tension of the rod: From elongation of the rod:

Therefore, A moves towards end D. σ

Ps δs  PL
A AE
30  2.5P 0.04 = 2.5P(4)(12)
7. Two wooden members of uniform cross section are joined 0.5
by the simple scarf splice shown. Knowing that the 0.5(29  106 )
maximum allowable tensile stress in the glued splice is 75 P = 6 kips P = 4833.33 lb
psi, determine Therefore, the largest load that can be applied is 4833 lb
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10. The 500-lb load is applied along the centroidal axis of the 480
member which has a square section of side 6 inches. FS   3.6
133.333
a. Determine the normal and shear stresses in the member
at section a-a. 13. A bar of metal 25 mm in diameter is tested on a length of
b. Determine the normal and shear stresses in the member 250 mm. In tension, the following results were recorded.
at section b-b. Load (kN) 10.4 31.2
Extension (mm) 0.039 0.089
a. What is the strain due to the 10.4-kN load?
b. What is the strain due to the 31.2-kN load?
c. What is the Young’s Modulus?
Solution: Solution:
 0.039
a. At section a-a a.    0.156103
P 500 L 250
   2  13.889 psi  0.089
A 6 b.    0.356103
τ=0 L 250
b. At section b-b 10.4  10 3
c. 1  0.156 10 3 1   21.187 MPa
P 500 
  cos2   2 cos2 30  10.417 psi
A
252 10 6
6 4
P 500 31.2  10 3
 sin 2  sin 60  6.014 psi 2  0.356 10 3  2   63.560 MPa
2A 2(6 2 ) 
252 10 6
11. The pin is made of a material having a failure shear stress 4
of 100 MPa. Determine the minimum required diameter of    1 (63.56  21.187) 106
E 2   211.865 GPa
the pin to the nearest mm using a factor of safety of 2.5. 2  1 (0.356  0.156) 103

14. Two identical wires, AB and BC, support a load P = 225

80 kN kN. The distance between supports A and C is b = 1.0 m,
and the wires are at an angle θ = 55 to the horizontal. The
wires are made of high-strength steel and have axial rigidity
Solution: EA = 165 kN. Calculate the downward displacement of
100 point B due to the load P.
FS  ; τa = 40 MPa
 allowable PAB PBC
35° 35°
80 103
V 2
 ; 40 106 
A 
d 2 106 L
4
d = 36 mm 225 N
δ 35°
ΔB
12. Link BC is 6 mm thick, has a width w = 25 mm, and is
made of a steel with a 480-MPa ultimate strength in Solution:
tension. What is the safety factor used if the structure
0.5
shown was designed to support a 16-kN load P? cos55  ; L = 0.8717 m
L
From the FBD:
PAB = PBC = PB
PAB cos35  PBC cos35  225; PAB = 137.337 N
PBC
137.337(0.8717)
  0.767103 m = 0.767 mm
156103
From the deformation:
= 16 kN
 0.767
Solution: cos35   ; ΔB = 0.936 mm
B B
MA = 0] 480PBC – 600(16) = 0; PBC = 20 kN
20 103
 = 133.333 MPa
6(25) 106

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15. The column is constructed from high-strength concrete Total reaction at C = 1.333 P
and four A-36 steel reinforcing rods. If it is subjected to an From shearing of pin C:
axial force of 800 kN, determine the required diameter of
each rod so that one-fourth of the load is carried by the 1.333 P
steel and three-fourths by the concrete. Est = 200 GPa and 50 10 6  2 ; P = 2121.105 N

Ec = 25 GPa. (6) 2  10 6
4
Solution: From shearing of pin B:
Axial loads: 2.333 P
50  10 6  ; P = 1683.237 N

PS = ¼(800) = 200 kN (10) 2  10 6
4
PC = ¾(800) = 600 kN
From tension of BD:
2.333P
133.333106  ; P = 2743.242 N
(18  10)(6) 106
Max. P that can be applied is 1683.237 N

17. A cylindrical assembly consisting of a brass core and an

aluminum collar is compressed by a load P. The length of
P S L PC L the aluminum collar and brass core is 350 mm, the
 S  C ; 
AE AE diameter of the core is 25 mm, and the outside diameter of
the collar is 40 mm. The moduli of elasticity of the
200 600 aluminum and brass are 72 GPa and 100 GPa, respectively.


AS (200) 3002  AS (25)
;
 AS = 3600 mm2 a. If the length of the assembly decreases by 0.1% when
the load P is applied, what is the magnitude of the
  load?
AS  4 d 2  ; 3600  d 2
4  b. What is the maximum permissible load P if the
d = 33.85 mm allowable stresses in the aluminum and brass are
80 MPa and 120 MPa, respectively?
16. In the steel structure shown, a 6-mm-diameter pin is used
at C and 10-mm-diameter pins are used at B and D. The
ultimate shearing stress is 150 MPa at all connections, and
the ultimate normal stress is 400 MPa in link BD. Knowing
that a factor of safety of 3.0 is desired, determine the
largest load P that may be applied at A. Note that link BD
is not reinforced around the pin holes.

Solution:
a. δ = 0.001(350) = 0.35 mm
Load carried by aluminum:
PL Pa (350)
 ; 0.35 
AE 
(40  252 ) 106 (72 109 )
2
4
Pa = 55134.95 N
Load carried by brass:
Pb (350)
Solution: 0.35 
PBD 
Allowable stresses: (252 ) 106 (100 109 )
150 4
C CH   50 MPa Pb = 49087.39 N
3
400 Total load P = 104222.34 N
  133.333 MPa
P CV 3 b. If the allowable stresses are σa = 80 MPa and
σb = 120 MPa,
MC = 0] 280P – 120PBD = 0; PBD = 2.333 P
aL bL Eb 100
FV = 0] PBD – P – CV = 0 ; CV = 1.333 P  a  b ;  ; b  a  a
Ea Eb Ea 72
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If σa = 80 MPa, σa = 111.111 MPa < 120 MPa Solution:



Pa  80 402  252   61261.06 N From the FBD:

4 M C  0 350 P  225 PBD  0 ; PBD = 1.555P
 
Pb  111.111 (25) 2   54541.48 N M B  0 125 P  225 PCE 0; PCE = 0.555P
 4 
From the similarity of triangles,
Total load P = 115802.54 N
 A   CE  BD   CE
 ; 350 BD  125 CE  225 A
18. A rigid bar of weight W = 800 N hangs from three equally 350 225
spaced vertical wires, two of steel and one of aluminum. 14 BD  5 CE  9(0.35)  3.15
The wires also support a load P acting at the midpoint of
the bar. The diameter of the steel wires is 2 mm, and the 1.555 P(225) 0.555 P(150)
14 5  3.15
diameter of the aluminum wire is 4 mm. What allowable 240(105) 300(72)
load P can be supported if the allowable stress in the steel P = 14.744 kN
wires is 220 MPa and in the aluminum wire is 80 MPa?
Assume Es = 210 GPa and Ea = 70 GPa. 20. The aluminum and steel pipes shown in the figure are
fastened to rigid supports at ends A and B and to a rigid
PS PA PS plate C at their junction. The aluminum pipe is twice as
long as the steel pipe. Two equal and symmetrically placed
loads P act on the plate at C. Calculate the stresses for the
W following data: P = 50 kN, cross-sectional area of
aluminum pipe Aa = 6000 mm2, cross-sectional area of steel
pipe As = 600 mm2, modulus of elasticity of aluminum
Ea = 70 GPa, and modulus of elasticity of steel Es = 200
GPa. RA RA
RA
Solution:
SL  AL
S   A;  ;
ES EA PS
EA 70 1
A  S  S  S
ES 210 3
If σS = 220 MPa, σA= 73.333 MPa < 80 MPa
PA
 
PS  220 (2) 2   691.15 N
4 
  
PA  73.333 (4) 2   921.534 N
4  RB
Allowable load P = 2PS + PA – W Solution:
P = 2(691.15) + 921.534 – 800 = 1503.835 N Axial loads: PS = RA (tension)
PA = 2P – RA ; PA = 100 – RA (compression)
19. Link BD is made of brass (E = 105 GPa) and has a cross-
sectional area of 240 mm2. Link CE is made of aluminum RA L  100  R A 2 L
S  A  0 ;  0
(E = 72 GPa) and has a cross-sectional area of 300 mm2. 600(200) 6000(70)
Knowing that they support rigid member ABC, determine
RA = 36.364 kN
the maximum force P that can be applied vertically at point
A if the deflection of A is not to exceed 0.35 mm. PS = 36.364 kN
PBD PA = 100 – 36.364 = 63.636 kN
36364
S   60.607 MPa
600106
63636
A   10.606 MPa
6000106
PCE
21. A homogeneous bar with a cross-sectional area of 500
δCE mm2 is attached to rigid supports. It carries the axial loads
P1 = 25 kN and P2 = 50 kN, applied as shown in the figure.
δBD
ΔA Determine the stress in segment BC.
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x 
1
E
 
 x   y   z 

 y   y    x   z 
1
E

1
 
z   z    x   y
E

Tensile stresses and elongation are taken as positive.
Compressive stresses and contractions are taken as negative.
Modulus of Rigidity, G
G  E
2 1  ν  where: G = modulus of rigidity
Axial loads: PAB = R1 E = modulus of elasticity
PBC = R1 – 25 v = Poisson’s ratio
PCD = 75 – R1 Bulk Modulus of Elasticity or
From the compatibility equation, find R1: Modulus of Volume Expansion, K
AB + BC  CD = 0 – a measure of a resistance of a material to change in
volume without change in shape or form.
R1 (0.6) + ( R1  25) (1.2)  (75  R1 ) (0.9) = 0 E 
AE K 
31  2  V V
AE AE
0.6R1 + 1.2R1 – 25(1.2) – 75(0.9) + 0.9R1 = 0 where V is the volume and ∆V is change in volume. The
R1 = 36.111 kN ratio ∆V/V is called volumetric strain and can be expressed
as
V  31  2 
Stress in segment BC:
PBC = R1 – 25 = 11.111 kN  
V K E
BC = 11111 = 22.222 MPa Dilatation
500 10  6 – change of volume per unit volume
Lateral Strain Tangent modulus
– the rate of change of stress with respect to strain
Poisson’s Ratio Specific strength
– ratio of the lateral strain to the axial strain – the ratio of the ultimate or tensile strength to the specific
y weight
ν  
x Specific modulus
where: v = Poisson’s ratio – ratio of the Young’s modulus to the specific weight
x = the strain due only to stress in the x
direction Thermal Stress
y = the strain induced in the perpendicular Temperature changes cause the body to expand or to
direction contract. The amount of linear deformation, δT , is given by
Biaxial Deformation δT = α L(Tf - Ti) = α L ∆T
If an element is subjected simultaneously by tensile where: α = coefficient of thermal expansion
stresses σx and σy in the x and y directions, the strain in the x- L = length,
ΔT = change in temperature
direction is  x and the strain in the y-direction is  Y .
E E
Simultaneously, the stress in the y-direction will produce a Pressure Vessels
lateral contraction on the x-direction of the amount y or Thin-Walled Cylindrical Vessel
σ A tank or pipe carrying a fluid or
 v Y . The resulting strain in the x-direction will be
E gas under a pressure is subjected to
x y   y E tensile forces, which resist bursting,
x   or x 
x

E E 1  2 developed across longitudinal and

and transverse sections.
y x or   x E
y   y 
y

E E 1  2
Triaxial Deformation
If an element is subjected simultaneously by three
Tangential stress,  T 
pD (also known as hoop stress,
mutually perpendicular normal stresses σx, σy, and σz which 2t
are accompanied by strains x , y , and z , respectively, girth stress, and circumferential stress

