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•12–1. An A-36 steel strap having a thickness of 10 mm and
a width of 20 mm is bent into a circular arc of radius r = 10 m.
Determine the maximum bending stress in the strap.

Moment-Curvature Relationship:
M
1
=
r
EI

however,

M =

I
s
c

1

1
c s
=
r
EI

s =

0.005
c
E = a
b C 200 A 109 B D = 100 MPa
r
10

12–2. A picture is taken of a man performing a pole vault,
and the minimum radius of curvature of the pole is
estimated by measurement to be 4.5 m. If the pole is 40 mm
in diameter and it is made of a glass-reinforced plastic for
which Eg = 131 GPa, determine the maximum bending
stress in the pole.

r ϭ 4.5 m

Moment-Curvature Relationship:
M
1
=
r
EI


however,

M =

I
s
c

I
1
c s
=
r
EI

s =

0.02
c
E = a
b C 131 A 109 B D = 582 MPa
r
4.5

Ans.

883



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12–3. When the diver stands at end C of the diving board,
it deflects downward 3.5 in. Determine the weight of the
diver. The board is made of material having a modulus of
elasticity of E = 1.5(103) ksi.

B

A

3.5 in.

2 in.

C
9 ft

3 ft

Support Reactions and Elastic Curve. As shown in Fig. a.

Moment Functions. Referring to the free-body diagrams of the diving board’s cut
segments, Fig. b, M A x1 B is
a + ©MO = 0;
and M A x2 B is
a + ©MO = 0;

M A x1 B + 3Wx1 = 0

M A x1 B = -3Wx1

-M A x2 B - Wx2 = 0

M A x2 B = -Wx2

Equations of Slope and Elastic Curve.
EI

d2v
= M(x)
dx2

For coordinate x1,
EI

d2v1
dx1 2

= -3Wx1

d2v1

3
= - Wx1 2 + C1
dx1
2

(1)

1
EIv1 = - Wx1 3 + C1x1 + C2
2

(2)

EI

For coordinate x2
EI

EI

d2v2
dx2 2

= -Wx2

dv2
1
= - Wx2 2 + C3
dx2
2


EIv2 = -

(3)

1
Wx2 3 + C3x2 + C4
6

(4)

Boundary Conditions. At x1 = 0, v1 = 0. Then, Eq. (2) gives
1
EI(0) = - W A 03 B + C1(0) + C2
2

C2 = 0

At x1 = 3 ft, v1 = 0. Then, Eq. (2) gives
1
EI(0) = - W A 33 B + C1(3) + 0
2

C1 = 4.5W

884

18 in.



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12-3.

Continued

At x2 = 9 ft, v2 = 0. Then, Eq. (4) gives
1
EI(0) = - W A 93 B + C3(9) + C4
6
9C3 + C4 = 121.5W
Continuity Conditions. At x1 = 3 ft and x2 = 9 ft,

(5)
dv2
dv1
. Thus, Eqs. (1) and
= dx1
dx2

(3) give

1
3
- W A 32 B + 4.5W = - c - W A 92 B + C3 d
2
2

C3 = 49.5W

Substituting the value of C3 into Eq. (5),
C4 = -324W
Substituting the values of C3 and C4 into Eq. (4),
v2 =

1
1
a - Wx2 3 + 49.5Wx2 - 324Wb
EI
6

At x2 = 0, v2 = -3.5 in. Then,
-324W(1728)
-3.5 =

1.5 A 106 B c

1
(18) A 2 3 B d
12

W = 112.53 lb = 113 lb


Ans.

885


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*12–4. Determine the equations of the elastic curve using
the x1 and x2 coordinates. EI is constant.

