SECTION 9.9
601
Castigliano’s Theorem
Castigliano’s Theorem
The beams described in the problems for Section 9.9 have constant
flexural rigidity EI.
M0
A
Problem 9.9-1 A simple beam AB of length L is loaded at the left-hand
end by a couple of moment M0 (see figure).
Determine the angle of rotation A at support A. (Obtain the solution
by determining the strain energy of the beam and then using Castigliano’s
theorem.)
Solution 9.9-1
L
Simple beam with couple M0
STRAIN ENERGY
M0
A
B
x
M0
L
(downward)
M M0 RAx M0 U
M20
M2dx
2 EI
2 EI
L
¢1 0
M20 L
x 2
≤ dx L
6 EI
CASTIGLIANO’S THEOREM uA L
RA Solution 9.9-2
M0 x
L
P
a
D
x
L
Pa
L
MAD RA x Pbx
L
MDB RB x Pax
L
UAD 1
2 EI
x
UDB 1
2EI
a
¢
0
0
b
¢
M2dx
2EI
Pbx 2
P2a3b2
≤ dx L
6 EIL2
Pax 2
P2a2b3
≤ dx L
6 EIL2
U UAD UDB P2a2b2
6 LEI
CASTIGLIANO’S THEOREM
D b
L
B
dU Pa2b2
dP 3 LEI
B
D
a
STRAIN ENERGY U b
RB A
Simple beam with load P
P
Pb
L
(clockwise)
(This result agree with Case 7, Table G-2)
Problem 9.9-2 The simple beam shown in the figure supports a
concentrated load P acting at distance a from the left-hand support and
distance b from the right-hand support.
Determine the deflection D at point D where the load is applied.
(Obtain the solution by determining the strain energy of the beam and
then using Castigliano’s theorem.)
A
M0 L
dU
dM0 3 EI
x
≤
L
M0 ¢ 1 RA B
(downward)
602
CHAPTER 9
Deflections of Beams
Problem 9.9-3 An overhanging beam ABC supports a concentrated load
P at the end of the overhang (see figure). Span AB has length L and the
overhang has length a.
Determine the deflection C at the end of the overhang. (Obtain
the solution by determining the strain energy of the beam and then using
Castigliano’s theorem.)
Solution 9.9-3
A
B
L
1
2 EI
x
a
L
C
1
2 EI
UCB Pa
L
B
STRAIN ENERGY U UAB RA A
C
a
Overhanging beam
P
x
P
(downward)
L
¢
0
Pax 2
P2a2L
≤ dx L
6 EI
a
(Px) dx 6 EI
P2a3
2
0
P2a2
(L a)
6 EI
U UAB UCB Pax
MAB RA x L
MCB Px
M2dx
2 EI
CASTIGLIANO’S THEOREM
C dU Pa2
(L a)
dP 3EI
(downward)
q0
Problem 9.9-4 The cantilever beam shown in the figure supports
a triangularly distributed load of maximum intensity q0.
Determine the deflection B at the free end B. (Obtain the
solution by determining the strain energy of the beam and then
using Castigliano’s theorem.)
B
A
L
Solution 9.9-4
Cantilever beam with triangular load
q0
STRAIN ENERGY
P
U
B
A
x
M Px q0 x3
6L
L
¢ Px 0
q0 x3 2
≤ dx
6L
Pq0 L
q20 L5
PL
6 EI
30 EI
42 EI
2 3
L
P fictitious load corresponding to deflection B
M2dx
1
2 EI
2 EI
4
CASTIGLIANO’S THEOREM
q0 L4
0U PL3
(downward)
0P 3 EI 30 EI
(This result agrees with Cases 1 and 8 of Table G-1.)
B SET P 0:
B q0 L4
30 EI
SECTION 9.9
Problem 9.9-5 A simple beam ACB supports a uniform load of intensity
q on the left-hand half of the span (see figure).
Determine the angle of rotation B at support B. (Obtain the solution
by using the modified form of Castigliano’s theorem.)
