Lecture 5
Solution Method for Beam Deflection
Problem 5-1: Consider the clamped-clamped elastic beam loaded by a uniformly distributed line load q.
q
x
l
EI
a) Formulate the boundary conditions.
b) Find the deflected shape of the beam using the direct integration method.
c) Find the maximum deflection magnitude and location.
d) Determine the location and magnitude of the maximum stress in the beam.
Problem 5-1 Solution:
(a) Boundary conditions
w(0)
w '(0)
w(l) 0
w '(l) 0
(b) Find the deflected shape use direct integration
We use the 4th order differentiated equation:
EIwIV
q
ql 3
6
l2
2
Integrate 3 times
EIw ' l
0
C1
C2l C3
1
Use B.C.
w '(0)
w '(l)
0
We get
C3
EIw' l
ql 3
6
0
l2
C1
2
0
C2l
(1)
Integrate equation
qx3
6
EIw '
C1
x2
2
C2 x
We get
EIw
qx 4
24
x3
C1
6
x2
C2
2
C4
Use B.C.
w(0)
w(l)
0
We get
C4
EIw l
0
ql 4
24
0
l3
l2
C1
C2
6
2
(2)
Combine equations (1) and (2) to solve for C1 and C2
C1
ql
2
C2
ql 2
12
2
Finally
w( x )
qx 2
x2
24EI
or
ql 4
24EI
w(l )
4
x
l
2lx l 2
x
2
l
3
x
l
(c) The maximum deflection magnitude occurs at the mid-span
wmax
x
q
l
24EI 2
l
2
wmax
l
x
2
2
l
2
2
2l
l
2
l2
ql 4
384EI
(d) Determine location and magnitude of the maximum stress
Stress distribution in the beam is
Mz
I
The maximum stress locates where moment and z magnitude are the maximum:
max
M max zmax
I
Recall
M
EIw ''
Substitute w x into the above equation
M
q
x2
2
lx l 2
2 12
3
Let’s consider moment at x
M
L
2
l
x
2
and x 0
q
M
M
2
1 l
2 2
max
max
ql 2
24
ql 2
12
x 0
M
l2
12
l l
2 2
x 0
ql 2
12
ql 2 zmax
12 I
4
Problem 5-2:
Calculate the second moment of inertia of the beam cross section for:
a) Solid rectangular cross section of width b and height h.
b) Thin-walled square box section of width and height b.
c) Solid circular cross section of radius r.
Problem 5-2 Solution:
(a) Solid rectangular cross section of width b and height h.
Ix
A
y 2 dA
h
2
h
2
Ix
y 2bdy
bh3
12
(b) Thin-walled square box section of width and height b.
I square
s
s4
12
1 2
s
12
1
4ts
12
2t
4
12
s 2t
2
s2
s 2t
2
4t 2 2s 2 4ts 4t 2
5
Eliminate higher order terms of t
I square
1
4ts
12
1
2ts 3
3
4t 2
2s 2
4ts
4t 2
4t 2 s 2
2 3
tb
3
I square
(c) Thin-walled square box section of width and height b.
dA rd
Ix
A
y 2 dA
2
r
0
0
2
rdrd
r4
sin 2 drd
4
2
0
r4
4 2
r
4
r sin
1
sin 2
4
2
0
3
J
Ix
Ix
Iy
r3
2
Iy
r3
4
6
Problem 5-3: In wood construction building codes the beam deflections cannot exceed L/360 where L
is the length of the beam. Where do you think this requirement comes from? Choose a typical beam
example and state clearly the formulation and your assumptions on the boundary conditions and loading.
Using the deflection criteria estimate the fracture strain of the plaster board which is nailed directly to the
ceiling beams (joist) in single home construction.
Problem 5-3 Solution:
The L/360 constraint is required to prevent the plaster board from cracking
We assume a simply supported beam with a distributed load
The deflected shape and subsequently curvature can be simplified as
0
xx
xx
z
d
dx
0 z
L / 360
L/2
z /180
The maximum strain will be at the outmost fiber
h
at z
h
xx
2
360
Typically the wooden beams used in house construction are 2in×6in. Use those values
6
xx
2
360
1
60
7
The strain seen by the extreme fiber of the wood is also seen by the plaster board.
8
Problem 5-4:
Given a beam with a “T” section subjected to pure bending shown in Figure 1, calculate:
a) the location of the neutral axis
b) the second moment of inertia
c) Find the shear stress distribution in the “T” section.
Figure 1
Problem 5-4 Solution:
a)
The location of neutral axis is
yi Ai
Ai
y1 A1 y2 A2
A1 A2
1
L t L t t
L t
2
L t t Lt
t
tL
2
3L2 3tL t 2
4 L 2t
9
If we assume L
t
3L
4
b)
t ,and don’t consider stress variation over the thickness in the flange, the
If we assume L
second moment of inertia is
I
Ii
Ai di
tL3
12
Lt
L
tL3
12
Lt
3
L L
4
tL3
12
tL3
8
Lt 3
12
2
I
Lt
2
Lt
2
L
2
3
L
4
L
2
2
5tL3
24
Shear stress distribution
First, let’s define two parts of the cross section: Part I the horizontal part and Part II the vertical part
Part I
QI
z I dA
where z I is the moment are to the shaded region I
10
zI
t
constant
2
dA t dy
L
L
2
y
L
t L
t
2
QI
QI
t
2
tdy
L
2
y
Part II
QII
2QI
z II dA
y 0
where zII is the moment are to the shaded region II
zI
QI
QII
t L
2QI
t
2
z d A tdz
L t
y 0
z
L
2
t
2
check QII
=
z
ztdz
L t
z2
0, yes!