P
A

EI

d2v1
dx1 2

= M1 (x)


M1(x) = 0;

EI

EI

d v1
dx1

2

L

= 0

x3

dv1
= C1
dx1

(1)

EI v1 = C1x1 + C2

(2)

M2(x) = Px2 - P(L - a)
EI


EI

d2 v2
dx2 2

= Px2 - P(L - a)

dv2
P 2
=
x - P(L - a)x2 + C3
dx2
2 2

EI v2 =

(3)

P(L - a)x22
P 3
x2 + C3x2 + C4
6
2

(4)

Boundary conditions:
At x2 = 0,

dv2

= 0
dx2

From Eq. (3), 0 = C3
At x2 = 0, v2 = 0
0 = C4
Continuity condition:
At x1 = a, x2 = L - a;

dv1
dv2
= dx1
dx2

From Eqs. (1) and (3),
C1 = - c

P(L - a)2
- P(L - a)2 d ;
2

C1 =

P(L - a)2
2

At x1 = a, x2 = L - a, v1 = v2
From Eqs. (2) and (4),
a


P(L - a)3
P(L - a)3
P(L - a)2
b a + C2 =
2
6
2

C2 = -

Pa(L - a)2
P(L - a)3
2
3

From Eq. (2),
v1 =

P
[3(L - a)2x1 - 3a(L - a)2 - 2(L - a)3]
6EI

Ans.

For Eq. (4),
v2 =

B

x1

2

P
[x2 - 3(L - a)x33]
6EI 2

Ans.

886

L
2


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•12–5.

Determine the equations of the elastic curve for
the beam using the x1 and x2 coordinates. EI is constant.


P
A

B

x1
L

Moment Functions. Referring to the FBDs of the beam’s cut segments shown in
Fig. b and c,
1
P(x1) = 0
2

a + ©MO = 0;

M(x1) +

a + ©MO = 0;

-Px2 - M(x2) = 0

M(x1) = -

P
x
2 1

And


EI

M(x2) = -Px2

d2v
= M(x)
dx2

For coordinate x1,
EI

EI

d2v1
dx1 2

= -

P
x
2 1

dv1
P
= - x1 2 + C1
dx1
4

EI v1 = -


(1)

P 3
x + C1x + C2
12 1

(2)

For coordinate x2,
EI

EI

d2v2
dx2 2

= -Px2

dv2
P
= - x2 2 + C3
dx2
2

EI v2 = -

(3)

P 3
x + C3x2 + C4

6 2

(4)

At x1 = 0, v1 = 0. Then, Eq (2) gives
EI(0) = -

P
(0) + C1(0) + C2
12

C2 = 0

At x1 = L, v1 = 0. Then, Eq (2) gives
EI(0) = At x2 =

P
(L3) + C1L + 0
12

C1 =

PL2
12

L
, v2 = 0. Then Eq (4) gives
2
EI(0) = -


P L 3
L
a b + C3 a b + C4
6 2
2

C3L + 2C4 =

PL3
24

(5)

887

x2
L
2


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•12–5.

Continued

At x1 = L and x2 =

-

dv2
L dv1
,
= . Thus, Eqs. (1) and (3) gives
2 dx1
dx2

P 2
P L 2
PL2
= - c - a b + C3 d
AL B +
4
12
2 2
C3 =

7PL2
24

Substitute the result of C3 into Eq. (5)

C4 = -

PL3
8

Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq (4),
v1 =

P
A -x1 3 + L2x1 B
12EI

Ans.

v2 =

P
A -4x2 3 + 7L2x2 - 3L3 B
24EI

Ans.

888


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11:52 AM


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12–6. Determine the equations of the elastic curve for the
beam using the x1 and x3 coordinates. Specify the beam’s
maximum deflection. EI is constant.

P
A

Support Reactions and Elastic Curve: As shown on FBD(a).
L

Slope and Elastic Curve:

For M(x1) = -

x3

d2v
= M(x)
dx2

EI

P
x.

2 1
EI

d2y1
dx21

EI y1 = For M(x3) = Px3 -

= -

P
x
2 1

dy1
P
= - x21 + C1
dx1
4

EI

[1]

P 3
x + C1x1 + C2
12 1

[2]


3PL
.
2
EI

d2y3
dx23

= Px3 -

3PL
2

dy3
P 2
3PL
=
x3 x3 + C3
dx3
2
2

EI

EI y3 =

[3]

P 3
3PL 3

x x3 + C3x3 + C4
6 3
4

[4]

Boundary Conditions:
y1 = 0 at x1 = 0. From Eq. [2], C2 = 0
y1 = 0 at x1 = L. From Eq. [2].
0 = -

PL3
+ C1L
12

C1 =

PL2
12

y3 = 0 at x3 = L. From Eq. [4].
0 =

PL3
3PL3
+ C3L + C4
6
4

0 = -


7PL3
+ C3L + C4
12

[5]

Continuity Condition:
At x1 = x3 = L,

-

dy1
dy3
. From Eqs. [1] and [3],
=
dx1
dx3

PL2
PL2
PL2
3PL2
+
=
+ C3
4
12
2
2


From Eq. [5], C4 = -

B

x1

Moment Function: As shown on FBD(b) and (c).