Deformable Bodies mvbmunar 022022

 10

pD Solution:
Longitudinal stress,  L 
4t In the x-direction:
where: p = internal pressure – external pressure 1

x   x  y
E

Spherical Shell
If a spherical tank of diameter D and x 
1
125 0.33(0)  1.6892103
thickness t contains gas under a pressure 74 10 3

of p, the stress at the wall can be expressed In the y-direction:

as:
wall stress,   pD
1

y   y  x
E

4t y

1
0  0.33(125) ; Δy = –1.1149 × 10–3
Thick-Walled Cylinder 2 74 103
In thin-walled cylinders, the wall 2  0.0011149
thickness is very small compared to the Slope =  1.995514
tank diameter. If in the case of thick- 1  0.0016892
walls cylinders, the tangential stress ST
and radial stress SR at any distance r
from the center is given by the following 3. A high-strength steel rod with E = 200 GPa and  = 0.3 is
equations: compressed by an axial force P. When there is no axial load,
the diameter of the rod is exactly 50 mm. In order to
a 2 pi  b 2 p o a 2 b 2  pi  p o  maintain certain clearances, the diameter of the rod must not
R  
b2  a2 
b2  a2 r 2 exceed 50.025 mm under load. What is the largest
a 2 pi  b 2 p o a 2 b 2  pi  p o  permissible load P?
T  
b2  a2 
b2  a2 r 2  50 mm

P P
PROBLEMS
1. A 20-mm-diameter rod made of an experimental plastic is Solution:
subjected to a tensile force of magnitude P = 6 kN. Knowing In the direction of the diameter:
that an elongation of 14 mm and a decrease in diameter of y 
y   x
0.85 mm are observed in a 150-mm length, determine E E
a. the modulus of elasticity,
b. Poisson’s ratio for the material, 0.025 P
 0  0.3

c. and the modulus of rigidity. 50
502  200   103
4
Solution:
P = 654.5 kN
PL 6  10 3 (150)
a.   ; 14  ; E = 204.628 MPa
AE 
(20 2 )  10 6 E 4. For the axial loading shown, determine the change in height
4 and the change in diameter of the brass cylinder shown.
y
0.85
b.   20  0.455
14
150
E 204.628
c. G    70.319 MPa
2(1  v) 2(1  0.455)

2. An aluminum plate (E = 74 GPa, v = 0.33) is subjected to a

centric axial load that causes a normal stress σ. Knowing x
that, before loading, a line of slope 2:1 is scribed on the
plate, determine the slope of the line when σ = 125 MPa.
y Solution:
PL L  58106 (0.135)
x Change in height:   
AE E 105109
=  0.0746103 m
= 0.0746 mm (decrease)

Deformable Bodies mvbmunar 022022

 11

Change in diameter: in the x direction: 8. A steel rod of diameter 15 mm is held snugly (but without
x   x    y   z  ;
1
E
D 1
D

  x  v  y  z
E
  any initial stresses) between rigid walls by the arrangement
shown in the figure. Calculate the temperature drop ΔT, in

 
D 1 degrees Celsius, at which the average shear stress in the
 0  0.33(58106  0) 12-mm diameter bolt becomes 45 MPa. For the steel rod,
85 105109 use as = 12 × 10–6/°C and E = 200 GPa.
D  0.0155 mm (increase)

5. The rails of a railroad track are welded together at their ends

(to form continuous rails and thus eliminate the clacking
sound of the wheels) when the temperature is 60°F. What
compressive stress σ is produced in the rails when they are
heated by the sun to 120°F if the coefficient of thermal Solution:
expansion α = 6.5 × 10–6/°F and the modulus of elasticity Axial load in the bolt:
E = 30 × 106 psi? P
Solution: V 2
 ; 45  10 
6
; P = 10178.76 N
 P  T ;
L
 LT ;   ET
A 
4
12 10
2 6

E
L
  30 10 6 (6.5 10 6 )(120  60) = 11700 psi  P  T ;  LT ;   ET
E
6. An aluminum pipe has a length of 60 m at a temperature of P
 ET ;
10178 .76

 200  10 9 12  10 6 T 
10°C. An adjacent steel pipe at the same temperature is
5 mm longer than the aluminum pipe. At what temperature
A  2
4
 
15  10 6

(degrees Celsius) will the aluminum pipe be 15 mm longer ΔT = 24°C

than the steel pipe? Assume that the coefficients of thermal
expansion of aluminum and steel are αa = 23 × 10–6/°C and 9. A steel wire AB is stretched between rigid supports. The
as = 12 × 10–6/°C, respectively. initial prestress in the wire is 42 MPa when the temperature
Solution: is 20°C. What is the stress in the wire when the temperature
At 10°C: drops to 0°C? Assume as = 14 × 10–6/°C and E = 200 GPa.
60 m 5 mm
Solution:
L PL
aluminum
  T   P ;  LT 
E AE
 
steel

After the change in temperature:   ET   P  200 109 14 10 6 20  42 10 6
δa σ = 98 MPa
60 m 5 mm
10. A liquid storage tank consists of a vertical axis circular
aluminum
cylindrical shell closed at its lower end by a hemispherical
steel shell as shown in the figure. The weight of the system is
δS 15 mm
carried by a ring-like support at the top and the lower
 a  5   S  15 ;  a   S  20 extremity is unsupported. A liquid of specific weight
9 kN/m3 entirely fills the container. Consider H = 9 m,
LT a  LT S  20 10 3 R = 2.4 m, and plate thickness of 2 mm.
23 10 6 (60)T  12 10 6 (60.005)T  20 10 3 a. Find the peak stress in the hemispherical region.
b. Calculate the peak circumferential stress in the
ΔT = 30.31° cylindrical region.
Tf = 30.31 + 10 = 40.31°C c. Determine the peak meridional stress (longitudinal
stress) in the cylindrical region.
7. The wall thickness of a 4-ft diameter spherical tank is 5/16
in. Calculate the allowable internal pressure if the stress is
limited to 8000 psi.
Solution:
For a spherical tank,   pD
4t
p (4)(12)
8000  ; p = 208.333 psi
 5
4 
 16 

Deformable Bodies mvbmunar 022022

 12

Solution: Solution:
a. Pressure at the bottom of the tank: 60
 allow   15 ksi
p = γh = 9(9) = 81 kPa 4
p = γh
pD 81(4.8)
Stress:     48600 kPa = 48.6 MPa = 62.4h
4t 4(0.002)
b. Circumferential stress in the cylindrical region
p = γh = 9(9 – 2.4) = 59.4 kPa
pD 59.4(4.8) 62.4h(25)(12)
   71280 kPa = 71.28 MPa 
pD
; 15  10 3 (144) 
2t 2(0.002) 2t 3
2
c. Weight of liquid: 16
h = 43.27 ft
 
W  9 (2.4) 2 (9  2.4)   2.43   1335.453 kN
2
 3  13. A high-strength steel rod with E = 200 GPa and v = 0.3 is
A = 2πRt = 2π(2.4)(0.002) = 30.159×100 –3 m2 compressed by an axial force P. When there is no axial
load, the diameter of the rod is exactly 50 mm. In order to
W 1335.453103
  = 44.28 MPa maintain certain clearances, the diameter of the rod must
A 30.159103 not exceed 50.025 mm under load. What is the largest
permissible load P?
11. Calculate the increase in stress for each segment of the
compound bar shown if the temperature increases by 100°F. Given: Req’d:
Assume that the supports are unyielding and that the bar is E = 200 GPa max. P
suitably braced to prevent buckling. v = 0.3

Solution: y x
y  
E E
P
50.025  50 0  502 106
  0.3 4
50 E 200109
P = 654.498 kN
Solution:
TORSION

δP
 Tal   Tst   Pal   Pst
P(10) P(15)
12.8(10)(100)  6.5(15)(100)  
2(10) 1.5(29)
P = 26691.837 lb The general torsion formula is   T
J
26691.837
 al   13345.918 psi where:  = shearing stress
2 T = applied torque
 st 
26691.837
 17794.558 psi  = radial distance of any internal fiber from
1.5 the axis of the shaft
J = polar moment of inertia.
12. The unpressurized cylindrical storage tank shown has a
Tr
3/16-in. wall thickness and is made of steel having a 60- Maximum shearing stress is  max 
ksi ultimate strength in tension. Determine the maximum J
height h to which it can be filled with water if a factor of For a solid cylindrical shaft:
safety of 4.0 is desired. The specific weight of water is 62.4  4
J D
lb/ft3. 32

16T
 max 
D 3

Deformable Bodies mvbmunar 022022

 13

For hollow cylindrical shaft PROBLEMS


J (D  d )
4 4
1. A solid steel bar of circular cross section has diameter
32 d = 1.5 in., length L = 54 in., and shear modulus of elasticity
16TD G = 11.5 × 106 psi. The bar is subjected to torques T acting
 max 
 D 4  d 4 
at the ends. If the torques have magnitude T = 250 lb-ft,
a. what is the maximum shear stress in the bar?
b. What is the angle of twist between the ends?
Angle of Twist Given:
The angle θ, in radians, through which the bar of length d = 1.5 in G = 11.5 × 106 psi
L will twist is L = 54 in T = 250 lb-ft
TL where: T = torque Solution:

JG 16T 16(250)(12)
a.  max    4527.074 psi
L = length of shaft D 3  (1.5) 3
G = shear modulus
TL 250(12)(54)
J = polar moment of inertia b.    = 0.02834 rad =1.62°
JG 
Shear Strain (1.5) 4 (11.5  106 )
Due to pure torsion, the shear strain in the bar is 32

  where: γ = strain 2. A copper rod of length L = 18.0 in. is to be twisted by
L
ρ = radial distance of any internal torques T at its ends until the angle of rotation between the
fiber ends of the rod is 3.0°. If the allowable shear strain in the
θ = angle of twist copper is 0.0006 rad, what is the maximum permissible
L = length diameter of the rod?
r Solution:
 max 
L Given: Re’q:
L = 18.0 in. γ = 0.0006 rad max. d
Torsional Stiffness
θ = 3.0°
It is the torque required to produce a unit angle of
r
rotation.  max  ; where: r = d/2
JG (N-m) L
kT   
L  3 
180 60
Torsional Flexibility d 
It is the reciprocal of stiffness which is defined as the  
2  60 
angle of rotation produced by a unit torque. 0.0006  ; d = 0.413 in.
18
Power Transmitted by the Shaft
A shaft rotating with a constant angular velocity ω (in 3. A generator shaft in a small hydroelectric plant turns at
radians per second) is being acted on by a twisting moment 120 rpm and delivers 50 hp.
T. The power transmitted by the shaft is a. If the diameter of the shaft is d = 3.0 in., what is the
P = T ω = 2T f maximum shear stress τmax in the shaft?
b. If the shear stress is limited to 4000 psi, what is the
where: T = torque in N-m
minimum permissible diameter dmin of the shaft?
f = number of revolutions per second
P = the power in watts. Solution:
Given:
Riveted Connections with Eccentric Load f = 120 rpm = 2 rps
y
P = 50 hp = 50(550) = 27500 ft-lb/sec
Torsional force T:
My c a. P = 2πfT ; 27500 = 2π(2)T; T = 2188.38 ft-lb
fx 

x y 2 2
 
16T 16(2188.38)(12)
  4953.48 psi
x
Mx c d 3  (3) 3
fy 

 x2  y 2  b. 4000 
16(2188.38)(12)
; d = 3.22 in.
where: yc, xc = distance of outermost rivet d 3
from the center of the
rivet group
T  f x 2   f y 2
Deformable Bodies mvbmunar 022022
 14

4. Knowing that d = 1.2 in., determine the torque T that causes 7. An aluminum bar of solid circular cross section is twisted
a maximum shearing stress of 7.5 ksi in the hollow shaft by torques T acting at the ends. The dimensions and shear
shown. modulus of elasticity are as follows: L = 1.2 m, d = 30 mm,
Solution: and G = 28 GPa.
16TD a. Determine the torsional stiffness of the bar.