VQ
It
Finally, the shear stress distribution is showed as below picture
11
Problem 5-5: Continuity Condition
Solve the problem of a simply-simply supported beam loaded by a point force acting at eh symmetry
plane, but at a distance a from the left support
In the notes of lecture 5 the solution of this problem was outlined, but not completed,
(a) Complete the derivation by calculating all four integration constants
(b) Proof that all continuity conditions are satisfied at x=a
(c) Show that in the limiting case of a=L/2 the solution is identical to one that was derived in class
and you were asked to memorize
Problem 5-5 Solution:
(a) The reaction force are calculated from moment equilibrium
RA
P
b
l
P
RB
P 1
a
l
a
l
(1)
The corresponding bending moments and shear forces are
RA x
Pbx
l
RB l
x
M x
V x
Pb
l
Pa
l
Pa l x
l
0
x
a
a
x
L
0
x
a
a
x
L
(2)
12
Integrating governing equation
EI
d 2w
dx 2
M x
(3)
Combing with equation (2), we have
EIwI
EIw
II
Pbx3
C1 x C2
6l
Pa lx 2 x 3
C2 x C 4
l
2
6
0
x
a
(4)
a
x
L
Boundary conditions and continuity conditions
w 0
w l
0
wI a
w II a
dw I
dx
dw II
dx
x a
(5)
x a
Substitute (4) into (5) , we have
C2
0
Pal 3
C3l C4 0
3
Pa la 2
Pba 3
C1a
6l
l
2
Pba 2
2l
C1
Pa
la
l
a3
6
a2
2
C3 a C4
(6)
C3
Solve for (6), we get the four unknown integration constants
C1
C2
C3
C4
Pa
a l 2l a
6l
0
Pa 2
2
3
Pa
6
Pl l a 2l a a
6l
13
(b) Check continuity condition
(i)
Check continuity of moment M at x=a, referring equation (2)
M
(ii)
M
Pa l a
l
Pba
l
0
Check continuity of shear force V at x=a, referring equation (2)
V
(iii)
M
Check continuity of angle
dw I
dx
x a
II
dw
dx
x a
V
V
Pb
l
Pa
l
P
at x=a
Pba 2
2l
Pa
a l 2l a
6l
a2
2
Pa
la
l
Pa 2
2
P l a 2l a a
6l
0
(iv)
Check continuity of deflection w
w
w
w
Pba 3 Pa 2
a l 2l a
6l
6l
2
Pa 3 P l a 2l a a
Pa la a 3
l 2 6
2
6l
w
w
Pa 3
6
0
All continuity conditions are satisfied.
(c) Referring to equation (4)
wI
a
L
,b
2
1 Pbx 3
EI 6l
C1 x C2
L
2
wI
wo
Px
3l 2 4 x 2
48 EI
PL3
I
w L
x
48 EI
2
They are the same as was derived in lecture notes.
14
Problem 5-6:
Another problem to test your knowledge on continuity conditions.
A system of two identical beams shown in the figure below is a statically determined problem. The beams
are rigidly welded, so that the angle remains 90 degrees.
Determine
(a) The distribution of bending moment and shear forces in both segments.
(b) Find the deflected shape of both beams and make sketch.
(c) Find the relation between the tip-- load and the vertical displacement of the tip.(Hint: assume
the rotations to be very small.)
(Note: Do not solve this problem using the Castigliano’s theorem, which is much simpler, but has not
been covered yet.)
Problem 5-6 Solution:
(a) Free body diagram as below, in which Rxa , Rxb and R ya are reaction forces
15
Moment equilibrium about point a
Rxb 2 L PL 0
P
2
Rxb
Horizontal and vertical forces balance
Rya
P
Rxa
Rxb
P
2
Beam 1
Let’s look at cross section O+
Where
V
M
P
2
P
L
2
Also, let’s look at cross section O-
16
Where
V
M
P
2
P
L
2
Distribution of shear force
Distribution of moment
17
Beam 2
At the left cross-section
V
M
P
PL
Distribution of shear force
Distribution of moment
18
(b) Deflection is anti-symmetric for beam 1, at point O, w
0 , deflection shapes are
(c) From moment distribution of beam 1, we know
M1
P
L
2
d 2 w1
dy 2
M1
y
0
y
L
Governing equation
EI
P
L y
2
0
y
L
0
y
Integrate twice
w1
P Ly 2
2 EI 2
y3
6
C1 y C2
L
Apply boundary conditions
w1
y 0
C1
w1
y L
PL2
, C2
6EI
0
0
19
P Ly 2
2 EI 2
w1
y3
6
PL2
y
6EI
0
y
L
At y=0
PL2
6EI
Vertical displacement of point O
o
uo
x L
P
dy
L EA
o
dy
x
PL
EA
Integrate governing equation of beam 2
d 2 w2
EI
dy 2
P L x
M2
We have
EI
dw2
dx
EIw2
P
x2
2
P Lx
Lx 2
2
x3
6
C1
C1 x C2
Use angle continuity condition
EI
dw2
dx
PL2
6
C1
w2
1 x 0
x 0
PL2
6EI
x 0
PL
EA
PIL
A
u0
C2
Finally
w2
P Lx 2
EI 2
x3
6
PL2
x
6EI
PL
EA
20
At x=L, the tip deflection is
PL3
2EI
wp
PL
EA
Now, let’s compare the magnitude of the two terms in w p , for a square cross-section
wp
II
wp I
II
I
For h L 1 20 , wp wp
PL
EA
PL3
2EI
2I
LA
h3
12
Lh 2
2
1 h
6 l
2
0.25 10 3 . wp II , which is the deflection caused by compression,
I
is three orders of magnitudes smaller than wp , which is the deflection caused by bending.
21
MIT OpenCourseWare
http://ocw.mit.edu
2.080J / 1.573J Structural Mechanics
Fall 2013
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.