Castigliano’s Theorem
q
C
A
B
L
—
2
Solution 9.9-5
Simple beam with partial uniform load
q
MODIFIED CASTIGLIANO’S THEOREM (EQ. 9-88)
M0
C
A
uB B
x
x
L
—
2
L
—
2
3 qL M0
8
L
RB qL M0
8
L
BENDING MOMENT AND PARTIAL DERIVATIVE FOR
SEGMENT AC
MAC RA x qx
qx
3 qL M0
¢
≤x 2
8
L
2
L
¢0 x ≤
2
0MAC x
0M0
L
BENDING MOMENT AND PARTIAL DERIVATIVE FOR
CB
MCB RBx M0 ¢
0MCB
x
1
0M0
L
qL M0
≤ x M0
8
L
L
¢0 x ≤
2
≤¢
0M
≤ dx
0M0
L2
B¢
0
1
EI
1
EI
2
SEGMENT
1
EI
M
qx2
3qL M0
x
≤x R B R dx
8
L
2
L
L2
B¢
0
qL M0
x
≤ x M0 R B 1 R dx
8
L
L
SET FICTITIOUS LOAD M0 EQUAL TO ZERO
uB 2
¢
M0 fictitious load corresponding to angle of
rotation B
RA EI
L2
¢
0
1
EI
0
3qLx qx2 x
≤ ¢ ≤ dx
8
2
L
L2
¢
qLx
x
≤ ¢ 1 ≤ dx
8
L
qL3
qL3
7qL3
(counterclockwise)
128 EI 96 EI 384 EI
(This result agrees with Case 2, Table G-2.)
603
L
—
2
604
CHAPTER 9
Deflections of Beams
Problem 9.9-6 A cantilever beam ACB supports two concentrated loads
P1 and P2, as shown in the figure.
Determine the deflections C and B at points C and B, respectively.
(Obtain the solution by using the modified form of Castigliano’s
theorem.)
Solution 9.9-6
P1
C
L
—
2
L
—
2
B
B 1
EI
x
L
MCB P2 x ¢ 0 x ≤
2
0MCB
0MCB
0
x
0P1
0P2
BENDING MOMENT AND PARTIAL DERIVATIVES FOR
SEGMENT AC
MAC P1 ¢ x L
≤ P2x
2
0MAC L
x
0P1
2
0MAC
x
0P2
¢
L
x L≤
2
MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION C
(MCB ) ¢
1
EI
0
L2
0
1
EI
0MCB
≤ dx
0P1
L
(MAC ) ¢
L2
L
—
2
MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION B
BENDING MOMENT AND PARTIAL DERIVATIVES FOR
SEGMENT CB
L
—
2
B
P2
1
EI
P2
C
Cantilever beam with loads P1 and P2
A
C P1
A
0MAC
≤ dx
0P1
L
B P1 ¢ x L2
L3
(2 P1 5 P2 )
48 EI
L
L
≤ P2x R ¢ x ≤ dx
2
2
L2
(MCB ) ¢
0
1
EI
0MCB
≤ dx
0P2
L
(MAC ) ¢
L2
0MAC
≤ dx
0P2
L2
1
EI
1
EI
(P2x) (x) dx
0
L
L2
B P1 ¢ x L
≤ P2x R (x) dx
2
P2L3
L3
(5 P1 14 P2 )
24 EI 48 EI
L3
(5P1 16P2 )
48 EI
(These results can be verified with the aid of Cases 4
and 5, Table G-1.)
SECTION 9.9
Problem 9.9-7 The cantilever beam ACB shown in the figure is
subjected to a uniform load of intensity q acting between points A and C.
Determine the angle of rotation A at the free end A. (Obtain the
solution by using the modified form of Castigliano’s theorem.)