 D4  d 4  b. If the angle of twist of the bar is 4°, what is the maximum
shear stress?
16T (1.6) c. What is the maximum shear strain (in radians)?
7.5 

 1.6 4  1.2 4  Solution:
T = 4.12334 in-k Given:
T = 343.612 ft-lb L = 1.2 m G = 28 GPa
d = 30 mm

0.034 28109 
5. Knowing that each of the shafts AB, BC, and CD consists of
a solid circular rod, determine the magnitude of the JG 32
maximum shearing stress that occurs in the shaft. a. kT    1855.503 N-m
L 1.2
60 N-m
TL  T (1.2)
b.   ; 4 
144 N-m JG 180 
32

0.034 28 10 9 
48 N-m
T = 129.539 N-m
16T 16(129.539)
Applied torques:  3   24.435 MPa
TAB = 48 N-m d  0.033
TBC = 96 N-m   
0.034 
TCD = 156 N-m d  180  0.8727103 rad
Solution: c.   
2L 2(1.2)
Max. T = 156 N-m
16T 16(156) 8. The propeller shaft of a large ship has outside diameter 18
   85.79 MPa
d 3  0.0213 in. and inside diameter 12 in. The shaft is rated for a
maximum shear stress of 4500 psi. If the shaft is turning at
100 rpm, what is the maximum horsepower that can be
6. The solid rod AB has a diameter dAB = 60 mm and is made transmitted without exceeding the allowable stress?
of a steel for which the allowable shearing stress is 85 MPa.
The pipe CD, which has an outer diameter of 90 mm and a Given: Req’d:
wall thickness of 6 mm, is made of aluminum for which the do = 18 in. τ = 4500 psi P
allowable shearing stress is 54 MPa. Determine the largest di = 12 in. f = 100 rpm = 5/3 rps
torque T that can be applied at A.
Solution:
16Td o 16T (18)

Applied torques: 
 do  di4 4
 ; 4500 

 184  124 
TAB = T
T = 4135121.33 in-lb
TBC = T
= 344593.444 ft-lb
5
Solution: P  2fT  2  344593.444  3608574.109 ft-lb
 3
From shearing of rod AB: = 6561.04 hp
16T 16T
  3 ; 85106 
D  0.0603 9. Four gears are attached to a circular shaft and transmit the
torques shown in the figure. The allowable shear stress in
T = 3604.978 N-m the shaft is 10,000 psi.
From shearing of pipe CD: a. What is the required diameter d of the shaft if it has a
16TD 16T (0.09) solid cross section?
 54 106 

 D d
4 4
;
 
 0.094  0.0784  b. What is the required outside diameter d if the shaft is
hollow with an inside diameter of 1.0 in.?
T = 3368.762 N-m
Max. T = 3368.762 N-m

Deformable Bodies mvbmunar 022022

 15

11. Two solid steel shafts (G = 77.2 GPa) are connected to a

coupling disk B and to fixed supports at A and C. For the
loading shown, determine
a. the reaction at each support,
b. the maximum shearing stress in shaft AB,
c. the maximum shearing stress in shaft BC.

Applied torques:
TAB = 8000 lb-in
TBC = 11000 lb-in
TCD = 7000 lb-in
Solution:
16T 16(11000)
a.   ; 10000 ; d = 1.776 in
d 3
d 3 Solution:
16Td o 16(11000 )d o
b.  
 ; 10000 
  
a.
 do  di
4 4
 d o 4  14 TAB  TA
TBC  1.4  TA
d o 4  17.6d o    0
do = 1.832 in

10. A stepped shaft ACB having solid circular cross sections  AB   BC

with two different diameters is held against rotation at the
TA (200) (1.4  TA )(250)
ends as shown in the figure. If the allowable shear stress in 
 
the shaft is 6000 psi, what is the maximum torque that may (50) 4 G (38) 4 G
be applied at section C? 32 32
TA = 1.105 kN-m
TB  1.4  1.105 = 0.295 kN-m
16(1105)
b.  AB   45.022 MPa
 (0.05) 3
16(295)
Solution: c.  BC   27.38 MPa
 (0.038) 3

12. From the given bracket shown, which consists of two plates
riveted to the column by 4-22-mm-diameter rivets which is
subjected to an eccentric load of 150kN,
a. What is the shearing stress in the bolts due to axial
Assume the stress in AC = 6000 psi load only?
16T 16TA b. What is the shearing stress in the bolts due to moment
  3 ; 6000 
d  (0.75) 3 alone?
c. What is the maximum shearing stress in the bolts?
TA = 497.01 lb-in
The rotations of AC and CB are equal, hence,
 AC   CB
TA (6) TB (15)

 4 
(0.75) G (1.5) 4 G
32 32
TB = 3180.863 lb-in
Check the stress in CB:
16(3180.863)
 CB   4800 psi < 6000 psi
 (1.5) 3 a) Due to axial load,  P  37.5 
Therefore, 4 
TO  TA  TB  497.01 3180.863  y
37500
 49.325 MPa
= 3677.873 lb-in 
2 (22) 2
4
Deformable Bodies mvbmunar 022022
 16

b) due to moment only Helical Springs

T = 150(200) = 30000 N-m

 x 2
 y2    100 2

 100 2 4  80000 mm2
30000 (0.1)
x   49.325 MPa

2 (22) 2 (80000 )  10 12
4
30000 (0.1)
y   49.325 MPa

2 (22) 2 (80000 )  10 12
4
Maximum shearing stress in the spring:   16 PR 1  d 
  49.325 2  49.325 2  69.756 MPa d 3  4 R 
3
c) Maximum shearing stress Spring deflection:   64PR n
 x  49.325 MPa
4
Gd
where: τ = shearing stress
 y  49.325  49.325  98.65 MPa P = axial load
R = spring mean radius
max.    x 2   y 2  49.3252  98.652 d = diameter of spring wire
n = number of coils of spring
= 110.294 MPa
G = modulus of rigidity
13. A plate is connected to a column by four bolts and carries
PROBLEMS
an eccentric load as shown in figure. The load P = 27.5 kN
with an eccentricity e = 200 mm. The dimension
1. Determine the maximum shearing stress and elongation in a
a = b = 150 mm. The diameter of the bolts is 25 mm.
helical spring composed of 20 turns of 20-mm-diameter
a. For this problem, assume that the load P acts through
wire on a mean radius of 90 mm when the spring is
the centroid of the bolts. What is the stress on the bolts
supporting a load of 1.5 kN. Use G = 83 GPa.
in MPa?
b. For this problem, assume that the only load acting on Given: Req’d:
the connection is a moment equal to the load P n = 20 P = 1.5 kN τ and δ
multiplied by its eccentricity e. What is the distance d = 20 mm G = 83 GPa
from the centroid of the bolts in mm, where, if a bolt R = 90 mm
is placed, the stress will be equal and opposite to the
Solution:
stress obtained in the first problem above?
c. What is the eccentricity, in mm of the load P, measured Shearing stress:
from the instantaneous center of the bolts? 16PR  d  16(1500 )( 0.090)  20 
 1   1  
d 
3 4 R   (0.2 )
3
 4(90) 
= 90.718 MPa
Elongation:
64 PR 3 n 64(1500)(0.0903 )(20)
 
Gd 4 83109 (0.2 4 )
= 0.105 m
Solution:
2. A helical spring is fabricated by wrapping wire ¾ in. in
27500
a.    14.006 MPa diameter around a forming cylinder 8 in. in diameter.

4 (25) 2  10 6 Compute the number of turns required to permit an
4 elongation of 4 in. without exceeding a shearing stress of 18
b. M = 27500(0.2) = 5500 N-m ksi. Use G = 12  106 psi.
P Md Given: Req’d:

4 
(x2  y 2 ) d = ¾ in
R = 4 in
τ = 18 ksi
G = 12 × 106 psi
n
27500 5500d

 0.75  δ = 4 in
4 2
 4  0.752  4
Solution:
d = 56.25 mm Shearing stress:
e = 200 + d = 200 + 56. 25 = 256. 25 mm 16PR  d 
 1  
d  4 R 
3

Deformable Bodies mvbmunar 022022

 17

16P(4)  0.75  4. A rigid bar, pinned at O, is supported by two identical

18000  1  ; P = 356.067 lb
 (0.753 )  4(4) 
springs as shown in the figure. Each spring consists of 20
turns of ¾-in-diameter wire having a mean diameter of 6 in.
Elongation: Determine the maximum load W that may be supported if
64 PR 3 n the shearing stress in the springs is limited to 20 ksi.

Gd 4
64(356.067 )( 4 3 )n
4 ; n = 10.413 turns
12  10 6 (0.75 4 )

3. Two steel springs arranged in series as shown support a load

P. The upper spring has 12 turns of 25-mm-diameter wire
on a mean radius of 100 mm. The lower spring consists of
10 turns of 20-mm-diameter wire on a mean radius of 75
mm. If the maximum shearing stress in either spring must Solution:
not exceed 200 MPa, compute the maximum value of P and Draw the free-body-diagram and the deformation
the total elongation of the assembly. Use G = 83 GPa. diagram of the bar:
Solution:
Upper spring:
16PR  d 
 1  
d 
3 4 R
16P(0.1)  25 
200 106  3 
1  
 (0.025 )  4(100) 
P = 5775 N From the FBD of the bar, the critical spring is the one subjected
Lower spring: to tension, therefore, the working stress in it is equal to the
allowable:
16PR  d 
 1   16PR  d 
d  4 R 
3
 1  
16P(0.075)  20  d 
3 4 R
200 106  1    3 
 (0.02 )  75) 
16P2 (3)  
3 4(
20000  1 4  ; 𝑃2 = 519.75 lb
 3   4(3) 
P = 3927 N 3
    
Maximum P = 3927 N 4
Total elongation of the spring: From the deformation diagram:
64 PR n 3 1 2
  ;  2  21
4 2 4
Gd
64(3927)(0.13 )(12) 64(3927)(0.0753 )(10) 64P2 R 3 n 64P1 R 3 n
  2 ; 𝑃1 = 259.875 lb
83 109 (0.0254 ) 83 109 (0.024 ) Gd 4 Gd 4
= 0.173 m
M O 0  2P1  4P2  7W  0
2(259.875) + 4(519.75) – 7W = 0
W = 371.25 lb

Deformable Bodies mvbmunar 022022

 18

SHEAR AND MOMENT IN BEAMS Curvature (k) – is defined as the reciprocal of the radius of
A beam is a bar subject to forces or couples that lie in a curvature which is a measure of how sharply a beam is bent.
plane containing the longitudinal axis of the bar.
Shearing Stress Formula
Shear and Moment Diagrams
A
Relationship Between Load, Shear, and Moment
t
dV
w The rate of change of the shear with NA
dx
respect to x is the load, or the load is the
slope of the shear diagram. VQ
fv 
V = (Area)load The change in shearing force between It
sections two represents the area under the where: fv = shearing stress
load diagram between these positions. V = shear force
Q  Ay = static moment of area
dM t = width of fiber under consideration
V The rate of change of the moment with
dx Rectangular section
respect to x is the shearing force, or the 6M
shear is the slope of the moment diagram. f b max 
bd 2
M = (Area)shear The change in bending moment between NA d
any two sections is equal to the area of the 3V
f v max 
shear diagram for this interval. 2bd
b
STRESSES IN BEAMS Circular section
Pure bending – refers to flexure of a beam under a constant
32M
bending moment; hence occurs only in f b max 
regions of a beam where the shear force is NA d d 3
zero. 4V
f v max 
Nonuniform bending – refers to flexure in the presence of 3A
shear forces, meaning that the bending Triangular section
moment changes as we move along the axis
of the beam. Mc
f b max 
Flexural rigidity (EI) – is a measure of the resistance of a beam d I
to bending, NA 3
where: I  bd
Radius of Curvature 36
EI Ec b
 
M fb
where: ρ = radius of curvature d
E = modulus of elasticity 3V
f v max 
I = centroidal moment of inertia 2A
M = bending moment
b
c = distance of outermost fiber from the
neutral axis Built-up Beams – beams that are fabricated from two or more
fb = bending stress pieces of material joined together to form a
single beam.
Flexure Formula
Shear Flow (q) – represents the longitudinal force per unit
My
fb  length transmitted across a section at a
I level y1 from the neutral axis.
where: y = distance of fiber under consideration VQ
from the neutral axis q  fv t 
I
Mc M
f b max   Spacing of Rivets or Bolts
I S
RI
where: M = maximum bending moment occurring s
VQ
in the beam
where R is the total shearing force to be resisted by the
I = section modulus
S bolts and is equal to the allowable shearing stress  area
c  number of bolts in the group.