C3 =

5PL2
6

PL3
4

889

L
2


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12–6.

Continued

The Slope: Substitute the value of C1 into Eq. [1],
dy1
P
=
A L2 - 3x21 B
dx1
12EI
dy1
P
= 0 =
A L2 - 3x21 B
dx1
12EI

x1 =

L
23

The Elastic Curve: Substitute the values of C1, C2, C3, and C4 into Eqs. [2] and [4],
respectively.
y1 =


Px1
A -x21 + L2 B
12EI

yO = y1 |x1 =

y3 =

L
23

=

PA

L
23

B

12EI

Ans.
a-

0.0321PL3
L3
+ L2 b =
3

EI

P
A 2x33 - 9Lx23 + 10L2x3 - 3L3 B
12EI

Ans.

yC = y3 |x3 = 32 L
=

2
P
3 3
3
3
c2 a L b - 9La Lb + 10L2 a L b - 3L3 d
12EI
2
2
2

= -

PL3
8EI

Hence,
ymax =


PL3
8EI

Ans.

890


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12–7. The beam is made of two rods and is subjected to
the concentrated load P. Determine the maximum
deflection of the beam if the moments of inertia of the rods
are IAB and IBC , and the modulus of elasticity is E.
EI

P
B
A

C

l

d2y
= M(x)
dx2

L

M1(x) = - Px1
EIBC

EIBC

d2y1
dx1 2

= - Px1

dy1
Px21
= + C1
dx1
2

EIBC y1 = -

(1)

Px31
+ C1x1 + C2

6

(2)

M2(x) = - Px2
EIAB

EIAB

d2y2
dx2 2

= - Px2

dy2
P
= - x2 2 + C3
dx2
2

EIAB y2 = -

(3)

P 3
x + C3x2 + C4
2 2

(4)


Boundary conditions:
At x2 = L,

0 = -

dy2
= 0
dx2

PL2
+ C3;
2

C3 =

PL2
2

At x2 = L, y = 0
0 = -

PL3
PL3
+
+ C4;
6
2

C4 = -


PL3
3

Continuity Conditions:
At x1 = x2 = l,

dy1
dy2
=
dx1
dx2

891


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12–7.

Continued


From Eqs. (1) and (3),
1
PI 2
1
PI 2
PL2
cc+ C1 d =
+
d
EIBC
2
EIAB
2
2
C1 =

IBC
PL2
Pl2
Pl2
c+
d +
IAB
2
2
2

At x1 = x2 = l, y1 = y2
From Eqs. (2) and (4),
IBC

PL2
Pl2
1
Pl3
Pl2
e+ c
a+
b +
dl + C2 f
EIBC
6
IAB
2
2
2
=

1
PL2l
PL3
Pl3
c+
d
EIAB
6
2
3

C2 =


IBC PL3
IBC Pl3
Pl3
IAB 3
IAB 3
3

Therefore,
y1 =

Px1 3
IBC
1
Pl2
PL2
Pl2
e+ c
a+
b +
dx1
EIBC
6
IAB
2
2
2
+

IBC PL3
IBC Pl3

Pl3
f
IAB 3
IAB 3
3

At x1 = 0, y1 |x = 0 = ymax
ymax =

=

IBC Pl3
IBC PL3
IAB 3
I
Pl3
P
e
f =
e l3 - L3 - a
bl f
EIBC IAB 3
IAB 3
3
3EIAB
IBC
IAB 3
P
e a1 b l - L3 f
3EIAB

IBC

Ans.

892


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*12–8. Determine the equations of the elastic curve for
the beam using the x1 and x2 coordinates. EI is constant.