Castigliano’s Theorem
q
C
A
B
L
—
2
L
—
2
Solution 9.9-7
Cantilever beam with partial uniform load
MODIFIED CASTIGLIANO’S THEOREM (EQ. 9-88)
q
M0
C
A
B
x
L
—
2
L
—
2
uA M0 fictitious load corresponding to the angle of
rotation A
BENDING MOMENT AND PARTIAL DERIVATIVE FOR
SEGMENT AC
2
qx
MAC M0 2
0MAC
1
0M0
¢0
x
L
≤
2
BENDING MOMENT AND PARTIAL DERIVATIVE FOR
CB
MCB M0 qL
L
¢x ≤
2
4
¢
L
x L≤
2
¢
1
EI
M
≤¢
0M
≤ dx
0M0
L2
¢ M0 0
1
EI
qx2
≤ (1)dx
2
L
B M0 L2
qL
L
¢ x ≤ R (1)dx
2
4
SET FICTITIOUS LOAD M0 EQUAL TO ZERO
uA SEGMENT
EI
605
1
EI
0
L2
qx2
1
dx 2
EI
L
L2
¢
qL
L
≤ ¢ x ≤ dx
2
4
qL3
qL3
48 EI 8 EI
7qL3
(counterclockwise)
48 EI
(This result can be verified with the aid of Case 3,
Table G-1.)
0MCB
1
0M0
Problem 9.9-8 The frame ABC supports a concentrated load P at point C
(see figure). Members AB and BC have lengths h and b, respectively.
Determine the vertical deflection C and angle of rotation C at end C
of the frame. (Obtain the solution by using the modified form of
Castigliano’s theorem.)
b
B
C
P
h
A
606
CHAPTER 9
Solution 9.9-8
Deflections of Beams
Frame with concentrated load
b
B
C
M0
MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION C
C x
P
h
C A
P concentrated load acting at point C
(corresponding to the deflection C)
M0 fictitious moment corresponding to the angle
of rotation C
BENDING MOMENT AND PARTIAL DERIVATIVES FOR
MEMBER AB
0MAB
b
0P
(0 x h)
0MAB
1
M0
BENDING MOMENT AND PARTIAL DERIVATIVES FOR
MEMBER BC
0MBC
x
0P
M
0M
≤ dx
0P
h
1
EI
≤¢
(Pb M0 )(b)dx 0
(0 x b)
1
EI
h
Pb2dx 0
Pb2
(3h b)
3 EI
1
EI
b
(Px M )(x)dx
0
0
1
EI
b
Px dx
2
0
(downward)
MODIFIED CASTIGLIANO’S THEOREM FOR ANGLE OF
ROTATION C
uC EI
¢
1
EI
M
≤¢
0M
≤ dx
0M0
h
(Pb M0 )(1)dx 0
1
EI
b
(Px M )(1) dx
0
0
Set M0 0:
uC MBC Px M0
¢
Set M0 0:
x
MAB Pb M0
EI
1
EI
h
Pb dx 0
Pb
(2h b)
2EI
1
EI
b
Px dx
0
(clockwise)
0MBC
1
0M0
Problem 9.9-9 A simple beam ABCDE supports a uniform load of
intensity q (see figure). The moment of inertia in the central part of the
beam (BCD) is twice the moment of inertia in the end parts (AB and DE).
Find the deflection C at the midpoint C of the beam. (Obtain the
solution by using the modified form of Castigliano’s theorem.)
q
B
A
C
I
L
—
4
D
I
2I
L
—
4
E
L
—
4
L
—
4
SECTION 9.9
Solution 9.9-9
MODIFIED CASTIGLIANO’S THEOREM (EQ. 9-88)
P
Integrate from A to C and multiply by 2.