Deformable Bodies mvbmunar 022022

 19

Members Made of Different Materials PROBLEMS

When members are made of several materials, the neutral 1. Determine the shear force V and bending moment M at a
axis generally does not pass through the centroid of the cross section located 16 ft from the left-hand end A of the
composite cross section. Using the ratio of the moduli of beam with an overhang shown in the figure.
elasticity of the materials, we obtain a transformed section
corresponding to an equivalent member made entirely of one
material. To obtain the transformed section, the width of each
element of the lower portion will be multiplied by the factor n,
RA
where n = E2 .
E1
Reaction at A:
 
M B  0 400(10)(15)  200(6)(9)  20RA  0
RA = 2460 lb
Shear force at section 16 ft from A:
VA  RA  400(10)  2460 4000  1540 lb
Bending moment at section 16 ft from A:
Combined Axial and Flexure
A member subject to bending moment M and an axial load
M A  2460(16)  400(10)(11)  4640 lb-ft
P causes a flexural stress of fb = My/I at any point from its
neutral axis, and a normal stress of P/A which is uniformly 2. The simple beam ACB shown in the figure is subjected to a
distributed over its entire area, respectively. The combination triangular load of maximum intensity 180 lb/ft.
of these stresses results if the member is eccentrically loaded. a. Locate the point where maximum moment occurs.
The combined stress developed is b. Determine the magnitude of the maximum moment.
x
f  f a  fb
P My y
f   y 180

A I x 6
where M is = Pe, e is the eccentricity of P from the
RA y  30x
neutral axis.

Reaction at A:

M B 0
1
2
 1 2
(180)(6)(3)  (180)(1)( )  7 R A  0
2 3
RA = 240 lb
a. Maximum moment occurs at the point of zero shear
Shear equation at section:
UNSYMMETRIC BENDING 1 1
V  RA  xy  240  x(30x)  240  15x 2
When the bending couple acts in a different plane or when 2 2
the member does neutral not possess any plane of symmetry, 0  240 15x 2 ; x = 4 ft
the member will not bend in the plane of the couple. The couple
exerted on the section is assumed to act in a vertical plane. The b. Maximum moment:
axis of the cross section will coincide with the axis of the couple 1 4
M A  240(4)  (4)(30)(4)( )  640 lb-ft
M representing the forces acting on that section if, and only if, 2 3
the couple vector M is directed along one of the principal
centroidal axes of the cross section. The distribution of the 3. A laminated wood beam on simple supports is built up by
stresses caused by the original couple M is obtained by gluing together three 2 in. × 4 in. boards (actual dimensions)
superposing the stress distributions caused by the couple to form a solid beam 4 in. × 6 in. in cross section, as shown
components Mz and My and we have in the figure. The allowable shear stress in the glued joints
M y M yz is 65 psi and the allowable bending stress in the wood is
x   z  1800 psi. If the beam is 6 ft long, what is the allowable load
Iz Iy P acting at the midpoint of the beam? Disregard the weight
the angle ϕ that the neutral of the beam.
axis forms with the z axis is
defined by the relation
I
tan   z tan 
Iy
Deformable Bodies mvbmunar 022022
 20

Solution: Point of zero shear:

P PL P(6) 70
Max. V = ; Max. M = = = 1.5P x  0.875 m
2 4 4 80
From shearing:
1
P M max  1670(1.5)  70(1.5)  (0875)(70)
 (4)( 2)( 2) 2
; 50    3
VQ 2
 = 2640.625 lb-ft
Ib 4( 6)
( 4) 6M 6(2640.625)(12)
12 fb  2
  958.412 psi
P = 1800 lb bd 1.5(11.5) 2
From bending:
6M 6 (1.5P )(12) 5. A prefabricated wood I-beam serving as a floor joist has
  2 1500 
bd (4)(6) 2 the cross section shown in the figure. The allowable load
in shear for the glued joints between the web and the
P = 2000 lb
flanges is 65 lb/in. in the longitudinal direction. Determine
Maximum allowable P = 1800 lb. the maximum allowable shear force Vmax for the beam.

4. For the beam and loading shown, determine the maximum

normal stress due to bending.

Solution:
1 1
I  (5)(9.5) 3  (4.375)(8) 3  170.573 in4
12 12
At the glued joints:
Solution: Q  5(0.75)( 4.375)  16.406 in3
Reactions: VQ Vmax (16.406)
f allow  ; 65 
M B 0 1600 (10.5)  80(9)( 4.5)  12 R A  0 I 170.573
Vmax  675.8 lb
R A  1670 lb

M A 0 12 R A  1600 (1.5)  80(9)(7.5)  0 6. A simply supported wood beam of rectangular cross section
and span length 1.2 m carries a concentrated load P art
RB  650 lb midspan in addition to its own weight. The cross section has
width 140 mm and height 240 mm. The weight density of
the wood is 5.4 kN/m3. Calculate the maximum permissible
value of the load P if the allowable bending stress is 8.5 MPa
and the allowable shear stress is 0.8 MPa.

Solution:
Uniformly distributed weight of the beam:
w  5400(0.14)(0.24)  181.44 N/m
a. Max. P from bending moment:
PL wL2 P(1.2) 181.44(1.2) 2
M max    
4 8 4 8
M max  0.3P  32.659 N-m
6M 6(0.3P  32.659)
 2; 8.5 106 
bd (0.14)(0.24) 2
P = 37971.137 kN
Deformable Bodies mvbmunar 022022
 21

b. Max. P from shearing stress: 9. A wood box beam is constructed of two boards, each
P wL P 181.44(1.2) 40 × 180 mm in cross section, that serve as flanges and two
Vmax     plywood webs, each 15 mm thick. The total height of the
2 2 2 2
Vmax  0.5P  108.864 N beam is 280 mm. The plywood is fastened to the flanges by
wood screws having an allowable load in shear of F = 800
3V 3(0.5P  108.864)
 ; 0.8  10 6  N each. If the shear force V acting on the cross section is
2bd 2(0.14)( 0.24) 10.5 kN, determine the maximum permissible longitudinal
P = 35622.272 kN spacing s of the screws.

Max. allowable load P = 35.622 kN

7. A built-up wooden beam section is made of two 50 mm ×

100 mm sections glued on the lower sides by a 50 mm × 200
mm section as shown. Calculate the maximum bending
stress if the 5 m simple beam is loaded with 2.4 kN/m.
A1  150(100)
= 15000
A2  50(100) I NA 
1 1
(210)(280) 3  (180)(200) 3  264.16  106 mm
4
12 12
= 5000
At  20000 Q  40(180)(120)  864  10 3 mm3
FI 800(2)(264.16 106 )
Location of neutral axis: s   46.6 mm
15000(50)  5000(150)
VQ 10500(864106 )
y  75 mm
20000
1 1
10. A wood beam with cross-sectional dimensions
I NA  (150)(100) 3  15000(25) 2  (50)(100) 3  5000(75) 2 200 mm × 300 mm is reinforced on its sides by steel plates
12 12
12 mm thick. The moduli of elasticity for the steel and
 54.167106 mm4 wood are Es = 204 GPa and Ew = 8.5 GPa, respectively.
wL2 2400(5) 2 Also, the corresponding allowable stresses are
M max    7500 N-m σs = 130 MPa and σw = 8.0 MPa. Calculate the maximum
8 8
Mc 7500(0.125) permissible bending moment Mmax when the beam is bent
fb    17.308 MPa about the z axis.
I 54.167106

8. A beam of T cross section is formed by nailing together two

boards having the dimensions shown in the figure. If the
total shear force V acting on the cross section is 1600 N and 204
each nail may carry 750 N in shear, what is the maximum n  24
allowable nail spacing? 8.5

Transformed section:

A1  50(200)  10000 A2  200(50)  10000

At  20000
Location of neutral axis:
10000(100)  10000(225) 1
y  162.5 mm I NA  (776)(300) 3  1746106 mm4
20000 12
6M 6M
I NA 
1 1
(50)(200) 3  10000(62.5) 2  (200)(50) 3  10000(62.5) 2 For wood:   2 ; 8 106 
12 12 bd 0.776(0.3) 2
 113.5417106 mm4 M = 93120 N-m
F VQ FI 6M
 ; s ; Q  10000 (62.5)  625  10 3 mm3 For steel: 13010  6
(24)
s I VQ 0.776(0.3) 2
750(113.5417106 ) M = 63050 N-m
s  85.16 mm
1600(625106 ) Therefore, max. M = 63050 N-m
Deformable Bodies mvbmunar 022022
 22

11. A simple beam on a 10 ft span supports a uniform load of

intensity 800 lb/ft. The beam consists of a wood member (4
in. × 11.5 in. in cross section) that is reinforced by 0.25 in.
thick steel plates on top and bottom. The moduli of
elasticity for the steel and wood are Es = 30 × 106 psi and
Ew = 1.5 × 106 psi, respectively. Calculate the maximum
bending stresses in the steel plates and in the wood member
due to the uniform load.