P

Referring to the FBDs of the beam’s cut segments shown in Fig. b and c,
a + ©MO = 0;

M(x1) +

PL
- Px1 = 0

2

M(x1) = Px1 -

PL
2

x1
x2

And
a + ©MO = 0;
EI

L
2

M(x2) = 0

d2v
= M(x)
dx2

For coordinate x1,
EI

EI

d2v1
dx21


= Px1 -

PL
2

dv1
P 2
PL
=
x x + C1
dx1
2 1
2 1

EI v1 =

(1)

P 3
PL 2
x x + C1x1 + C2
6 1
4 1

(2)

For coordinate x2,
EI


EI

d2v2
dx22

= 0

dv2
= C3
dx2

(3)

EI v2 = C3x2 = C4
At x1 = 0,

(4)

dv1
= 0. Then, Eq.(1) gives
dx1

EI(0) =

PL
P 2
(0 ) (0) + C1
2
2


C1 = 0

At x1 = 0, v1 = 0. Then, Eq(2) gives
EI(0) =
At x1 = x2 =

PL 2
P 3
(0 ) (0 ) + 0 + C2
6
4

C2 = 0

dv2
L dv1
=
,
. Thus, Eqs.(1) and (3) gives
2 dx1
dx2

P L 2
PL L
a b a b = C3
2 2
2
2
Also, at x1 = x2 =


C3 = -

PL2
8

L
, v = v2. Thus, Eqs, (2) and (4) gives
2 1

PL L 2
PL2 L
P L 3
a b a b = ab a b + C4
6 2
4 2
8
2

C4 =

PL3
48

Substitute the values of C1 and C2 into Eq. (2) and C3 and C4 into Eq (4),
v1 =

P
A 2x31 - 3Lx21 B
12EI


Ans.

v2 =

PL2
(-6x2 + L)
48EI

Ans.

893

L
2


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•12–9.

Determine the equations of the elastic curve using

the x1 and x2 coordinates. EI is constant.
EI

d2y
= M(x)
dx2

M1 =

EI

EI

P

A

x1

Pb
x
L 1

d2y1
dx21

=

a


EI y3 =

L

(1)

Pb 3
x + C3x1 + C2
6L 1

(2)

Pb
x - P(x2 - a)
L 2

But b = L - a. Thus
M2 = Paa1 -

EI

EI

d2y2
dx2 2

x2
b
L


= Pa a1 -

x2
b
L

dy2
x22
= Paax2 b + C3
dx2
2L

EI y2 = Pa a

(3)

x22
x22
b + C3x2 + C4
2
6L

(4)

Applying the boundary conditions:
y1 = 0 at x1 = 0
Therefore,C2 = 0,
y2 = 0 at x2 = L
0 =


b

x2

Pb
x
L 1

dy1
Pb 2
=
x + C1
dx1
2L 1

M2 =

B

Pa L2
+ C3L + C4
3

(5)

Applying the continuity conditions:
y1 |x1 = a = y2 |x2 = a
Pb 3
a2
a3

a + C1a = Pa a
b + C3a + C4
6L
2
6L

(6)

dy1
dy2
2
2
=
dx1 x1 = a
dx2 x2 = a
a2
Pb 2
a + C1 = Pa a a b + C3
2L
2L

(7)

894


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•12–9.

Continued

Solving Eqs. (5), (6) and (7) simultaneously yields,
C1 = C4 =

Pb 2
A L - b2 B ;
6L

C3 = -

Pa
A 2L2 + a2 B
6L

Pa3
6

Thus,
EIy1 =


Pb 3
Pb 2
x A L - b 2 B x1
6L 1
6L

or
v1 =

Pb
A x3 - A L2 - b2 B x1 B
6EIL 1

Ans.

and
EIy2 = Pa a
y2 =

x22
x32
Pa
Pa3
b A 2L2 + a2 B x2 +
2
6L
6L
6

Pa

C 3x22 L - x32 - A 2L2 + a2 B x2 + a2L D
6EIL

Ans.

895


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12–10. Determine the maximum slope and maximum
deflection of the simply supported beam which is subjected
to the couple moment M0 . EI is constant.

