9
Deflections of Beams
Differential Equations of the Deflection Curve
The beams described in the problems for Section 9.2 have constant
flexural rigidity EI.
Problem 9.2-1 The deflection curve for a simple beam AB (see figure)
is given by the following equation:
y
B
A
q0x
v
(7L4 10L2x 2 3x4)
360LEI
L
Describe the load acting on the beam.
Solution 9.2-1
v
Probs. 9.2-1 and 9.2-2
Simple beam
q0 x
(7L4 10 L2x 2 3x 4 )
360 LEI
Take four consecutive derivatives and obtain:
q0
q0 x
v–– LEI
From Eq. (9-12c): q EIv–– q0 x
L
The load is a downward triangular load of maximum
intensity q0.
Problem 9.2-2 The deflection curve for a simple beam AB (see figure)
is given by the following equation:
q0L4
x
v
sin L
4EI
(a) Describe the load acting on the beam.
(b) Determine the reactions RA and RB at the supports.
(c) Determine the maximum bending moment Mmax.
L
x
548
CHAPTER 9
Solution 9.2-2
Deflections of Beams
Simple beam
q0 L4
x
sin
L
4EI
3
q0 L
x
v¿ 3 cos
L
EI
2
q0 L
x
v– 2 sin
L
EI
q0 L
x
v‡ cos
EI
L
q0
x
v‡¿ sin
EI
L
(b) REACTIONS (EQ. 9-12b)
v
q0 L
x
cos
L
q0 L
V RA q0 L
q0 L
V RB ; RB V EIv‡ At x 0:
At x L:
(c) MAXIMUM BENDING MOMENT (EQ. 9-12a)
M EIv– (a) LOAD (EQ. 9-12c)
q0 L2
x
2 sin
L
q0 L2
L
For maximum moment, x ; Mmax 2
2
x
q EIv–– q0 sin
L
The load has the shape of a sine curve, acts
downward, and has maximum intensity q .
0
q0
L
Problem 9.2-3 The deflection curve for a cantilever beam AB
(see figure) is given by the following equation:
y
A
q0x 2
v
(10L3 10L2x 5Lx 2 x 3)
B
x
120LEI
L
Describe the load acting on the beam.
Probs. 9.2-3 and 9.2-4
Solution 9.2-3
Cantilever beam
v
q0
q0 x2
(10 L3 10 L2x 5 L x2 x3 )
120 LEI
Take four consecutive derivatives and obtain:
v–– q0
(L x)
LEI
L
From Eq. (9-12c):
q EIv–– q0 ¢ 1 x
≤
L
The load is a downward triangular load of maximum
intensity q .
0
SECTION 9.2
Differential Equations of the Deflection Curve
Problem 9.2-4 The deflection curve for a cantilever beam AB
(see figure) is given by the following equation:
q x2
360L EI
0
v (45L4 40L3x 15L2x 2 x 4)
2
(a) Describe the load acting on the beam.
(b) Determine the reactions RA and MA at the support.
Solution 9.2-4
v
Cantilever beam
q0 x2
(45L4 40 L3x 15 L2x2 x4 )
360 L2EI
q0
v¿ (15 L4x 20 L3x2 10 L2x3 x5 )
60 L2EI
q0
v– (3 L4 8 L3x 6 L2x2 x4 )
12 L2EI
q0
v‡ 2 (2 L3 3 L2x x3 )
3 L EI
v‡¿ q0
(L2 x2 )
L2EI
V EIv‡ At x 0:
q0
(2 L3 3 L2x x3 )
3L2
V RA M EIv– At x 0:
2q0 L
3
q0
(3 L4 8 L3x 6 L2x2 x4 )
12L2
M MA q0 L2
4
NOTE: Reaction RA is positive upward.
Reaction MA is positive clockwise (minus means
MA is counterclockwise).
(a) LOAD (EQ. 9-12c)
q EIv–– q0 ¢ 1 (b) REACTIONS RA AND MA (EQ. 9-12b AND EQ. 9-12a)
x2
≤
L2
The load is a downward parabolic load of maximum
intensity q .