800(10) 2
M max   10000lb-ft
8
Solution:
Transformed section: a. Maximum compressive stress due to vertical load

M = Pe = 3000 (0.1)
30 = 300 N-m
n  20
1.5 A
4


3002  2882 
A = 5541.769 mm2
I

64

3004  2884 
1 1  I  59.901106 mm4
I NA  (4)(11.5) 3  (80)(0.25) 3  80(0.25)(5.875) 2 
12 12 
= 1887.7919 in4 P Mc 3000 300(0.15)
fc    

Mc 10000(12)(5.75) A I 5541.96910 6
59.901106
For wood: w    365 psi f c  1.293 MPa
I 1887.7919
For steel: b. Maximum tensile stress due to vertical and lateral loads
Mc 10000(12)(6)
s  (20)  (20)  7627.96 psi
I 1887.7919
M = 3000 (0.1) + 450 (3)
12. A hollow circular pole, 6 mm thick and 300 mm outside = 1650 N-m
diameter and 3 m high, weighs 150 N/m. The pole is 3000 1650 (0.15)
ft   
subjected to a vertical load P = 3 kN at an eccentricity 5541 .969  10 6 59.901  10 6
e = 100 mm from the centroid of the section, and a lateral
force H = 0.45 kN at the top of the pole. f t  3.590
a. Determine the maximum compressive stress at the base
due to the vertical load.
b. Determine the maximum tensile stress at the base due to
the vertical and lateral loads.
c. If the hollow pole is replaced by a solid wood pole of c. Shear stress at the base
250 mm diameter, determine the maximum shear stress
at the base. 
A (250) 2
4
A = 49087.385 mm2
V = 450 N
4V 4(450)
fv    12.223 kPa
3 A 3(49087.385106 )

Deformable Bodies mvbmunar 022022

 23

13. A wood beam of rectangular cross section is simply Solution:

supported on a span of length L. The longitudinal axis of
the beam is horizontal, and the cross section is tilted at an
angle α. The load on the beam is a vertical uniform load of
intensity q acting through the centroid C. Determine the
orientation of the neutral axis and calculate the maximum
tensile stress if b = 80 mm, h = 140 mm, L = 1.75 m,
α = 22.5°, and q = 7.5 kN/m.

a. Maximum stress
M y  1600 sin 30  800 lb-in

Solution: M z  1600cos30  1385.641 lb-in

Bending moments: 1
Iy  (3.5)(1.5) 3  0.9844 in4’
wL 7500sin 22.5(1.75)
2 2 12
My    1098.72 N-m
8 8 1
I z  (1.5)(3.5) 3  5.3594 in4
wL2 7500cos22.5(1.75) 2 12
Mz    2652.545 N-m
8 8 b h
My Mz
Moments of inertia: 2 2
 max 
hb3 140(80) 3 Iy Iz
Iy    5.973 106 mm4
12 12 800(0.75) 1385.641(1.75)
 max    1061.96 psi
bh3 80(140) 3 0.9844 5.3594
Iz    18.293106 mm4
12 12 b. angle of inclination of neutral axis
b h
M y  Mz 
 max  2   2   1098.72(0.04)  2652.545(0.07)
Iy Iz 5.973 106 18.293 106 Iz 5.3594
tan   tan   tan 30
 max  17.508 MPa Iy 0.9844
Orientation of neutral axis: θ = 72.35°

Iz 18.293
tan   tan   tan 22.5
Iy 5.973
θ = 51.8°

14. A 1600 lb-in. couple is applied to a wooden beam, of

rectangular cross section 1.5 by 3.5 in., in a plane forming
an angle of 30° with the vertical. Determine
a. the maximum stress in the beam,
b. the angle that the neutral surface forms with the
horizontal plane.

Deformable Bodies mvbmunar 022022

 24

Transformation of Stress and Strains

  x y
2

Assume that a state of plane stress occurs at point Q (with  max      xy 2
σz = τzx = τzy = 0) and defined by the stress components σx, σy  2 
and τxy. The stress components σx’, σy’ and τx’y’ associated with and the corresponding value of the normal stresses is
the element after it has been rotated through an angle θ about  x  y
    ave 
the z-axis, in terms of σx, σy, τxy and θ, are 2
 x  y  x  y Mohr’s Circle for Plane stress
 x   cos 2   xy sin 2
2 2
The circle defined in the preceding section to derive some
 x  y  x  y of the basic formulas relating to the transformation of plane
 y   cos 2   xy sin 2
2 2 stress was first introduced by the German engineer Otto Mohr
 x  y and is known as Mohr’s circle for plane stress.
 xy   sin 2   xy cos 2
2 Construction of the Mohr’s Circle
1. Given the state of stress, plot X of coordinates (σx, –
τxy) and point Y of coordinates (σy, +τxy).
2. Draw the circle with diameter XY.

The abscissas of the points of intersection A and B of the circle

with the horizontal axis represent the principal stresses, and xthe
angle of rotation bringing the diameter XY into AB is twice the
angle θp defining the principal planes, with both angles having
the same sense. The diameter DE defines the maximum
shearing stress and the orientation of the corresponding plane.
Principal Stresses
The value θp of the angle of rotation which correspond to
the maximum and minimum values of the normal stress at point
Q can be determined from the relation
x’
2 xy
tan 2 p 
 x  y

while the corresponding value of the shearing stress is zero. The

value of θs of the angle θ for which the largest value of the
shearing stress occurs is
 x  y
tan 2 s  
2 xy

The two values obtained for θs are 90º apart and the planes (c)
of maximum shearing stress are at 45º to the principal planes.
The maximum value of the shearing stress for a rotation in the
plane of stress is

Deformable Bodies mvbmunar 022022

 25

PROBLEMS 3. An element in plane stress is subjected to stresses as shown

in the figure. Determine the stresses acting on an element
1. For the state of plane stress shown in the figure, determine
oriented at an angle θ = 30° from the x axis, where θ is
a. the principal planes,
positive when counterclockwise.
b. the principal stresses,
c. the maximum shearing stress and the corresponding
Given:
normal stress.
 x  100 MPa
 y  80 MPa
 xy  28 MPa

Solution:
Given:
Construct the Mohr’s circle:
 x  16 MPa  y  48 MPa  xy  60 MPa
2 xy 2(60)
a. tan 2 p    1.875 MPa
 x  y 16  48
2 p  61.93 ;  P  30.96, 59.04
2
 x  y  x  y 
b.  max, min       xy 2

2  2 
2
16  48  16  48 
      60 
2
2  2  1
R (100  80) 2  (28  28) 2  29.732
 16  68 2
 max  52 MPa;  min  84 MPa 28
  Arc tan  70.35
10
 x  y 
2
16  48 
2
    60  10.35
c.  max     xy 2      60 2
  2  At A: a  R cos   29.732 cos 10.35  29.25
 2 
b  R sin   29.732 sin 10.35  5.34
 max  68 MPa
 A  90  a  90  29.25  119.25 MPa
 x  y 16  48
 ave    16 MPa  A  90  a  90  29.25  60.75 MPa
2 2
 A  b  5.34 MPa
2. The figure shows an element in pure shear and the
corresponding Mohr’s circle. Points A and B represent the 4. A cantilever beam of rectangular cross section is subjected
stresses in the x face and on the y face, respectively. to a concentrated load P = 120 kN acting at the free end. The
a. What is the value of the normal axial stress? beam has a width b = 100 mm and h = 200 mm. Point A is
b. How much is the maximum in-plane shear stress? located at a distance 0.5 m from the free end and a distance
c. At what angle, in degrees, from the x-axis does the 150 mm from the bottom of the beam.
maximum in-plane shear stress occur? a. Compute the maximum principal stress at A.
b. Compute the minimum principal stress at A.
c. Compute the maximum shear stress at A.

a. Normal axial stress:   40 MPa

b. Max. in-plane shear stress:   40 MPa
c. Angle from the x-axis where the maximum in-plane
90
shear stress occur:  s   45
2
Deformable Bodies mvbmunar 022022
 26

Solution: Solution:

M  120(500)  60000 N-m

100(200) 3
I  66.667 106 mm4
12
60000(0.05)
x   45 MPa
66.667106
y 0
VQ
 xy  ; Q  100(50)(75)  375 10 3 mm3
Ib A1  150(25)  3750
120000(375106 ) A2  150(25)  3750
 xy   6.75 MPa
66.667106 (0.10) At  7500

Location of neutral axis:

At y  A1 (12.5)  A2 (100)  56.25 mm
1
I NA  (150)( 25) 3  3750 (43.75) 2 
12
1
(25)(150) 3  3750 (43.75) 2
12
I NA  21.582  10 6 mm 4

a. Maximum principal stress at A

T  7500cos60
2 = 3750 N
 45 
R    0   (6.75) 2
 2  Tl  7500sin 60
R = 23.491 = 6495.2 N
M l  3750(0.05625)
= 210.94 N-m
M A  210.94  6495.2(2)  13201.34 N-m
3720 13201.32(0.11875)
A  6
  72.141 MPa
a. Maximum principal stress at A: 750010 21.582106

 max  22.5  R  22.5  23.491 45.991 MPa  x  72.141 MPa

b. Minimum principal stress at A: y 0

 min  22.5  R  22.5  23.491  0.991 MPa  max  72.141 MPa
c.  A  R  23.491 MPa
b. Maximum principal stress at B
5. A cantilever beam of T section is loaded by an inclined force
P = 7.5 kN. The line of action of the force is inclined at an
angle of 60° to the horizontal and intersects the top of the
beam at the end of the cross-section. The beam is 2 m long
and the cross section of the
a. Determine the maximum principal stress at A. Q  150(25)( 43.75)
b. Determine the maximum principal stress at B.
c. Determine the maximum shear stress at B. = 164062.5 mm3

3720 13201.34(0.03125)
x  6

750010 21.582106
 x  19.61 MPa
y 0

Deformable Bodies mvbmunar 022022

 27

6495.2(164062.5 109 )
 xy    1.97 MPa
21.582106 (0.025) R  (9.805) 2  (1.97) 2  10
19.61
 ave   9.805 MPa
2
 max   ave  R  19.805 MPa

c.  max  R  10 MPa

Deformable Bodies mvbmunar 022022

 28

BOARD EXAM PROBLEMS

1. The figure below shows a beam ABCD supporting three
concentrated loads two of which are inclined to the
longitudinal axis of the beam. What is the value of the
largest normal (axial) force? 6. A steel rod having a length of 1m has a cross sectional area
A. 1 B. 3 C. 2 D. 4 of 100 mm2. If it is subjected to an axial tensile force,
compute the stiffness of the rod with a modulus of elasticity
of 200,000 MPa.
A. 0.0005 mm/kN B. 5  107 mm/kN
C. 20  103 kN/mm D. 20 kN/mm
2. A steel cylinder with 12 inches outside diameter and a wall
thickness of 1 inch is filled with concrete and used as a pier
to support an axial load in compression. If the allowable
stresses are 40,000 psi for steel and 3,500 psi for concrete 7. A round bar of length L tapers uniformly from a diameter
D at one end to a smaller diameter d at the other end.
and the Young’s modulus for steel is 30  106 psi and for
Determine the elongation caused by an axial tensile load P.
concrete is 2.8  106 psi, what is the allowable load on the
pier? P P
d D
A. 1,570,372 lb B. 1,480,220 lb
C. 1,364,564 lb D. 1,274,695 lb
L

8. A timber column, 3 m. high has a rectangular cross section,

100 mm  200 mm and is reinforced over its complete
length by two steel plates each 200 mm wide and 10 mm
thick attached to its 200 mm wide faces. The column is to
carry a load of 100 kN. Modulus of elasticity of wood is
15000 N/mm2 and that of steel is 200000 N/mm2. If the
3. A mild steel column is hollow and circular in cross-section failure stress of timber is 55 N/mm2 and that of steel is 380
with an external diameter of 350 mm and internal diameter N/mm2,
of 300 mm. It carries a compressive load of 2,000 kN. The a. Which of the following gives the actual stress of
shortening of the column if its initial height is 5 m and E = timber using a factor of safety of 3?
200,000 N/mm2 is b. Which of the following gives the actual stress of steel
A. 78.4 N/mm2 B. – 0.00039 using a factor of safety of 2?
C. 1.95 mm D. none in the list c. Is the section adequate, not adequate or not allowed by
the code?

