Lectures 19, 20 ENGR 240 Strength of Materials Fall 2015
2016
Combined Loading I & II
States of Stress
At any desired section of a member, the
internal reactions may be obtained by
applying the Statics equilibrium equations:
VyQ/(It)
My
P/A
A
Vy Myx/Iy
Vx Mx
Tr/J
P
ΣFx = 0 to find Vx
ΣFy = 0 to find Vy
ΣFz = 0 to find P
ΣMx = 0 to find Mx
ΣMy = 0 to find My B
ΣMz = 0 to find T
VxQy/(Iyty)
Then
P/A
σ = EP/AEMxy/IxEMyx/Iy Tr/J
Mxy/Ix
Axial Load Flexure
τ = EVxQy/(Iyty)EVyQx/(Ixtx)ETr/J
Flexure Torsion
(circular
members only)
Lecture 21 ENGR 240 Strength of Materials Fall 2015
2016
Deflection of Beams
w
We have previously covered the
relationship:
1/ρ = M/(EI) or ρ = EI/M L
m = dy/dx = tan θ ≈ θ, if θ is very small.
y=0@x=0
Arc length: dx = ρdθ y=0@x=L
y’ = 0 @ x = L/2
1/ρ = dθ/dx = d2y/dx2 = M/(EI)
P
M = EIy’’ a
∫EIy’’ = ∫M, where M is a function of x.
EIy’ = ∫M
L
y’ = 1/(EI)∫M
For 0 ≤ x ≤ a,
= slope of the elastic curve, i.e. the EIy1’’ = M1(x)
deflected shape of the neutral axis of y1 = 0 @ x = 0
the beam.
For a ≤ x ≤ L,
∫y’ = 1/(EI)∫∫M EIy2’’ = M2(x)
y2=0@x=L
y = 1/(EI)∫∫M
Also need matching boundary conditions.
= equation of the elastic curve; value y1 ’ = y2 ’ @ x = a
of the deflection (upward is y1 = y2 @ x = a
positive).
Four boundary conditions are needed to
Double integration yields two constants of solve for C1, C2, C3, and C4.
integration, C1 and C2. They must be
determined from the physical boundary
conditions.
P
y’ = 0 @ x = 0
y=0@x=0
Lecture 22 ENGR 240 Strength of Materials Fall 2015
2016
Indeterminate Beams by Integration M
Statically indeterminate beams are
characterized by the fact that not enough
PL/2 V
equilibrium equations exist to solve for the A
unknown reactions. At least one redundant
will exist. (A redundant is a reaction that is ΣMcut = 0
present but unnecessary for equilibrium.) M – Ax + PL/2 = 0
As before, we consider the deflection to M = Ax – PL/2
help us analyze the problem. The boundary
conditions put restraints on the system such EIy’’ = M = Ax – PL/2
that we can solve for all unknowns. EIy’ = Ax2/2 – PLx/2 + C1
EIy = Ax3/6 – PLx2/4 + C1x + C2
1. Determine the number of unknowns,
2. Write the applicable equilibrium From the boundary conditions:
equations: y = 0 @ x = 0, therefore C2 = 0
ΣM = 0 ΣFy = 0 y’ = 0 @ x = L, therefore
3. Find the deflection equation by 0 = AL2/2 –PL2/2 + C1
integrating the internal moment equation C1 = PL2/2 – AL2/2
twice. y = 0 @ x = L, therefore
4. Apply the physical boundary conditions 0 = AL3/6 – PL3/4 + PL3/2 – AL3/2 +0
to solve for the constants of integration and AL3/3 = PL3/4
redundants. Known values of y or y’ will A = 3P/4
provide the equations that equilibrium did
not.
Example:
The beam shown is fixed at the right end
and supported by a roller near the left end.
Determine the reaction at the left end in
terms of P. Let the origin be at the roller
support as shown.
L/4
L
A B
PL/2
x
Make a FBD between A and B.
Lecture 23A ENGR 240 Strength of Materials Fall 2015
2016
Beam Deflection by Superposition Plus P
We have solved several deflection problems
using the integration method. We can now
y2
tabulate our results of the common
problems and use these solutions rather
than starting from scratch each time.
Appendix D is a table compiled by the
authors, giving y’ = y1’ + y2’
1. The equation of the elastic curve. y = y1 + y2
2. The maximum deflection, and
3. Slopes at the ends.
The method of superposition may be used
for two or more loadings: “If several causes
act simultaneously on a system and each
effect is directly proportional to its cause,
then the total effect is the sum of the
individual effects when considered
separately, provided Hooke’s Law is valid
within the range of these effects either
individually or combined.” The stresses
must remain in the elastic region and the
deformation resulting from any given load
must be small and not affect the condition
of application of the other loads.