0
q0
L
549
550
CHAPTER 9
Deflections of Beams
Deflection Formulas
Problems 9.3-1 through 9.3-7 require the calculation of deflections
using the formulas derived in Examples 9-1, 9-2, and 9-3. All beams
have constant flexural rigidity EI.
q
Problem 9.3-1 A wide-flange beam (W 12 35) supports a uniform
load on a simple span of length L 14 ft (see figure).
Calculate the maximum deflection max at the midpoint and the
angles of rotation at the supports if q 1.8 k/ft and E 30 106 psi.
Use the formulas of Example 9-1.
Solution 9.3-1 Simple beam (uniform load)
W 12 35
L 14 ft 168 in.
q 1.8 kft 150 lbin. E 30 106 psi
I 285 in.4
h
L
Probs. 9.3-1, 9.3-2 and 9.3-3
ANGLE OF ROTATION AT THE SUPPORTS
(EQs. 9-19 AND 9-20)
u uA uB MAXIMUM DEFLECTION (EQ. 9-18)
qL3
(150 lbin.)(168 in.) 3
24 EI 24(30 106 psi)(285 in.4 )
0.003466 rad 0.199º
5 qL4
5(150 lbin.)(168 in.) 4
max 384 EI 384(30 106 psi)(285 in.4 )
0.182 in.
Problem 9.3-2 A uniformly loaded steel wide-flange beam with simple
supports (see figure) has a downward deflection of 10 mm at the midpoint
and angles of rotation equal to 0.01 radians at the ends.
Calculate the height h of the beam if the maximum bending stress is
90 MPa and the modulus of elasticity is 200 GPa. (Hint: Use the formulas
of Example 9-1.)
Solution 9.3-2 Simple beam (uniform load)
A B 0.01 rad
max 10 mm
max 90 MPa
E 200 GPa
Calculate the height h of the beam.
Eq. (9-18): max 5 qL4
384 EI
or q 384 EI
5 L4
qL3
24 EIu
or q Eq. (9-19): u uA 24 EI
L3
Equate (1) and (2) and solve for L: L Mc Mh
Flexure formula: s I
2I
16 5u
(1)
(2)
(3)
Maximum bending moment:
qL2h
qL2
∴ s
M
8
16 I
16 Is
Solve Eq. (4) for h: h qL2
Substitute for q from (2) and for L from (3):
32s
h
15Eu2
Substitute numerical values:
h
32(90 MPa)(10 mm)
96 mm
15(200 GPa)(0.01 rad) 2
(4)
(5)
SECTION 9.3
551
Deflection Formulas
Problem 9.3-3 What is the span length L of a uniformly loaded simple
beam of wide-flange cross section (see figure) if the maximum bending
stress is 12,000 psi, the maximum deflection is 0.1 in., the height of
the beam is 12 in., and the modulus of elasticity is 30 106 psi?
(Use the formulas of Example 9-1.)
Solution 9.3-3 Simple beam (uniform load)
max 12,000 psi max 0.1 in.
h 12 in. E 30 106 psi
Solve Eq. (2) for q:
Calculate the span length L.
Equate (1) and (2) and solve for L:
Eq. (9-18): max Flexure formula: s 5qL4
384 EI
or q 384 EI
5L4
L
16 Is
L2h
24 Eh
B 5s
L2 (2)
Substitute numerical values:
24(30 106 psi)(12 in.)(0.1 in.)
L2 14,400 in.2
5(12,000 psi)
L 120 in. 10 ft
Problem 9.3-4 Calculate the maximum deflection max of a uniformly
loaded simple beam (see figure) if the span length L 2.0 m, the intensity
of the uniform load q 2.0 kN/m, and the maximum bending stress
60 MPa.
The cross section of the beam is square, and the material is aluminum
having modulus of elasticity E 70 GPa. (Use the formulas of Example 9-1.)
Solution 9.3-4 Simple beam (uniform load)
L 2.0 m q 2.0 kNm
E 70 GPa
max 60 MPa
Solve for b3: b3 b4
12
S
b3
6
L = 2.0 m
3qL2
4s
(4)
5Ls 4Ls 13
¢
≤
24E 3q
(The term in parentheses is nondimensional.)