4. A short deep cantilever carries a vertically downward load

at its free end. Assume the shear stress is uniformly
distributed over the cross-section of the beam. Use
G = 25000 N/mm2. If the shear is 25 N/mm2 and the length 9. A steel rod having a length of 1 m has a cross-sectional area
of the beam is 0.50 m., the deflection due to shear at the of 100 mm2. It is subjected to an axial tensile force of
free end is: 27 kN. Modulus of elasticity is 200000 MPa.
a. Which of the following gives the deformation of the
steel bar?
b. Which of the following gives the stiffness of the rod?
c. Which of the following gives the flexibility of the rod?
5. A metal specimen with 36 mm  has a length L = 360 mm.
A force of 300 kN elongates the length by 1.2 mm. What is 10. A plate with width of 300 mm and thickness of 20 mm is
the elastic modulus? to be connected to two plates of the same material with half
A. 198763 MPa B. 176839 MPa the thickness, by 25 mm diameter rivets, as shown in the
C. 199999 MPa D. 88419 MPa figure. The rivet holes have a diameter 2 mm larger than
the rivet diameter. The plate is A36 steel with yield
Deformable Bodies mvbmunar 022022
 29

strength Fy = 248 MPa, allowable tensile stress of 0.60Fy, b. Which of the following gives the stress of the
and allowable bearing stress of 1.35Fy. The rivets are bronze?
A502, Grade 2, hot-driven rivets with allowable shear c. Which of the following gives the elongation of the
stress of 150 MPa. rods?
a. Which of the following most nearly gives the max.
load in kN that can be applied to the connection
without exceeding the allowable shear stress in rivets?
A. 700 C. 640
B. 590 D. 550
b. Which of the following most nearly gives the max.
load in kN that can be applied to the connection
without exceeding the allowable bearing stress
between the plates and the rivets?
A. 620 C. 650
B. 700 D. 670 13. An 8-mm diameter steel rod, 20 m long supports a 10 kN
c. Which of the following most nearly gives the max. weight. Find the maximum tensile stress in the rod. The
load in kN that can be applied to the connection unit weight of steel is 77 kN/m3.
without exceeding the allowable tensile strength of the
plates. 14. To stiffen the foot bridge, a short post BD, supported by a
A. 732 C. 620 steel cable ADC, is added. The maximum tension in the
B. 650 D. 590 cable is 2 kN.
a. What is the maximum weight W of a person can the
footbridge carry?
b. If W = 800 N, what is the resulting force in the post
BD.
c. If the area of the cable is 113 mm2, how much is the
resulting elongation of the steel cable due to the
maximum tension of 2 kN? Es = 200 GPa.

15. For the given riveted connection shown, having the

11. The lower ends of the three bars in the figure are at the allowable stresses,
same level before the rigid homogeneous 18 Mg block is Fy = 248 MPa, Shear stress, Fy = 150 MPa ,
attached. Each steel bar has an area of 600 mm2 and Tensile stress = 0.60Fy, Bearing stress = 1.35Fy
E = 200 GPa. For the bronze bar, the area is 900 mm2 and Rivet hole has a diameter 2 mm greater than the rivet
E = 83 GPa. diameter.
a. Which of the following gives the stress developed in The main plate has a width of 300 mm and a thickness of
the bronze bar? 20 mm
b. Which of the following gives the stress developed in a. Compute the diameter of rivets such that the shear
the steel bar? capacity of the rivets is equal to its bearing capacity.
c. Which of the following gives the elongation of the b. Compute the maximum value of P so as not to exceed
bars? the allowable shearing stress.
c. Compute the maximum value of P so as not to exceed
the allowable tensile stress of plates.

12. A rigid block weighing 176.4 kN is attached as shown on 16. A hollow cast-iron pole has an outside diameter of 450 mm
the figure. If the block is to remain horizontal, and an inside diameter of 350 mm. It is subjected to a
a. Which of the following gives the ratio of the load compressive force of 1200 kN (weight included)
carried by each load? throughout its length of 1.2 m. The pole is braced to
prevent bending and buckling. E = 100 GPa.
Deformable Bodies mvbmunar 022022
 30

a. What is the resulting stress due to the compression 19. Four cables are used to lift a precast concrete slab with
force? dimensions 3 m × 3 m and 76 mm thick. The cables are
b. Calculate the total contraction of the member due to attached to a hook above the center point of the slab.
the compressive force. Given: h = 2 m; area of each cable = 79 mm2; concrete
c. Find the load that would result to a total compressive unit weight = 24 kN/m3.
strain of 0.0003 mm/mm. a. What is the tensile force in each of the cables?
b. Determine the normal stress in each of the cables.
c. If the cables elongated by 1 mm, what is the vertical
displacement of the precast slab?

20. The rigid bar AB is hinged at A and supported by a steel

plate hanger, designated as d. The hanger is fixed at C and
17. A steel pipe column is supported on a circular steel base
bolted at D with two plates.
plate and a concrete pedestal. Column ends are hinged and
Given: L1 = 2.0 m, L2 = 1.2 m, H = 3.0 m, width of the
sidesway is prevented.
plate hanger = 40 mm; thickness of plate hanger = 10 mm;
Given:
bolt diameter = 20 mm; allowable bolt shear stress = 68
Axial load = 800 kN
MPa; allowable bolt bearing stress = 240 MPa.
Column outside diameter = 260 mm
a. What is the allowable stress in the hanger based on bolt
Column unsupported length = 3 m
capacity in double shear at D?
Allowable compressive stress = 55 MPa
b. If the minimum tensile stress in the hanger is 138 MPa,
Allowable bearing stress on the pedestal = 10 MPa
find the allowable load W.
a. What is the minimum required thickness of the column
c. If the load W = 60 kN, find the vertical displacement at
based on the allowable compressive stress?
B.
b. Find the minimum required diameter of the base plate.
c. If the thickness of the column is 10 mm, calculate the
slenderness ratio.

21. The antenna tower shown is attached to its foundation by

a ball-and-socket joint. It is supported in its upright
position by four high strength steel wires, namely, AB, AC,
AD, and AE.
Given: Height of tower, h = 20 m
18. A rigid bar AB is hinged at A and is supported by a rod CD
Diameter of the tower, d = 150 mm
at C. The rod is pin connected at D, as shown. Neglect
Steel modulus of elasticity = 200 GPa
deflections of the bar due to bending. Use E = 200 GPa.
Diameter of steel wires = 6.5 mm
a. What is the displacement of the loaded end B of the
a. Find the stress in wire AD if a pulling force F = 5 kN
bar?
acts along the x axis at the tip of the tower.
b. Determine the tensile stress induced in rod CD by the
b. Due to the pulling force F at the tip of the tower, the
80 kN load.
resulting stress in wire AD is 220 MPa. What would be
c. If the allowable stress in rod CD is 124 MPa, what
the elongation of wire AD?
weight W can be safely applied?
c. How far would the tip of the tower move from its
original position if wire AD elongates by 22 mm due to
the pulling force F?

Deformable Bodies mvbmunar 022022

 31

22. A steel bar carries an ore bucket of weight W hanging at

its lower end. Given: bar diameter = 12 mm; length = 18
m; steel modulus of elasticity = 200 GPa. Neglect weight
of bar.
a. Calculate the weight W that can be safely carried so as
not to exceed the allowable steel tensile stress of 138
MPa.
b. Calculate the weight W that can be safely applied so as
not to exceed the allowable elongation of 10 mm.
c. Calculate the total strain if the applied load W = 20 kN.
26. A horizontal beam AD supported by an inclined strut CB
carries a load W = 2.4 kN at the position shown. The strut
23. The angle bracket shown, 75 mm × 75 mm × 15 mm thick,
which consists of one bar is connected to the beam by a
is 200 mm long. It is connected to the 16 mm thick flange
bolt passing through the three bars meeting at joint C. BC
of the column support with 2-16 mm diameter bolts. The
is 75 mm wide × 10 mm thick strut while AD is 100 mm
bracket is subjected to a load of 42 kN from beam A.
wide by 10 mm thick rigid bar.
Assume that the load is applied concentrically at the
a. What is the minimum required bolt diameter in mm at
connection.
C if the allowable bolt shear stress is 68 MPa?
a. Find the critical bolt bearing stress in MPa.
b. Determine the stress in strut BC.
b. What is the resulting bolt shear stress in MPa?
c. Calculate the bolt stress at A if its diameter is 20 mm.
c. What is the bearing pressure, in MPa, on the bracket?

27. Six steel cables are supporting a circular heavy moulding

of diameter 2 m from an overhead point. If the moulding
weighs 2.5 kN/m and the attachment point is 3 m above it,
determine the following:
a. Find the tension in each steel wire.
24. From the given figure:
b. What is the diameter of the wire that will not exceed
W = 2.4 kN, L1 = 1.2 m, L2 = 1.2 m, L3 = 1.2 m, BC is 75
the allowable stress of 124 MPa?
mm wide × 10 mm thick strut, AD is 100 mm wide × 10
c. If the wire is 10 mm in diameter, find the vertical
mm thick rigid bar.
displacement of the moulder.
a. What is the minimum required bolt diameter at C if the
allowable bolt shear stress is 68 MPa?
b. Determine the stress in strut BC.
c. Calculate the bolt stress at A if its diameter is 20 mm.

28. The tensile member shown, 50 mm × 75 mm in cross

section is subjected to a load P = 200 kN. The plane A-A
25. A rigid bar AB is hinged at A and is supported by a rod at makes an angle of 15° with the x-axis.
C. The rod is pin-connected at D, as shown. Neglect a. What is the tensile stress at section A-A?
deflection of the bar due to bending. Use E = 250 GPa. b. Determine the shear stress on plane A-A.
a. If the displacement of the loaded end B of the bar is 3 c. At what angle of plane A-A is the shear stress
mm, determine the weight W. maximum?
b. Determine the tensile stress induced in bar CD by the
load W = 100 kN.
c. If the allowable stress in rod CD is 124 MPa, what
weight W can be safely applied?

Deformable Bodies mvbmunar 022022

 32

32. A square hollow steel strut with a wall thickness t1 = 10

mm is pin connected to two gusset plates having a
thickness t2 = 12 mm, which are welded to the base plate
having a thickness of 12 mm and fastened to a concrete
29. From the figure shown, P = 60 kN base by 4 – 16 mm diameter anchor bolts. Diameter of pin
Allowable shearing stress = 82 MPa is 16 mm, compressive load P = 48 kN, θ = 30°.
Allowable bearing stress = 138 MPa a. Calculate the bearing stress between the strut and the
Thickness of the yoke = 12 mm pin in MPa.
What is the minimum bolt diameter b. Calculate the shear tress in the pin in MPa.
required in mm? c. Calculate the shear stress in the anchor bolt in MPa.

33. A square steel bar with a side of 20 mm is 1.7 m long. It

has a material yield point of 275 MPa and a modulus of
elasticity of 200 MPa. An applied axial load gradually
builds up from zero to a value such that the elongation of
the bar is 15 mm after which the load is removed. Assume
30. From the figure shown:
an elastic, perfectly plastic behavior.
a. Calculate the tensile stress in the body of the bolt.
a. Determine the value of the force applied to the bar until
b. Find the tensile stress at the root of the threads.
the end of the bar has moved down by 15 mm.
c. Find the compressive stress at the head as the bolt bears
b. Determine the residual strain after complete removal of
on the surface to resist the tensile load.
the load.
c. Determine the permanent set of the bar after the force has
been removed.

33. A bar of metal 25 mm in diameter is tested on a length of

250 mm. In tension, the following results were recorded.
Load (kN) 10.4 31.2
Extension (mm) 0.039 0.089
a. Which of the following gives the strain due to the
10.4-kN load?
b. Which of the following gives the strain due to the
31.2-kN load?
c. Which of the following gives the Young’s Modulus?
31. A tension member made up of a pair of angles is connected
a shown with 4 – 25 mm diameter bolts in standard holes.
35. The following data are taken from a stress-strain diagram
All structural steel is A36. Assume that the connection
test for a structural steel bar. The curve is linear from the
between the angles and the structural tee is satisfactory.
origin and the first point.
Allowable bolt shear = 117 MPa
Stress (MPa) Strain (mm/mm)
Allowable tensile stress = 150 MPa
0 0
Allowable bearing stress = 480 MPa
229 0.0008
a. Find the value of P by shear and tension.
314 0.0012
b. Find the value of T by bearing.
341 0.0016
c. What is the diameter of the bolts if P = 360 kN?
355 0.0020
368 0.0024
a. Which of the following gives the modulus of resilience
of this material?
b. Which of the following gives the modulus of elasticity?
c. Which of the following gives the modulus of toughness?