C
B
I
C 2
E
D
I
2I
2¢
¢
MAC 0MAC
≤¢
≤ dx
EI
0P
1
≤
EI
L
P fictitious load corresponding to the deflection C
at the midpoint
qL P
RA 2
2
2¢
BENDING MOMENT AND PARTIAL DERIVATIVE FOR THE
LEFT-HAND HALF OF THE BEAM (A TO C)
qLx qx2 Px
L
¢0 x ≤
2
2
2
2
0MAC x
L
¢0 x ≤
0P
2
2
2
EI
MAC C 1
≤
2 EI
¢
0
qLx qx2 Px x
≤ ¢ ≤ dx
2
2
2 2
L2
¢
L4
LOAD
qLx qx2 Px x
≤ ¢ ≤ dx
2
2
2 2
P EQUAL TO ZERO
qLx qx2 x
≤ ¢ ≤ dx
2
2
2
L2
L4
¢
qLx qx2 x
≤ ¢ ≤ dx
2
2
2
13 qL4
67 qL4
6,144 EI 12,288 EI
31qL4
4096 EI
(downward)
MA
A
B
a
B
L
C
BENDING MOMENT AND PARTIAL DERIVATIVES FOR
SEGMENT BC
MA Pa
(downward)
L
L
MAx Pax
(0 x L)
MBC RC x L
L
0MBC
x 0MBC
ax
0MA
L
0P
L
Reaction at support C: RC a
x
C
Overhanging beam ABC
P
A
¢
0
L4
1
EI
Problem 9.9-10 An overhanging beam ABC is subjected to a couple MA
at the free end (see figure). The lengths of the overhang and the main span
are a and L, respectively.
Determine the angle of rotation A and deflection A at end A.
(Obtain the solution by using the modified form of Castigliano’s
theorem.)
MA
L4
SET FICTITIOUS
C Solution 9.9-10
607
Nonprismatic beam
q
A
Castigliano’s Theorem
L
x
MA couple acting at the free end A (corresponding
to the angle of rotation A)
P fictitious load corresponding to the deflection A
BENDING MOMENT AND PARTIAL DERIVATIVES FOR
SEGMENT AB
MAB MA Px (0 x a)
0MAB
0MAB
1
x
0MA
0P
608
CHAPTER 9
Deflections of Beams
MODIFIED CASTIGLIANO’S THEOREM FOR ANGLE OF
ROTATION A
uA M 0M
¢ ≤¢
≤ dx
EI 0MA
1
EI
1
EI
EI
A ¢
a
(MA Px)(1) dx
0
1
EI
L
¢
0
MAx Pax
x
≤ ¢ ≤ dx
L
L
L
a
M dx EI 1
A
0
0
MA
(L 3a)
3 EI
L
¢
M
≤¢
0M
≤ dx
0P
a
1
EI
(M Px)(x)dx
1
EI
A
0
L
¢
0
MAx Pax
ax
≤ ¢ ≤ dx
L
L
L
Set P 0:
Set P 0:
uA MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION A
1
EI
A MAx x
≤ ¢ ≤ dx
L
L
(counterclockwise)
a
MAxdx 0
MAa
(2L 3a)
6EI
0
¢
MAx ax
≤ ¢ ≤ dx
L
L
(downward)
B
A
A
L
P
STRAIN ENERGY OF THE SPRING (EQ. 2-38a)
C
US a
L
k
x
R2B P2 (L a) 2
2k
2 kL2
x
STRAIN ENERGY OF THE BEAM (EQ. 9-80a)
RA
RB
UB Pa
(downward)
L
P
RB (L a) (upward)
L
RA 2 EI
M2dx
TOTAL STRAIN ENERGY U
U UB US BENDING MOMENT AND PARTIAL DERIVATIVE FOR
SEGMENT AB
MAB RAx Pax
L
dMAB
ax
dP
L
(0 x L )
BENDING MOMENT AND PARTIAL DERIVATIVE FOR
BC
SEGMENT
dMBC
x
dP
(0 x a)
C
k
Beam with spring support
B
MBC Px
L
P
Problem 9.9-11 An overhanging beam ABC rests on a simple support
at A and a spring support at B (see figure). A concentrated load P acts
at the end of the overhang. Span AB has length L, the overhang has
length a, and the spring has stiffness k.