M0

A

B
L


Support Reactions and Elastic Curve: As shown on FBD(a).
Moment Function: As shown on FBD(b).
Slope and Elastic Curve:
EI

d2y
= M(x)
dx2

EI

M0
d2y
=
x
2
L
dx

EI

M0 2
dy
=
x + C1
dx
2L

EI y =


[1]

M0 3
x + C1x + C2
6L

[2]

Boundary Conditions:
y = 0 at x = 0. From Eq. [2].
0 = 0 + 0 + C2

C2 = 0

y = 0 at x = L. From Eq. [2].
0 =

M0 3
A L B + C1 (L)
6L

C1 = -

M0L
6

The Slope: Substitute the value of C1 into Eq. [1],
M0
dy
=

A 3x2 - L2 B
dx
6LEI
M0
dy
= 0 =
A 3x2 - L2 B
dx
6LEI
uB =

x =

23
L
3

M0L
dy
2
= dx x = 0
6EI

umax = uA =

M0L
dy 2
=
dx x = L
3EI


Ans.

The Elastic Curve: Substituting the values of C1 and C2 into Eq. [2],
y =

ymax occurs at x =

M0
A x3 - L2x B
6LEI

23
L,
3
ymax = -

23M0L2
Ans
27EI

The negative sign indicates downward displacement.

896


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12–11. Determine the equations of the elastic curve for the
beam using the x1 and x2 coordinates. Specify the beam’s
maximum deflection. EI is constant.

P

Referring to the FBDs of the beam’s cut segments shown in Fig. b and c,
2
M(x1) - Px1 = 0
3

a + ©M0 = 0;

A
B

x1

2P
x
M(x1) =
3 1


a

2a
x2

And
1
P
P(3a - x2) - M(x2) = 0 M(x2) = Pa x
3
3 2

a + ©M0 = 0;
EI

d2y
= M(x)
dx2

For coordinate x1,
EI

EI

d2y1
dx21

=

2P

x
3 1

dy1
P 2
=
x + C1
dx1
3 1

EI y1 =

(1)

P 3
x = C1x1 + C2
9 1

(2)

For coordinate x2,
EI

EI

d2y2
dx2

2


= Pa -

P
x
3 2

dy2
P 2
= Pax2 x + C3
dx2
6 2

EI y2 =

(3)

Pa 2
P 3
x x + C3x2 + C4
2 2
18 2

(4)

At x1 = 0, y1 = 0. Then, Eq (2) gives
EI(0) =

P 3
A 0 B + C1(0) + C2
9


C2 = 0

At x2 = 3a, y2 = 0. Then Eq (4) gives
EI(0) =

Pa
P
(3a)2 (3a)3 + C3(3a) + C4
2
18
C3(E2
3E1I1
8E1I1

TAC =

B

A

TAC L31
3E1I1

(+ T)

w

3wA 2E2L41


8 C 3E1I1L2 + A 2E2L31 D

Ans.

1021


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Page 1022

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–130. Determine the reactions at A and B. Assume the
support at A only exerts a moment on the beam. EI is
constant.

P

A
2

(uA)1 =

PL

;
8EI

(uA)2 =

B

MAL
EI

L
–
2

By superposition:

L
–
2

0 = (uA)1 - (uA)2
0 =

MAL
PL2
8EI
EI

MA =


PL
8

Ans.

Equilibrium:
a + ©MB = 0;
MB =

-

PL
PL
+
- MB = 0
8
2

3PL
8

Ans.

+ ©F = 0 ;
:
x

Bx = 0

Ans.


+ c ©Fy = 0 ;

By = P

Ans.

12–131. The beam is supported by the bolted supports at its
ends. When loaded these supports do not provide an actual
fixed connection, but instead allow a slight rotation a before
becoming fixed. Determine the moment at the connections
and the maximum deflection of the beam.

P

u - u¿ = a
L
—
2

ML
ML
PL2
= a
16EI
3EI
6EI
ML = a
M = a


PL2
- a b(2EI)
16EI

2EI
PL
ab
8
L

¢ max = ¢ - ¢ ¿ =

Ans.

M(L2 )
PL3
- 2c
C L2 - (L>2)2 D d
48EI
6EIL

¢ max =

PL3
L2 PL
2EIa
a
b
48EI
8EI

8
L

¢ max =

aL
PL3
+
192EI
4

Ans.