5qL4
Maximum deflection (Eq. 9-18): 384 EI
5qL4
Substitute for I: 32 Eb4
Flexure formula with M Substitute for S: s q = 2.0 kN/m
Substitute b into Eq. (2): max CROSS SECTION (square; b width)
I
3qL2
4b3
qL2
:
8
s
(3)
(1)
Mc Mh
I
2I
Maximum bending moment:
qL2h
qL2
∴ s
M
8
16I
24 Eh
5s
q
Substitute numerical values:
(1)
5Ls 5(2.0 m)(60 MPa)
1
1
m
mm
24E
24(70 GPa)
2800
2.8
(2)
¢
M qL2
S
8S
4(2.0 m)(60 MPa) 13
4 Ls 13
≤ B
R 10(80) 13
3q
3(2000 Nm)
max (3)
10(80) 13
mm 15.4 mm
2.8
552
CHAPTER 9
Deflections of Beams
Problem 9.3-5 A cantilever beam with a uniform load (see figure)
has a height h equal to 1/8 of the length L. The beam is a steel wideflange section with E 28 106 psi and an allowable bending stress
of 17,500 psi in both tension and compression.
Calculate the ratio /L of the deflection at the free end to the length,
assuming that the beam carries the maximum allowable load. (Use the
formulas of Example 9-2.)
Solution 9.3-5
h 1
L 8
E 28 106 psi
17,500 psi
qL
8 EI
qL3
L 8EI
(1)
2
qL
Mc
¢
I
2
d
s L
¢ ≤
L 2E h
Substitute numerical values:
qL2
:
2
17,500 psi
1
(8) L 2(28 106 psi)
400
2
≤¢
qL h
h
≤
2I
4I
Problem 9.3-6 A gold-alloy microbeam attached to a silicon wafer
behaves like a cantilever beam subjected to a uniform load (see
figure). The beam has length L 27.5 m and rectangular cross
section of width b 4.0 m and thickness t 0.88 m. The total
load on the beam is 17.2 N.
If the deflection at the end of the beam is 2.46 m, what is
the modulus of elasticity Eg of the gold alloy? (Use the formulas
of Example 9-2.)
Solution 9.3-6
t
b
L
Substitute numerical values:
Eq 3(17.2 mN)(27.5 mm) 3
2(4.0 mm)(0.88 mm) 3 (2.46 mm)
80.02 109 Nm2 or Eq 80.0 GPa
Determine Eq.
bt3
12
q
Gold-alloy microbeam
Cantilever beam with a uniform load.
L 27.5 m b 4.0 m t 0.88 m
qL 17.2 N max 2.46 m
I
(3)
Substitute q from (3) into (2):
(2)
Flexure formula with M Eq. (9-26):
L
Solve for q:
4Is
q 2
Lh
4
Maximum deflection (Eq. 9-26): max s
h
Cantilever beam (uniform load)
Calculate the ratio L.
∴
q
Eq qL4
8 EqI
3 qL4
2 bt3max
or
Eq qL4
8 Imax
SECTION 9.3
Problem 9.3-7 Obtain a formula for the ratio C / max of the deflection
at the midpoint to the maximum deflection for a simple beam supporting
a concentrated load P (see figure).
From the formula, plot a graph of C / max versus the ratio a /L that
defines the position of the load (0.5 a /L 1). What conclusion do you
draw from the graph? (Use the formulas of Example 9-3.)
553
Deflection Formulas
P
A
B
a
b
L
Solution 9.3-7
Simple beam (concentrated load)
Pb(3L2 4b2 )
(a b)
48EI
2
2 32
Pb(L b )
(a b)
Eq. (9-34): max 93 LEI
c
(33L)(3L2 4b2 )
(a b)
max
16(L2 b2 ) 32
Eq. (9-35):
C GRAPH OF c max VERSUS aL
Because a
b, the ratio versus from 0.5 to 1.0.
Replace the distance b by the distance a by
substituting L a for b:
c
(33L)(L2 8ab 4a2 )
max
16(2aL a2 ) 32
Divide numerator and denominator by L2:
a
a2
(33L) ¢ 1 8 4 2 ≤
c
L
L
2 32
max
a a
16L ¢ 2 2 ≤
L L
a
a2
(33) ¢ 1 8 4 2 ≤
c
L
L
max
a a2 32
16 ¢ 2 2 ≤
L L
c
(33)(1 8b 4b2 )
max
16(2b b2 ) 32
c
max
0.5
0.6
0.7
0.8
0.9
1.0
1.0
0.996
0.988
0.981
0.976
0.974
NOTE: The deflection c at the midpoint of the beam
is almost as large as the maximum deflection max.