Determine the downward displacement C of the end of the
overhang. (Obtain the solution by using the modified form of
Castigliano’s theorem.)
Solution 9.9-11
1
EI
M2dx P2 (L a) 2
2 EI
2 kL2
APPLY CASTIGLIANO’S THEOREM (EQ. 9-87)
dU
d M2dx
d P2 (L a) 2
B
R
dP dP 2 EI
dP
2 kL2
d M2dx P(L a) 2
dP 2 EI
kL2
C a
SECTION 9.9
DIFFERENTIATE UNDER THE INTEGRAL SIGN (MODIFIED
CASTIGLIANO’S THEOREM)
C 1
EI
¢
P(L a)
M dM
≤¢
≤ dx EI dP
kL2
L
¢
0
1
EI
2
609
Castigliano’s Theorem
a
(Px)(x)dx 0
P(L a) 2
kL2
Pa L Pa3 P(L a) 2
3 EI 3 EI
kL2
2
Pax
ax
≤ ¢ ≤ dx
L
L
C Pa2 (L a) P(L a) 2
3 EI
kL2
q
Problem 9.9-12 A symmetric beam ABCD with overhangs at both ends
supports a uniform load of intensity q (see figure).
Determine the deflection D at the end of the overhang. (Obtain the
solution by using the modified form of Castigliano’s theorem.)
A
B
L
—
4
D
C
L
—
4
L
Solution 9.9-12 Beam with overhangs
P
q
A
B
x
D
C
x
x
L
¢0 x ≤
4
1
EI
M
≤¢
L
≤
4
0M
≤ dx
0P
L4
¢
0
qx2
≤ (0) dx
2
L
B
0
L
1
L
≤ R B ¢ x ≤ R RB x
4
2
4
2
(0 x L)
L
1
EI
1
EI
q
3 qL P
L
¢x ≤ ¢
≤x
2
4
4
4
0MBC
x
0P
4
x
3qL P
q
L 2
¢x ≤ ¢
≤x R
2
4
4
4
L4
¢
0
qx2
Px ≤ (x)dx
2
SET P 0:
D SEGMENT BC
¢
x
1
B R dx 4
EI
BENDING MOMENTS AND PARTIAL DERIVATIVES
MBC B q ¢ x EI
1
EI
3 qL 5P
RC 4
4
0MAB
0
0P
¢0
0MCD
x
0P
D P fictitious load corresponding
to the deflection D
L
length of segments AB and CD
4
L length of span BC
SEGMENT AB
qx2
MAB 2
qx2
Px
2
MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION D
q intensity of uniform load
3 qL P
RB 4
4
MCD SEGMENT CD
q
L 2 3qL
x
¢x ≤ x R B R dx
2
4
4
4
B
0
L4
¢
0
qx2
≤ (x) dx
2
5 qL4
qL4
37 qL4
768 EI 2048 EI
6144 EI
(Minus means the deflection is opposite in direction
to the fictitious load P.)
∴ D 37 qL4
6144 EI
(upward)
610
CHAPTER 9
Deflections of Beams
Deflections Produced by Impact
The beams described in the problems for Section 9.10 have constant
flexural rigidity EI. Disregard the weights of the beams themselves,
and consider only the effects of the given loads.
Problem 9.10-1 A heavy object of weight W is dropped onto the
midpoint of a simple beam AB from a height h (see figure).
Obtain a formula for the maximum bending stress max due to the
falling weight in terms of h, st, and st, where st is the maximum
bending stress and st is the deflection at the midpoint when the weight
W acts on the beam as a statically applied load.
Plot a graph of the ratio max /st (that is, the ratio of the dynamic stress
to the static stress) versus the ratio h /st. (Let h /st vary from 0 to 10.)