1022

L
—
2


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


*12–132. The beam is supported by a pin at A, a spring
having a stiffness k at B, and a roller at C. Determine the
force the spring exerts on the beam. EI is constant.

w

A
B

Method of Superposition: Using the table in appendix C, the required
displacements are
5w(2L)4
5wL4AC
5wL4
=
=
384EI
384EI
24EI

T

Fsp (2L)3
Fsp L3
PL3AC
=
=
yB – =
48EI

48EI
6EI

c

yB ¿ =

Using the spring formula, ysp =

Fsp
k

L

C
k
L

.

The compatibility condition requires
ysp = yB ¿ + yB –

(+ T)

3

Fsp
k


=

Fsp =

Fsp L
5wL4
+ ab
24EI
6EI
5wkL4

4 A 6EI + kL3 B

Ans.

•12–133.

The beam is made from a soft linear elastic
material having a constant EI. If it is originally a distance
¢ from the surface of its end support, determine the
distance a at which it rests on this support when it is
subjected to the uniform load w0 , which is great enough to
cause this to happen.

w0
⌬

a
L


The curvature of the beam in region BC is zero, therefore there is no bending
moment in the region BC, The reaction F is at B where it touches the support. The
slope is zero at this point and the deflection is ¢ where
¢ =

R(L - a)3
w0(L - a)4
8EI
3EI

u1

=

w0(L - a)3
R(L - a)2
6EI
2EI

Thus,
1

8¢EI 4
R = a
b
9w30

Ans.
1


72¢EI 4
L - a = a
b
w0
1

72¢EI 4
a = L - a
b
w0

Ans.

1023


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Page 1024

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–134. Before the uniform distributed load is applied on
the beam, there is a small gap of 0.2 mm between the beam
and the post at B. Determine the support reactions at A, B,

and C. The post at B has a diameter of 40 mm, and the
moment of inertia of the beam is I = 875(106) mm4. The
post and the beam are made of material having a modulus
of elasticity of E = 200 GPa .

30 kN/m

A

6m

Equations of Equilibrium. Referring to the free-body diagram of the beam, Fig. a,
+ ©F = 0;
:
x

Ax = 0

+ c ©Fy = 0;

A y + FB + Cy - 30(12) = 0

a + ©MA = 0;

Ans.
(1)

FB(6) + Cy(12) - 30(12)(6) = 0

(2)


Method of superposition: Referring to Fig. b and the table in the Appendix, the
necessary deflections are

(vB)1 =

(vB)2 =

5(30) A 12 4 B
8100kN # m3
5wL4
=
=
T
384EI
384EI
EI
FB A 12 3 B
36FB
PL3
=
=
48EI
48EI
EI

c

The deflection of point B is
vB = 0.2 A 10 - 3 B +


FB(a)
FBLB
= 0.2 A 10 - 3 B +
AE
AE

T

The compatibility condition at support B requires

A+TB

vB = (vB)1 + (vB)2
0.2 A 10 - 3 B +

FB (1)
36FB
8100
=
+ ab
AE
EI
EI

0.2 A 10 - 3 B E +
FB

p
A 0.04 2 B

4

+

FB
36FB
8100
=
A
I
I
36FB

875 A 10 - 6 B

8100

=

875 A 10 - 6 B

-

C
1m

0.2 A 10 - 3 B C 200 A 109 B D
1000

FB = 219.78 kN = 220 kN


Ans.

Substituting the result of FB into Eqs. (1) and (2),
A y = Cy = 70.11 kN = 70.1 kN

Ans.

1024

B

0.2 mm
6m


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–135. The 1-in.-diameter A-36 steel shaft is supported by
unyielding bearings at A and C. The bearing at B rests on a
simply supported steel wide-flange beam having a moment

of inertia of I = 500 in4. If the belt loads on the pulley are
400 lb each, determine the vertical reactions at A, B, and C.