The greatest difference is only 2.6% and occurs when
the load reaches the end of the beam ( 1).
1.0
c
max
0.95
0.5
ALTERNATIVE FORM OF THE RATIO
a
Let b L
0.974
0.75
1.0
a
=
L
554
CHAPTER 9
Deflections of Beams
Deflections by Integration of the Bending-Moment Equation
Problems 9.3-8 through 9.3-16 are to be solved by integrating the second-order
differential equation of the deflection curve (the bending-moment equation).
The origin of coordinates is at the left-hand end of each beam, and all beams
have constant flexural rigidity EI.
Problem 9.3-8 Derive the equation of the deflection curve for a cantilever
beam AB supporting a load P at the free end (see figure). Also, determine
the deflection B and angle of rotation B at the free end. (Note: Use the
second-order differential equation of the deflection curve.)
Solution 9.3-8 Cantilever beam (concentrated load)
BENDING-MOMENT EQUATION (EQ. 9-12a)
EIv– M P(L x)
B.C.
y
P
A
B
x
L
v(0) 0 ∴ C1 0
Px2
(3L x)
6EI
Px
v¿ (2L x)
2EI
PL3
B v(L) 3EI
PL2
uB v¿(L) 2EI
(These results agree with Case 4, Table G-1.)
v
Px2
C1
2
B.C. v¿(0) 0 ∴ C2 0
PLx 2 Px3
EIv C2
2
6
EIv¿ PLx Problem 9.3-9 Derive the equation of the deflection curve
for a simple beam AB loaded by a couple M0 at the left-hand
support (see figure). Also, determine the maximum deflection
max. (Note: Use the second-order differential equation of the
deflection curve.)
y
M0
B
A
L
Solution 9.3-9
Simple beam (couple M0)
BENDING-MOMENT EQUATION (EQ. 9-12a)
x
EIv– M M0 ¢ 1 ≤
L
EIv¿ M0 ¢ x EIv M0 ¢
x2
≤ C1
2L
x2 x3
≤ C1x C2
2 6L
B.C.
v(0) 0 ∴ C2 0
B.C.
v(L) 0 ∴ C1 MAXIMUM DEFLECTION
M0
v¿ (2 L2 6 Lx 3 x2 )
6 LEI
Set v¿ 0 and solve for x:
3
x1 L ¢ 1 ≤
3
Substitute x1 into the equation for v:
max (v) xx1
M0L
3
M0x
v
(2L2 3Lx x2 )
6 LEI
M0 L2
93EI
(These results agree with Case 7, Table G-2.)
x
SECTION 9.3
555
Deflections by Integration of the Bending-Moment Equation
Problem 9.3-10 A cantilever beam AB supporting a triangularly
distributed load of maximum intensity q0 is shown in the figure.
Derive the equation of the deflection curve and then obtain
formulas for the deflection B and angle of rotation B at the free
end. (Note: Use the second-order differential equation of the
deflection curve.)
y
q0
x
B
A
L
Solution 9.3-10
Cantilever beam (triangular load)
BENDING-MOMENT EQUATION (EQ. 9-12a)
q0
EIv– M (L x) 3
6L
q0
EIv¿ (L x) 4 C1
24L
B.C.
v¿(0) 0 ∴ c2 EIv q0 L3
24
q0L3x
q0
(L x) 5 C2
120L
24
B.C.
v(0) 0 ∴ c2 q0 L4
120
q0 x2
(10 L3 10 L2x 5 Lx2 x3 )
120 LEI
q0 x
v¿ (4 L3 6 L2x 4 Lx2 x3 )
24 LEI
v
B v(L) q0 L4
30 EI
uB v¿(L) q0 L3
24 EI
(These results agree with Case 8, Table G-1.)
y
Problem 9.3-11 A cantilever beam AB is acted upon by a uniformly
distributed moment (bending moment, not torque) of intensity m
per unit distance along the axis of the beam (see figure).