Solution 9.10-1
W
h
A
B
L
—
2
L
—
2
Weight W dropping onto a simple beam
MAXIMUM DEFLECTION (EQ. 9-94)
max st (2st 2hst ) 12
MAXIMUM BENDING STRESS
For a linearly elastic beam, the bending stress is
proportional to the deflection .
smax max
2h 12
∴
1 ¢1 ≤
sst
st
st
smax sst B 1 ¢ 1 2h 12
≤ R
st
h
st
smax
sst
0
2.5
5.0
7.5
10.0
2.00
3.45
4.33
5.00
5.58
WL3
for a simple beam with a load
48 EI
at the midpoint.
NOTE: st GRAPH OF RATIO max/st
6
max
st
4
2
0
2.5
5.0
7.5
10.0
h
st
Problem 9.10-2 An object of weight W is dropped onto the midpoint of a
simple beam AB from a height h (see figure). The beam has a rectangular
cross section of area A.
Assuming that h is very large compared to the deflection of the beam
when the weight W is applied statically, obtain a formula for the
maximum bending stress max in the beam due to the falling weight.
W
h
A
B
L
—
2
L
—
2
SECTION 5.5
Solution 9.10-2
Weight W dropping onto a simple beam
Height h is very large.
sst M WL
S
4S
st WL3
48 EI
MAXIMUM DEFLECTION (EQ. 9-95)
max 2hst
For a linearly elastic beam, the bending stress is
proportional to the deflection .
smax max
2h
∴
sst
st
B st
2 hs2st
B st
W2L2
16 S2
s2st 3 WEI
2
st
SL
(2)
(3)
Substitute (2) and (3) into (1):
smax 18 WhE
B AL
(1)
Problem 9.10-3 A cantilever beam AB of length L 6 ft is
constructed of a W 8 21 wide-flange section (see figure). A
weight W 1500 lb falls through a height h 0.25 in. onto the
end of the beam.
Calculate the maximum deflection max of the end of the
beam and the maximum bending stress max due to the falling
weight. (Assume E 30 106 psi.)
Solution 9.10-3
s2st For a RECTANGULAR BEAM (with b, depth d):
bd 3
bd 2 I
3
3
I
S
2
12
6
bd A
S
MAXIMUM BENDING STRESS
smax 611
Method of Superposition
W = 1500 lb
W 8 21
h = 0.25 in.
A
B
L = 6 ft
Cantilever beam
DATA: L 6 ft 72 in. W 1500 lb
h 0.25 in. E 30 106 psi
W 8 21
I 75.3 in.4 S 18.2 in.3
MAXIMUM BENDING STRESS
MAXIMUM DEFLECTION (EQ. 9-94)
Equation (9-94) may be used for any linearly elastic
structure by substituting st W/k, where k is the
stiffress of the particular structure being considered. For
instance: Simple beam with load at midpoint:
48 EI
k 3
L
Cantilever beam with load at the free end: k Equation (9-94):
max st (2st 2 h st ) 12 0.302 in.
3 EI
Etc.
L3
Consider a cantilever beam with load P at the free
end:
Mmax PL
PL3
smax max S
S
3 EI
smax 3 EI
Ratio:
2
max
SL
3 EI
∴ smax 2 max 21,700 psi
SL
For the cantilever beam in this problem:
(1500 lb)(72 in.) 3
WL3
st 3 EI 3(30 106 psi)(75.3 in.4 )
0.08261 in.
Problem 9.10-4 A weight W 20 kN falls through a height h 1.0
mm onto the midpoint of a simple beam of length L 3 m (see figure).
The beam is made of wood with square cross section (dimension d on
each side) and E 12 GPa.
If the allowable bending stress in the wood is allow 10 MPa,
what is the minimum required dimension d?