Deformable Bodies mvbmunar 022022

 33

36. A rod with cross-sectional area of 100 sq mm is stretched c. If the circumferential stress is limited to 5 MPa, what is
between two fixed points such that the initial tensile load is the maximum height of water to which the tank may be
10 kN. The rod has an initial modulus of elasticity (E) equal filled?
to 180 GPa and a coefficient of linear expansion (α) equal
to 0.00001 mm/degree centigrade. The initial tension is 42. An 8 mm thick steel tank has an outside diameter of 600
designed such that the rod will have a zero stress when there mm and a length of 3 m. It is subjected to an internal
is a certain change in temperature of the rod. Which of the pressure of 2 MPa.
following most nearly gives the value of this temperature a. Determine the circumferential stress in the tank.
change T: b. Find the longitudinal stress in the tank.
A. 55.6 ºC B. –66.6 ºC c. To what value could the internal pressure be increased
C. –55.6 ºC D. –76.6 ºC if the allowable design stress is 120 MPa?

37. A 25-m steel rod with a cross-sectional area of 1250 mm2 43. A water tank 3 m in diameter and 6 m high is made from a
is secured between rigid supports at both ends. There is no steel having a thickness of 12 mm. When the tank is filled
stress at 20°C; E = 200000 MPa. with water, determine the circumferential stress.
a. Compute the stress when the temperature drops to 0°C.
If it has a coefficient of thermal expansion of 44. A welded steel cylindrical drum made of 10-mm plate has
0.0000117 m/(m-°C). an internal diameter of 1.20 m. By how much will the
b. Compute the strain of the rod if the rod supports yield diameter be changed by an internal pressure of 1.5 MPa?
and move together a distance of 0.5 mm as temperature Assume that Poisson’s ratio is 0.30 and E = 200 GPa.
drops.
c. Compute the stress in the rod if the supports yield and 45. A 50-mm diameter steel tube with a wall thickness of 2 mm
move together a distance of 0.5 mm as the temperature just fits in a rigid hole. Find the tangential stress if an axial
drops. compressive load of 10 kN is applied. Assume v = 0.30 and
E = 200 × 109 N/m2. Neglect the possibility of buckling.

46. A 50-mm diameter solid metal bar is subjected to a

compressive load of 178 kN. Transverse and longitudinal
changes are measured with the use of an electronic strain
gages and the strains are determined to be 0.0012
longitudinally and 0.0004 transversely.
a. Compute the Poisson’s ratio.
b. Compute the modulus of elasticity.
c. Compute the modulus of rigidity.

38. A thin-walled cylindrical shell has an internal diameter of 47. A large spherical tank contains gas with an internal
2 m and is fabricated from plates 20 mm thick. Calculate pressure of 3.75 MPa. Diameter of the tank is 19 m. Yield
the safe pressure in the shell if the tensile strength of the stress of the tank is 570 MPa with a factor of safety of 3.0.
plate is 400 kN/mm2 and the factor of safety is 6. Determine the required thickness of the tank.
A. 1.33 N/mm2 B. 13.3 N/mm2
2
C. 133.3 N/mm D. 0.13 N/mm2 48. Two circular shafts, one hollow and one solid, are made of
the same material and have the diameters shown below. If
39. A thin walled cylindrical shell has an internal diameter of Th is the twisting moment that the hollow shaft can resist
2 m and is fabricated from plates 20 mm thick. The tensile and Ts is the twisting moment that the solid shaft can resist,
strength of the plates is 400 N/mm2. The factor of safety is the ratio of Th to Ts is:
6, Young’s modulus is 200,000 N/mm2, and its Poisson’s 1 9 1 15
A. B. C. D.
ratio v  0.3 . Calculate the safe pressure in the shell. 4 16 2 16
A. none of the list B. 1.33 N/mm2
C. 0.33 N/mm2 D. 3.33 N/mm2

40. Which of the following gives the ratio of shear modulus G

to the modulus of elasticity of steel if it has a poisons ratio
of 0.25?
49. The hollow steel shaft (G = 12  106 psi) must transmit a
41. A water tank 3 m in diameter and 6 m high is made from torque of 300,000 in-lb. The total angle of twist must not
steel having a thickness of 12 mm.
exceed 3 per 100 ft. The maximum shearing strength must
a. When the tank is filled with water, determine the
not exceed 16,000 psi. Find the inside diameter d if the
circumferential stress.
outside diameter is equal to 12 inches.
b. Determine the longitudinal stress at the bottom of the
A. 11.05″ B. 6.0″ C. 1.0″ D. 3.0″
tank when it is filled with water.

Deformable Bodies mvbmunar 022022

 34

50. The maximum allowable torque in kN-m. for a 50 mm equal and opposite to the stress obtained in the first
diameter steel shaft when the allowable shearing stress is problem above?
81.5 MPa is: c. Which of the following most nearly gives the
A. 3.0 B. 1.0 C. 4.0 D. 2.0 eccentricity in mm of the load P, measured from the
instantaneous center of the bolts?

51. The rotation or twist in degrees of a shaft, 800 mm. long,

subjected to a torque of 80 N-m, 20 mm in diameter, and
shear modulus G of 80,000 MPa is:
A. 3.95 B. 2.92 C. 1.48 D. 3.94

56. Compound shaft shown is attached to rigid supports.

For the bronze segment AB: For the steel segment BC:
Diameter = 75mm Diameter = 50mm
G = 35 GPa G = 83 GPa
52. A hollow steel tube has an outside diameter of 220 mm and τ = 60 MPa τ = 80 MPa
inside diameter of 200 mm. Which of the following gives a. Which of the following gives the ratio of b/a so that
the max. torque that it could carry if the allowable shearing each material will be stressed to its max. permissible
stress is 75.5 MPa? limit?
A. 50 kN-m B. 60 kN-m b. Which of the following gives the reaction at A?
C. 40 kN-m D. 30 kN-m c. Which of the following gives the max. angle of twist
when a = 1 m?

53. From the given bracket shown, which consists of two plates
riveted to the column by 4-22 mm Ø rivets which is
subjected to an eccentric load of 150kN.
a. Which of the following gives the shearing stress due to
axial load only? 57. A steel shaft with a constant diameter of 60 mm is loaded
b. Which of the following gives the shearing stress due to by torques applied as shown. Using G = 83 GPa.
moment alone? a. Which of the following gives the torque applied on the
c. Which of the following gives the max. shearing stress? right side if the allowable shearing stress is not to exceed
23.58 MPa?
54. From the given figure shows a shaft of 60 mm diameter b. Which of the following gives the reaction at the support?
with value of G = 83000 MPa. c. Which of the following gives the angle of twist of the
a. Which of the following gives the reaction at the fixed shaft?
end?
b. Which of the following gives the maximum torsional
stress?
c. Which of the following gives the angle of twist of the
shaft? 58. A riveted bracket is subjected to a load P acting at an angle
of θ from the vertical. There are 4 – 25 mm diameter rivets,
two on each column flange and two plates, one on each
column flange.
55. A plate is connected to a column by four bolts and carries a. Which of the following gives the max. shear stress if
an eccentric load as shown in figure. For this problem, the P = 40 kN, θ = 45°, b = 100 mm, a = 150 mm, and
load P = 32.5 kN with an eccentricity e = 250 mm. The e = 250 mm?
dimensions a = 150 mm and b = 200 mm. The diameter of b. Which of the following gives the maximum value of P if
the bolts is 20 mm. e = 250 mm and the allowable shearing stress is 120
a. For this problem, assume that the load P acts through MPa?
the centroid of the bolts. Which of the following most c. Which of the following gives the minimum eccentricity
nearly gives the stress on the bolts in MPa? e when P = 40 kN if the allowable shearing stress is 120
b. For this problem, assume that the only load acting on MPa?
the connection is a moment equal to the load P
multiplied by its eccentricity e. Which of the following
most nearly gives the distance from the centroid of the
bolts in mm, where, if a bolt is placed, the stress will be
Deformable Bodies mvbmunar 022022
 35

torque of 3 kN-m at its free end. What is the resulting angle

of twist, in degrees, of the bar. G = 78 GPa.
63. A solid post is subjected to pure torsion.
Given:
Post diameter = 75 mm
Length = 2.5 m
Shear modulus of elasticity = 78 GPa
a. What is the torsional rigidity of the post in kN-m2?
b. Determine the torsional stiffness of the post.
59. A plate is connected to a column by four bolts and carries c. Determine the maximum shear stress in the post if it is
an eccentric load as shown in figure. For this problem, the subjected to a torque of 540 N-m.
load P = 27.5 kN with an eccentricity e = 200 mm. The
dimension a = b = 150 mm. The diameter of the bolts is 25 64. A simple beam of length L has a concentrated load of P at
mm. a distance a from the left support and b from the right
a. For this problem, assume that the load P acts through the support. The maximum moment of the beam is:
2
centroid of the bolts. Which of the following most nearly A. Pa B.
Pab
gives the stress on the bolts in MPa? L L
b. For this problem, assume that the only load acting on the Pa 2
b
C. D. None of the above
connection is a moment equal to the load P multiplied by L2
its eccentricity e. Which of the following gives the
distance from the centroid of the bolts in mm, where, if a 65. A beam 8.5 m long is fixed at the left end and supported by
bolt is placed, the stress will be equal and opposite to the a roller at a point 1.5 m from the right end. There is an
stress obtained in the first problem above? internal hinge at a point 2.5 m from the fixed end. The beam
c. Which of the following most nearly gives the eccentricity is loaded by a clockwise couple equal to 50 kN- m at a point
in mm of the load P, measured from the instantaneous 1 m to the right of the fixed end and a concentrated load
center of the bolts? equal to 12 kN at a point 4.5 m to the right of the fixed end
and a downward uniform vertical load equal to 5 kN/m
starting at a point 4.5 m from the fixed end up to the right
end of the beam. Which of the following most nearly gives
the maximum moment?
A. 21.9 kN-m B. 52.9 kN-m
C. 32.3 kN.m D. 72.2 kN-m

60. The compound shaft shown is attached to rigid supports.

For the bronze segment AB: For the steel segment BC:
Diameter = 75mm Diameter = 50mm
G = 35 GPa G = 83 GPa 66. A simply-supported beam has the shear diagram shown in
τ = 60 MPa τ = 80 MPa the figure.
length of steel = 1m a. Determine the maximum positive bending moment, in
a. Which of the following gives the ratio of the lengths of kN-m.
the bronze and steel so that each material will be A. 20 B. 30 C. 40 D. 10
stressed to its maximum permissible limit? b. Determine the maximum negative bending moment, in
b. Which of the following gives the maximum torque kN-m.
carried by the steel? A. 45 B. 25 C. 15 D. 36
c. Which of the following gives the relative angle of twist c. Determine the location of the inflection point from the
with respect to the support if maximum torsion is left support, in meters.
applied at B? A. 4.2 B. 4.6 C. 3.2 D. 3.6