Derive the equation of the deflection curve and then obtain
formulas for the deflection B and angle of rotation B at the free end.
(Note: Use the second-order differential equation of the deflection curve.)
m
B
A
x
L
Solution 9.3-11 Cantilever beam (distributed moment)
mx2
BENDING-MOMENT EQUATION (EQ. 9-12a)
v
(3L x)
6 EI
EIv– M m(L x)
mx
v¿ (2L x)
x2
2EI
EIv¿ m ¢ Lx ≤ C1
2
mL3
B v(L) B.C. v¿(0) 0 ∴ C1 0
3 EI
Lx2 x3
mL2
EIv m ¢
≤ C2
u
v¿(L)
B
2
6
2 EI
B.C.
v(0) 0 ∴ C2 0
Problem 9.3-12 The beam shown in the figure has a roller support at A
and a guided support at B. The guided support permits vertical movement
but no rotation.
Derive the equation of the deflection curve and determine the
deflection B at end B due to the uniform load of intensity q. (Note:
Use the second-order differential equation of the deflection curve.)
y
q
A
B
L
x
556
CHAPTER 9
Deflections of Beams
Solution 9.3-12 Beam with a guided support
REACTIONS AND DEFLECTION CURVE
y
BENDING-MOMENT EQUATION (EQ. 9-12a)
EIv– M qLx q
qL2
MB =
2
x
A
B
EIv¿ B.C.
L
EIv y
A
x
B
B
qLx2 qx3
C1
2
6
v(L) 0
RA = qL
qx2
2
∴ C1 qL3
3
qLx3 qx4 qL3x
C2
6
24
3
v(0) 0 C2 0
qx
v (8L3 4Lx2 x3 )
24 EI
B.C.
B v(L) 5 qL4
24 EI
y
Problem 9.3-13 Derive the equations of the deflection curve
for a simple beam AB loaded by a couple M0 acting at distance
a from the left-hand support (see figure). Also, determine the
deflection 0 at the point where the load is applied. (Note: Use
the second-order differential equation of the deflection curve.)
M0
B
A
a
b
L
Solution 9.3-13
Simple beam (couple M0)
BENDING-MOMENT EQUATION (EQ. 9-12a)
EIv– M M0 x
L
M0 x2
EIv¿ C1
2L
(0
x
B.C.
(0
x
a)
4 (v)Left (v)Right
∴ C4 M0 L ¢ a L
≤ C1L
3
at x a
(a
M0
x2
¢ Lx ≤ C2
L
2
x
(a
L)
x
M0 x3
C1x C3
6L
(0
x
M0 a
2
M0
(2L2 6aL 3a2 )
6L
M0 x
v
(6aL 3a2 2L2 x2 )
6 LEI
C1 L)
(0
x
a)
M0
(3a2L 3a2x 2L2x 3Lx2 x3 )
6 LEI
(a x L)
M0 a(L a)(2a L)
0 v(a) 3 LEI
M0 ab(2a L)
3LEI
v
a)
2 v(0) 0 C3 0
M0 x2 M0 x3
C1x M0 ax C4
EIv 2
6L
(a x L)
B.C.
B.C.
∴ C4 1 (v¿ )Left (v¿ )Right at x a
C2 C1 M0a
EIv 3 v(L) 0
2
M0
EIv– M (L x)
L
EIv¿ a)
B.C.
NOTE: 0 is positive downward. The pending results
agree with Case 9, Table G-2.
x
SECTION 9.3
557
Deflection by Integration of the Bending-Moment Equation
Problem 9.3-14 Derive the equations of the deflection curve
for a cantilever beam AB carrying a uniform load of intensity
q over part of the span (see figure). Also, determine the
deflection B at the end of the beam. (Note: Use the
second-order differential equation of the deflection curve.)
y
q
x
B
A
a
b
L
Solution 9.3-14
Cantilever beam (partial uniform load)
BENDING-MOMENT EQUATION (EQ. 9-12a)
q
q
EIv– M (a x) 2 (a2 2ax x2 )
2
2
(0 x a)
q 2
x3
2
EIv¿ ¢ a x ax ≤ C1 (0 x a)
2
3
1 v¿ (0) 0
EIv– M 0
EIv¿ C2
B.C.