3 ft
5 ft

A

2 ft
5 ft

B

For the shaft:
(¢ b)1 =

(¢ b)2 =

400
lb

800(3)(5)
13200
A -52 - 32 + 102 B =
6EIs(10)
EIs
By A 103 B
48EIs

C


5 ft

20.833By
=

EIs

For the beam:

¢b =

By A 103 B
48EIb

20.833By
=

EIb

Compatibility condition:
+ T ¢ b = (¢ b)1 - (¢ b)2
20.833By
EIb
Is =

=

20.833By
13200

EIs
EIs

p
(0.5)4 = 0.04909 in4
4

20.833By (0.04909)
500

400
lb

= 13200 - 20.833By

By = 634 lb

Ans.

Form the free-body digram,
A y = 243 lb

Ans.

Cy = 76.8 lb

Ans.

1025



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Page 1026

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*12–136. If the temperature of the 75-mm-diameter post
CD is increased by 60°C, determine the force developed in
the post. The post and the beam are made of A-36 steel, and
the moment of inertia of the beam is I = 255(106) mm4.

3m

3m

A
B

C
3m

D

Method of Superposition. Referring to Fig. a and the table in the Appendix, the

necessary deflections are
(vC)1 =

FCD A 33 B
9FCD
PLBC 3
c
=
=
3EI
3EI
EI

(vC)2 = (uB)2LBC =

3FCD (3)
9FCD
MOLAB
c
(LBC) =
(3) =
3EI
3EI
EI

The compatibility condition at end C requires

A+cB

vC = (vC)1 + (vC)2

=

9FCD
9FCD
18FCD
c
+
=
EI
EI
EI

Referring to Fig. b, the compatibility condition of post CD requires that
dFCD + vC = dT
dFCD =

(1)

FCD (3)
FCD LCD
=
AE
AE

dT = a¢TL = 12 A 10 - 6 B (60)(3) = 2.16 A 10 - 3 B m
Thus, Eq. (1) becomes
3FCD
18FCD
+
= 2.16 A 10 - 3 B

AE
EI
3FCD

p
A 0.0752 B
4

+

18FCD

255 A 10 - 6 B

= 2.16 A 10 - 3 B C 200 A 109 B D

FCD = 6061.69N = 6.06 kN

Ans.

1026


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Page 1027


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

•12–137.

The shaft supports the two pulley loads shown.
Using discontinuity functions, determine the equation of
the elastic curve. The bearings at A and B exert only vertical
reactions on the shaft. EI is constant.

x
A

12 in.

B

12 in.
70 lb

180 lb

M = -180 6 x - 0 7 - (-277.5) 6 x - 12 7 - 70 6 x - 24 7
M = -180x + 277.5 6 x - 12 7 - 70 6 x - 24 7
Elastic curve and slope:
EI

d2v
= M = -180x + 277.5 6 x - 12 7 - 70 6 x - 24 7

dx2

EI

dv
= -90x2 + 138.75 6 x - 12 7
dx

EIv = -30x3 + 46.25 6 x - 12 7

3

2

- 35(x - 24 7

2

- 11.67 6 x - 24 7

+ C1
3

+ C1x + C2 (1)

Boundary conditions:
v = 0

at


x = 12 in,

From Eq. (1)
0 = -51,840 + 12C1 + C2
12C1 + C2 = 51 840
v = 0

at

(2)

x = 60 in.

From Eq.(1)
0 = -6 480 000 + 5 114 880 - 544 320 + 60C1 + C2
60C1 + C2 = 1909440

(3)

Solving Eqs. (2) and (3) yields:
C1 = 38 700
v =

C2 = -412 560

1
[-30x3 + 46.25 6 x - 12 7
EI

3


- 11.7 6 x - 24 7

3

+ 38 700x - 412 560]

Ans.

1027

36 in.


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12–138. The shaft is supported by a journal bearing at A,
which exerts only vertical reactions on the shaft, and by a
thrust bearing at B, which exerts both horizontal and
vertical reactions on the shaft. Draw the bending-moment
diagram for the shaft and then, from this diagram, sketch

the deflection or elastic curve for the shaft’s centerline.
Determine the equations of the elastic curve using the
coordinates x1 and x2 . EI is constant.