W
h
A
d
B
d
L
—
2
L
—
2
612
CHAPTER 9
Solution 9.10-4
Deflections of Beams
Simple beam with falling weight W
DATA: W 20 kN h 1.0 mm L 3.0 m
E 12 GPa
allow 10 MPa
SUBSTITUTE (2) AND (3) INTO EQ. (1)
2smaxd 3
8hEd 4 12
1 ¢1 ≤
3 WL
WL3
CROSS SECTION OF BEAM (SQUARE)
d dimension of each side
d4
d3
I
S
12
6
SUBSTITUTE NUMERICAL VALUES:
MAXIMUM DEFLECTION (EQ. 9-94)
1000 3
1600 4 12
d 1 B1 d R
9
9
max st (2st
2 h st )
2(10 MPa)d3
8(1.0 mm)(12 GPa)d 4 12
1 B1 R
3(20 kN)(3.0 m)
(20 kN)(3.0 m) 3
12
SQUARE BOTH SIDES, REARRANGE, AND SIMPLIFY
MAXIMUM BENDING STRESS
For a linearly elastic beam, the bending stress is
proportional to the deflection .
smax max
2h 12
∴
1 ¢1 ≤
(1)
sst
st
st
STATIC TERMS st AND st
M
WL 6
3 WL
sst ¢
≤ ¢ 3≤ S
4
d
2d3
st WL3
WL3 12
WL3
¢ 4≤ 48 EI 48 E d
4 Ed 4
(2)
1000 2 3 1600
2000
≤ d d
0
9
9
9
2500d 3 36d 45 0 (d meters)
¢
SOLVE NUMERICALLY
d 0.2804 m 280.4 mm
For minimum value, round upward.
d 281 mm
(3)
Problem 9.10-5 A weight W 4000 lb falls through a height
h 0.5 in. onto the midpoint of a simple beam of length L 10 ft
(see figure).
Assuming that the allowable bending stress in the beam is
allow 18,000 psi and E 30 106 psi, select the lightest
wide-flange beam listed in Table E-1 in Appendix E that will be
satisfactory.
Solution 9.10-5
W = 4000 lb
h = 0.5 in.
A
B
L
— = 5 ft
2
L
— = 5 ft
2
Simple beam of wide-flange shape
DATA: W 4000 lb h 0.5 in.
L 10 ft 120 in.
allow 18,000 psi E 30 106 psi
MAXIMUM DEFLECTION (EQ. 9-94)
max st (2st 2hst ) 12
or
(d meters)
max
2h 12
1 ¢1 ≤
st
st
STATIC TERMS st AND st
WL3
48 EI
4 sallow S
smax
4S
sallow ¢
≤
sst
WL
WL
2h
48 EI
96 hEI
2h ¢
≤
st
WL3
WL3
sst M WL
S
4S
st SUBSTITUTE (2) AND (3) INTO EQ. (1):
MAXIMUM BENDING STRESS
For a linearly elastic beam, the bending stress is
proportional to the deflection .
smax max
2h 12
1 ¢1 ≤
∴
(1)
sst
st
st
4sallowS
96hEI 12
1 ¢1 ≤
WL
WL3
REQUIRED SECTION MODULUS
S
WL
96 hEI 12
B 1 ¢1 ≤ R
4sallow
WL3
(2)
(3)
SECTION 5.5
Trial
Actual
beam
I
S
W 8 35
127
31.2
W 10 45
248
49.1
W 10 60
341
66.7
W 12 50
394
64.7
W 14 53
541
77.8
W 16 31
375
47.2
Lightest beam is W 14 53
SUBSTITUTE NUMERICAL VALUES
20 3
5 I 12
in. ≤ B 1 ¢ 1 ≤ R
3
24
(S in.3; I in.4)
S¢
(4)
PROCEDURE
1. Select a trial beam from Table E-1.
2. Substitute I into Eq. (4) and calculate required S.
3. Compare with actual S for the beam.
4. Continue until the lightest beam is found.
Problem 9.10-6 An overhanging beam ABC of rectangular
cross section has the dimensions shown in the figure. A weight
W 750 N drops onto end C of the beam.
If the allowable normal stress in bending is 45 MPa,
what is the maximum height h from which the weight may
be dropped? (Assume E 12 GPa.)