B.C.
C1 0
(a
(a
x
x
L)
L)
2 (v¿ )Left (v¿ )Right at x a
∴ C2 EIv qa3
6
q a2x2 ax3 x4
¢
≤ C3 (0
2 2
3
12
C3 0
qa3x
EIv C2 x C4 C4 (a
6
B.C. 4 (v)Left (v)Right at x a
B.C.
3 v(0) 0
∴ C4 x
L)
qa4
24
v
qx2
(6a2 4ax x2 )
24 EI
(0
v
qa3
(4x a) (a
24 EI
L)
x
x
a)
qa3
(4L a)
24 EI
(These results agree with Case 2, Table G-1.)
B v(L) x
a)
Problem 9.3-15 Derive the equations of the deflection curve for a
cantilever beam AB supporting a uniform load of intensity q acting
over one-half of the length (see figure). Also, obtain formulas for the
deflections B and C at points B and C, respectively. (Note: Use
the second-order differential equation of the deflection curve.)
y
q
A
B
C
L
—
2
Solution 9.3-15 Cantilever beam (partial uniform load)
BENDING-MOMENT EQUATION (EQ. 9-12a)
B.C. 1 v¿ (0) 0 C1 0
q
qL
L
EIv– M (L2 2Lx x2 )
EIv– M (3L 4x) ¢ 0 x
≤
2
8
2
q
x3
qL
L
EIv¿ ¢ L2x Lx2 ≤ C2
2
EIv¿ (3Lx 2x ) C1 ¢ 0 x
≤
2
3
8
2
L
—
2
¢
L
2
¢
L
2
x
x
L≤
L≤
x
558
CHAPTER 9
B.C.
L
2 (v¿ )Left (v¿ )Right at x 2
qL3
∴ C2 48
EIv B.C.
Deflections of Beams
qL 3Lx2 2x3
¢
≤ C3
8
2
3
4 (v)Left (v)Right at x ∴ C4 ¢0
L
≤
2
x
qL4
384
L
2
qLx2
L
(9L 4x) ¢ 0 x
≤
48 EI
2
7qL4
L
C v ¢ ≤ 2
192EI
q
v
(16x4 64 Lx3 96 L2x2 8 L3x L4 )
384 EI
L
¢
x L≤
2
v
C3 0
3 v(0) 0
EIv B.C.
q L2x2 Lx3 x4
qL3
¢
≤
x C4
2 2
3
12
48
L
¢
x L≤
2
B v(L) 41qL4
384EI
y
Problem 9.3-16 Derive the equations of the deflection curve
for a simple beam AB with a uniform load of intensity q acting
over the left-hand half of the span (see figure). Also, determine
the deflection C at the midpoint of the beam. (Note: Use the
second-order differential equation of the deflection curve.)
q
B
A
C
L
—
2
Solution 9.3-16
Simple beam (partial uniform load)
2 v(0) 0 C3 0
qL2x2 qLx3
qL3x
EIv C1x C4
16
48
48
L
¢
x L≤
2
BENDING-MOMENT EQUATION (EQ. 9-12a)
2
EIv– M EIv¿ 3qLx qx
8
2
¢0
3qLx2 qx3
C1
16
6
2
EIv¿ B.C.
qL
qLx
8
8
¢
L
2
2
qL x qLx
C2
8
16
¢
2
EIv L
≤
2
x
L≤
3 v(L) 0
B.C.
4 (v)Left (v)Right at x x
L
2
3qL3
128
qx
v
(9L3 24Lx2 16x3 ) ¢ 0
384EI
L
2
¢0
qL4
48
∴ C1 qL
48
qLx3 qx4
C1x C3
16
24
∴ C4 C1L B.C.
L≤
x
1 (v¿ )Left (v¿ )Right at x ∴ C2 C1 x
L
2
B.C.
L
≤
2
x
¢0
2
EIv– M L
—
2
L
≤
2
v
x
qL
(8x3 24Lx2 17L2x L3 )
384EI
L
¢
x L≤
2
5qL4
L
≤
2
768EI
(These results agree with Case 2, Table G-2.)
C v ¢
L
≤
2
x