Thus, the slope and deflection at any point
is the sum of the individual effects as long
as Hooke’s Law applies.
Is equivalent to
w
y1
Lecture 23B ENGR 240 Strength of Materials Fall 2016
2015
Indeterminate Beams by Superposition Is equivalent to
PL/2
When a beam is statically indeterminate,
deformation equations may be written to
supplement those from equilibrium.
B
Note: The number of unknowns minus the
number of equilibrium equations
equals the number of redundants. yB1 = -Mx2/(2EI) = -(PL/2)(L)2/(2EI)
= -PL3/(4EI)
We may remove as many redundant
restraints as necessary to make the beam
statically determinate. Consider the Plus
statically determinate beam deflected by
the applied loads. The deformations at the
points where the redundant restraints were
removed can be evaluated. Then the
reactions can be applied back with such
magnitude so as to make the combined
deformation from the loads and restraints B
consistent with the boundary conditions of
the original beam.
yB2 = +BL3/(3EI)
Example: Determine the roller reaction at B.
Solve:
yB = yB1 + yB2
PL/2
Substitute known boundary conditions:
yB = 0 = -PL3/(4EI) + BL3/(3EI)
A B
B = 3P/4
L L/4
Known Boundary Conditions:
y=0@x=0
y=0@x=L
Lecture 24 ENGR 240 Strength of Materials Fall 2015
2016
Column Buckling P
M
When we analyzed short columns under
compression, we assumed that the columns
V
would fail by crushing. The state of stress is
σ = -P/A.
If there is any eccentricity to the load (P is P
off-center), a bending moment will be
created. Then σ = -P/A +/- My/I , where
M = Pe.
Make a cut and sum forces:
Short columns are defined by 0 < λ <60.
Where λ is the slenderness ratio equals the ΣMcut: = 0
column height divided by the radius of
gyration of the appropriate axis. M + Py = 0
h = column height
Remember from last week,
r = radius of gyration = e(I/A) EIy’’ = M, so EIy’’ + Py = 0
And EIy’’ + Py = 0, is an Ordinary Linear
Intermediate columns are 60 < λ < 120, and
Homogeneous Differential Equation with
long columns are 120 < λ < 300. constant coefficients. Its characteristic
equation is:
Buckling investigates the force P that will
cause a long slender column to become [D2 + P/(EI)]y = 0
unstable. (The column will suddenly bend
to a curved shape due to an axial load, D = Ee[P/(EI)] i
probably applied with a slight eccentricity.)
Let p = e[P/(EI)]
The critical load that can be put on the
column before it will buckle was derived by
the Swiss mathematician, Leonhard Euler.
So D = Epi.
P P
The general solution is:
y = A sin (px) + B cos (px)
Boundary conditions:
h y
y=0@x=0 ∴B=0
y = 0 @ x = h ∴ A sin (ph) = 0
So A = 0 or sin (ph) = 0
From Trigonometry:
P P
Lecture 24 ENGR 240 Strength of Materials Fall 2015
2016
The sine of a function is 0 at intervals of π.
So (ph) = nπ, where n =1,2,3,…
e[P/(EI)] (h) = nπ
P = n2π2EI/h2
The critical Euler Buckling Load (n = 1) for
a pin-connected column is:
Pcr = π2EI/h2
And buckling will occur about the Imin axis
(usually Iy).
For n =2:
P
P = 4Pcr
Each increase in n, will add one more bow
in the column.
Lecture 25 ENGR 240 Strength of Materials Fall 2015
2016
Column End Conditions Fixed-Fixed (K = 0.5)
(typical lateral frame column)
Euler’s formula: Pcr = π2EI/h2 was derived
for a column with pin-connected ends; that
is, the ends were allowed to rotate (M = 0)
but not to move. For columns with other
end conditions, this formula can be altered
rather than derive completely different one L Le = L/2
each time. In general,
Pcr = π2EI/Le2
Where Le is the effective length (height) of
the column. (Le is the length so that the
column would appear to be pin-connected.) The quarterpoints are the points of
So Le = KL, where K is the scalar to convert inflection. (M = 0)
the length to an analogous pin-connected
length.
Fixed-Pinned (K = 0.7)
Some possible end conditions:
Fixed-Free (K = 2)
(flagpole)
Le = 0.7L
L
Le = 2h
L
This one must be derived from a
differential equation.
Boundary Conditions:
y=0@x=0
y’ = 0 @ x = 0
y=0@x=L