Solution 9.10-6
40 mm
A
Required
S
41.6 (NG)
55.0 (NG)
63.3 (OK)
67.4 (NG)
77.8 (OK)
66.0 (NG)
W
h
C
B
1.2 m
613
Method of Superposition
2.4 m
40 mm
500 mm
Overhanging beam
DATA: W 750 N LAB 1.2 in. LBC 2.4 m
E 12 GPa allow 45 MPa
bd 3
1
(500 mm)(40 mm) 3
12
12
2.6667 106 mm4
2.6667 106 m4
bd 2 1
(500 mm)(40 mm) 2
6
6
133.33 103 mm3
133.33 106 m3
S
DEFLECTION C AT THE END OF THE OVERHANG
C
P
3 EI
2
C a (L a)
(1)
MAXIMUM DEFLECTION (EQ. 9-94)
I
B
Stiffness of the beam: k Equation (9-94) may be used for any linearly elastic
structure by substituting st W/k, where k is the
stiffness of the particular structure being considered.
For instance:
48 EI
Simple beam with load at midpoint: k 3
L
3 EI
Cantilever beam with load at free end: k 3 Etc.
L
For the overhanging beam in this problem (see Eq. 1):
P
A
st W Wa2 (L a)
k
3 EI
(2)
in which a LBC and L LAB:
L
a
P load at end C
L length of spear AB
a length of overhang BC
From the answer to Prob. 9.8-5 or Prob. 9.9-3:
Pa2 (L a)
C 3 EI
st W(L2BC )(LAB LBC )
3 EI
(3)
EQUATION (9-94):
max st (2st 2 h st ) 12
or
max
2h 12
1 ¢1 ≤
st
st
(4)
614
CHAPTER 9
Deflections of Beams
MAXIMUM BENDING STRESS
For a linearly elastic beam, the bending stress is
proportional to the deflection .
smax max
2h 12
(5)
∴
1 ¢1 ≤
sst
st
st
M WLBC
sst S
S
(6)
MAXIMUM HEIGHT h
Solve Eq. (5) for h:
smax
2h 12
1 ¢1 ≤
sst
st
¢
h
W(L2BC )(LAB LBC ) sallowS sallowS
¢
≤¢
2≤
6 EI
WLBC
WLBC
st smax smax
¢
≤¢
2≤
sst
2 sst
W(L2BC ) (LAB LBC )
0.08100 m
6 EI
sallow S 10
3.3333
WLBC
3
(7)
A
EI
R
Im
L
Rotating flywheel
NOTE: We will disregard the mass of the beam and
all energy losses due to the sudden stopping of the
rotating flywheel. Assume that all of the kinetic
energy of the flywheel is transformed into strain
energy of the beam.
KINETIC ENERGY OF ROTATING FLYWHEEL
1
kE Im 2
2
M 2dx
2 EI
M Rx, where x is measured from support A.
L
R2L3
1
(Rx) 2dx U
2 EI q
6 EI
STRAIN ENERGY OF BEAM U 10 10
≤¢
2 ≤ 0.36 m
3
3
or h 360 mm
Problem 9.10-7 A heavy flywheel rotates at an angular speed (radians per second) around an axle (see figure). The axle is rigidly
attached to the end of a simply supported beam of flexural rigidity EI
.and length L (see figure). The flywheel has mass moment of inertia Im
about its axis of rotation.
If the flywheel suddenly freezes to the axle, what will be the
reaction R at support A of the beam?
Solution 9.10-7
(8)
SUBSTITUTE NUMERICAL VALUES INTO E Q. (8):
h (0.08100 m) ¢
smax 2
smax
2h
≤ 2¢
≤11
sst
sst
st
h
Substitute st from Eq. (3), st from Eq. (6), and
allow for max:
CONSERVATION OF ENERGY
kE U
R
1
R2 L3
Im 2 2
6 EI
3 EI Im 2
B
L3
NOTE: The moment of inertia IM has units of kg m2
or N m s2