61. A cantilever hollow circular bar 5 mm thick with outside

diameter of 75 mm is subjected to a torque of 3 kN-m at its
67. A beam 25 m long is simply supported at the left end and
free end. Find the maximum shear stress in the bar.
at a distance of 20 m from the left end. It carries a uniform
62. A cantilever hollow circular bar is 1.5 m long, 5 mm thick, load of 5 kN/m between the two supports and a
and with outside diameter of 75 mm. It is subjected to a concentrated load of 20 kN at the other end. Compute the
distance of maximum moment from the left support.
Deformable Bodies mvbmunar 022022
 36

A. 9 m B. 8 m C. 10 m D. 18 m

68. From the given beam subjected to the loads as shown in the
figure, 73. The cantilever beam shown is loaded with a
a. Which of the following gives the reaction at the roller counterclockwise couple of 60 kN-m 1 m from the fixed
C? end and a uniformly distributed load of 5 kN/m over 3 m
A. 27 kN B. 22 kN C. 13 kN D. 16 kN from the free end.
b. Which of the following gives the max. shear? a. Which of the following gives the moment at A in kN-m?
A. 17 kN B. 12 kN C. 13 kN D. 15 kN A. –9 B. –7.5 C. –6 D. –8.5
c. Which of the following gives the max. negative b. Which of the following gives the shear at C in kN?
moment? A. 15 B. 20 C. 10 D. 25
A. –12 kN-m B. –10 kN-m c. Which of the following gives the moment at C in kN-
C. –16 kN-m D. –18 kN-m m?
A. –20.5 B. –18.5 C. –25 D. –22.5

69. A simple beam of length L has a concentrated load of P at

a distance a from the left support and b from the right 74. The figure shows a semi-circular arch subjected to equal
support. The maximum moment of the beam is: but oppositely directed forces at A and B.
a. Find the shear force at D.
70. From the given shear diagram shown. b. Find the moment acting at D.
a. Which of the following gives the max. concentrated c. Determine the axial force at D.
load?
b. Which of the following gives the max. negative
moment?
c. Which of the following gives the point of inflection from
the right end?

75. From the figure shown:

a. What is the moment at midspan if x1 = 2 m.
b. Find the distance x1 if the moment at midspan is zero.
c. Determine the distance x2 so that the maximum moment
in the beam is the least possible value.
71. A beam is loaded as shown on the figure.
a. Which of the following gives the reaction at A?
–10
b. Which of the following
–20 gives the maximum shear?
c. Which of the following gives the maximum moment?

76. A BW 775 × 287 steel I beam has the following

dimensions:
H 755mm total beam depth
tw 19mm web thickness
72. The beam shown in the figure carries a triangular load B 360mm flange width
which varies from 12 kN/m at A to 0 at B and a concentrated tf 32mm flange thickness
load of 20 kN at C. The minimum section modulus Sx is:
a. Which of the following gives the nearest value of the A. 9680 cm3 B. none in the list
reaction at B? C. 9480 mm3 D. 9365 cm3
b. Which of the following gives the nearest value of the
location of the point of inflection from A?
c. Which of the following gives the nearest value of the
maximum positive moment?

Deformable Bodies mvbmunar 022022

 37

mm. The structural steel section of the composite system

has the following properties:
beam depth, d 413 mm
flange width, bf 180 mm
cross sectional area, As 9480 mm2
moment of inertia 274.3 x 106 mm4
Where is the new location of the neutral axis from the slab
in mm? Use n =10 to transform the area of concrete to steel.
A. 142.59 B. 234.50
77. A tee section is made up of a 30 mm × 150 mm flange and C. 152.20 D. 562.43
a 30 mm × 160 mm web. The neutral axis above the lowest
fibers of the section is:
A. 155 mm B. 35 mm
C. 160 mm D. 125.97 mm

84. A wooden beam 150 mm. wide by 300 mm. deep is loaded
as shown. If the maximum allowable flexural stress is 8
MPa,
78. A strip of steel 1 mm thick is bent into an arc of circle of
1) Which of the following gives the max. moment capacity
radius 1.0 m, E is equal to 200 GPa. The maximum bending
of the beam?
stress is:
A. 20 kN-m B. 21 kN-m
A. 100 MPa B. 50 MPa
C. 19 kN-m D. 18 kN-m
C. 200 MPa D. none of the list
2) Which of the following gives the max. value of w if
x = 2 m?
79. The structural I-beam supporting a floor carries a floor load
A. 10 kN/m B. 9 kN/m
of 4.6 kN/m2. The beams span 6.0 m and are simply
C. 11 kN/m D. 12 kN/m
supported at their ends. Determine the center line spacing
3) Which of the following gives the max. value of P if
if the allowable stress in the beam is 120 MPa and the
x = 2m?
section modulus is 534 × 103 mm3.
A. 20 kN B. 19 kN
A. 3.75 m B. 4.0 m
C. 18 kN D. 21 kN
C. 3.45 m D. 3.0 m
80. A structural steel wide flange section has a flange section
150 mm × 12 mm and a web section 300 mm × 9 mm. It
carries a max. shear of 90 kN. If the average shearing stress
33.33 MPa, and INA =108 × 106 mm4, with a statistical
moment of area equal to 382,050 mm3, compute the max.
shearing stress.
85. A timber beam has a circular cross section having a
A. 35.375 MPa B. 42.276 MPa
diameter of 250 mm. It has a simple span of 4 m. NSCP
C. 28.375 MPa D. 52.365 MPa
specifications states that in a circular beam, the strength is
equal to the strength of square section having the same area.
81. A simple beam has a span of 5.0 m. The maximum moment
Allowable stresses of wood are as follows:
in the beam is 69.0 kN-m. The allowable bending stress is
Shearing stress parallel to the grain: fv = 1.73 MPa
138 MPa. The required section modulus is:
Bending stress: fb = 16.5 MPa
A. 2.0 × 103 mm3 B. 9.522 × 106 mm3
3 3 Allowable deflection = 1/240 of span
C. 250 × 10 mm D. 500 × 103 mm3
Modulus of elasticity of wood = 7.31 GPa
a. Which of the following gives the maximum uniform
82. A wooden log is to be used as a foot bridge to span 3 m.
load so that the allowable shear stress parallel to the
The log is required to support a concentrated load of 30 kN
grain is not exceeded?
at midspan. If the allowable stress in shear is 0.7 MPa, what
b. Which of the following gives the maximum uniform
is the diameter of the log that would be needed? Assume
load so that the allowable bending stress is not
the log is very nearly circular and the bending stresses are
exceeded?
adequately met.
c. Which of the following gives the maximum uniform
A. 146 B. 165
load so that the allowable deflection is not exceeded?
C. 157 D. 191

83. A W 410 × 74.4 supports a slab with a thickness of 150

mm. The effective width of slab was computed to be 2000
Deformable Bodies mvbmunar 022022
 38

88. A 4-m simple span beam having a cross-section of 300 mm

× 500 mm carries a uniformly distributed load throughout
its span. The beam is 80% stress grade Apitong with
allowable bending stress of 16.5 MPa, allowable shearing
stress of 1.73 MPa and modulus of elasticity of 7.31 GPa.
Allowable deflection is 1/240 of the span.
a. Which of the following most nearly gives the value of the
uniform load without exceeding the allowable bending
stress of the beam, in kN/m?
A. 118 B. 103 C. 132 D. 124
86. A composite beam is made up of a wide flange with two
b. Which of the following most nearly gives the value of the
cover plates of 320 mm × 12 mm placed at the top and at
uniform load without exceeding the allowable shearing
the bottom. The beam has a simple span of 6 m and carries
stress of the beam, in kN/m?
a total load of 360 kN/m.
A. 63 B. 54 C. 87 D. 75
Allowable bending stress: Fb = 150 MPa
c. Which of the following most nearly gives the value of the
Allowable shear stress: Fv = 100 MPa
uniform load without exceeding the allowable deflection,
in kN/m?
Allowable deflection:   L
360
Properties of wide flange section
A = 0.02 m2
tw = 0.018 m
d = 0.56 m
I = 0.001568 m4
E = 200 GPa
A. 102 B. 98 C. 124 D. 114

89. A simply supported beam spans 8 m and supports a

superimposed uniformly distributed load of 20 kN/m.
Beam properties are as follows:
A = 12320 mm2 Ix = 445 × 106 mm4
bt = 193 mm Iy = 23 × 106 mm4
tf = 19 mm Fy = 248 MPa
a. Which of the following gives the section modulus of the tw = 11 mm
section? d = 465 mm
b. Which of the following gives the deflection of the Consider bending about the x-axis.
beam? a. What is the maximum bending stress?
c. Is the beam b. How much is the maximum web shear stress?
A. adequate c. Calculate the maximum horizontal shear stress.
B. inadequate for flexure and shear
C. inadequate for flexure and deflection 90. The combination section shown consists of a wide flange
D. inadequate for flexure, shear, and deflection and a channel.
Properties of sections:
87. The beam is 150 mm wide and 300 mm deep. The W 600 mm × 125 kg/m
allowable bending stress on extreme fiber is 12 MPa. If x = A = 15935 mm2
2.5 m: d = 610 mm
a. Which of the following gives the maximum moment that tw = 12 mm
the beam can resist, in kN-m? bf = 229 mm
A. 24 B. 16 C. 18 D. 27 tf = 20 mm
b. Which of the following gives the maximum value of w, Ix = 986 × 106 mm4
in kN/m? Iy = 39 × 106 mm4
A. 12.5 B. 8.6 C. 9.5 D. 16.4
c. Which of the following gives the maximum value of P, C 300 mm × 31 kg/m
in kN? A = 3829 mm2
A. 27 B. 24 C. 16 D. 18 d = 300 mm
tw = 7 mm
bf = 75 mm
tf = 13 mm
Ix = 53.7 ×106 mm4
Iy = 1.6 × 106 mm4
x = 18 mm
Deformable Bodies mvbmunar 022022
 39

The beam is subjected to a uniformly distributed load 92. A footing is 2.40 m by 1.50 m. The total load P of 1,000
(beam weight included) of 20 kN/m throughout the simply kN on the footing has an eccentricity if 0.40 m in the
supported span of 8 m and a moving load P in kN. direction of the longer dimension. Determine the maximum
a. Calculate the resisting moment (kN-m) of the soil pressure at the base.
combination section if the allowable flexural stress A. 454 kN/m2 B. 556 kN/m2
2
Fbx = 186 MPa. C. 893 kN/m D. 652 kN/m2
b. Calculate the load P (kN) such that the allowable bending
stress, Fbx = 186 MPa. 93. A hollow circular pole, 6 m thick, with 300 mm outside
c. If the wide flange and the channel were joined by 2-20 diameter and height of 3 m, weighs 150 N/m. The pole is
mm diameter rivets spaced at 150 mm on center subjected to the following: vertical load P = 3 kN at an
longitudinally throughout its length, calculate the eccentricity e = 100 mm from the centroid of the section;
allowable load P (kN) if the allowable shear stress in the lateral force H = 0.45 kN at the top of the pole.
rivets, Fv = 100 MPa. a. Determine the maximum compressive stress at the base
due to the loads.
91. A rectangular footing 2 m. by 1.5 m carries a vertical load b. Determine the maximum tensile stress at the base due to
of 500 kN and moment of 100 kN-m in the longer direction the vertical and lateral loads.
of the footing. What is the maximum soil pressure under c. If the hollow pole is replaced by a solid wood pole of
the footing? 250 mm diameter, determine the maximum shear stress
A. 353 kN/m2 B. 267 kN/m2 at the base.
C. 3000 psf D. 533 kN/m2
94. A solid pole 3 m high and 250 mm in diameter is fixed at
the base. It is subjected to a compressive force of 3 kN
acting at an eccentricity of 100 mm from the centroidal axis
and a lateral load of 0.45 kN applied at the top. What is the
maximum compressive stress at the base of the pole?

Deformable Bodies mvbmunar 022022