80 lb
A

x1

EI

d2v1
dx21

= 26.67x1

dv1
= 13.33x21 + C1
dx1

(1)

EIv1 = 4.44x31 + C1x1 + C2

(2)

EI

For M2 (x) = -26.67x2
EI


d2v2
dx22

= -26.67x2

dv2
= -13.33x22 + C3
dx2

(3)

EIv2 = -4.44x32 + C3x2 + C4

(4)

EI

Boundary conditions:
v1 = 0

at

x1 = 0

at

x2 = 0

From Eq.(2)

C2 = 0
v2 = 0
C4 = 0
Continuity conditions:
dv1
dv2
= dx1
dx2

at

x1 = x2 = 12

From Eqs. (1) and (3)
1920 + C1 = -( -1920 + C3)
C1 = -C3
v1 = v2

(5)

x1 = x2 = 12

at

7680 + 12C1 = -7680 + 12C3
C3 - C1 = 1280

(6)

Solving Eqs. (5) and (6) yields:

C3 = 640

80 lb
12 in.

For M1 (x) = 26.67 x1

C1 = -640

v1 =

1
A 4.44x31 - 640x1 B lb # in3
EI

Ans

v2 =

1
A -4.44x32 + 640x2 B lb # in3
EI

Ans.

1028

B

4 in.

4 in.
x2
12 in.


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12–139. The W8 * 24 simply supported beam is subjected
to the loading shown. Using the method of superposition,
determine the deflection at its center C. The beam is made
of A-36 steel.

6 kip/ft
5 kipиft
A

B
C
8 ft

Elastic Curves: The elastic curves for the uniform distributed load and couple

moment are drawn separately as shown.
Method of superposition: Using the table in Appendix C, the required
displacements are

(¢ C)1 =

-5(6) A 164 B
2560 kip # ft3
-5wL4
=
=
T
768EI
768EI
EI

(¢ C)2 = -

= -

=

M0x
A L2 - x2 B
6EIL
5(8)
C (16)2 - (8)2 D
6EI(16)

80 kip # ft3

EI

T

The displacement at C is
¢ C = (¢ C)1 + (¢ C)2
=

80
2560
+
EI
EI

=

2640 kip # ft3
EI
2640(1728)

=

29 A 103 B (82.8)

= 1.90 in.

Ans.

T


1029

8 ft


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*12–140. Using the moment-area method, determine the
slope and deflection at end C of the shaft. The 75-mmdiameter shaft is made of material having E = 200 GPa.
Support Reactions and

B

A

M
Diagram. As shown in Fig. a.
EI

1m


1m

15 kN

= (1) B

=

ΗtC>A Η

1 7.5
1
3
1
a
b (2) R + c (2) d B a b(2) R
2 EI
3
2
EI

5.5 kN # m3
EI

= (1 + 1) B

1 7.5
1
1
3

a
b(2) R + c (2) + 1 d B a b(2) R
2 EI
3
2
EI
2
3
1
+ c (1) d B a b(1) R
3
2
EI

ΗuC>A Η

=

9 kN # m3
EI

=

1
3
1 7.5
a
b(2) + a b(3)
2 EI
2

EI

=

3 kN # m3
EI

Referring to the geometry of the elastic curve, Fig. b,

uA =

ΗtB>A Η
LAB

5.5
EI
2.75kN # m2
=
=
2
EI

uC = uC>A - uA =

=

3
2.75
EI
EI


0.25 kN # m2
=
EI

0.25 A 103 B

200 A 109 B c

p
A 0.03754 B d
4

= 0.805 A 10 - 3 B rad

Ans.

and
¢ C = Η tC>AΗ - Η tB>A ¢

=

9
5.5 3
a b
EI
EI 2

=


0.75 kN # m3
=
EI

LAC
≤
LAB

0.75 A 103 B

p
200 A 10 B c A 0.03754 B d
4

1m
3 kN

Moment Area Theorem. Referring to Fig. b,

ΗtB>A Η

C

= 0.002414 m = 2.41 mm c Ans.

9

1030



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•12–141. Determine the reactions at the supports. EI is
constant. Use the method of superposition.

w
A

wL
C L3 - 2(3L)L2 + (3L)3 D
24EI

¢B = ¢C =

D
B
L

4

11wL

12EI

=

Due to symmetry, By = Cy
By (L)(2L)
¢ BB = ¢ CC =

6EI(3L)

C (3L)2 - (2L)2 - L2 D

4By L3
=

9EI
By (L)(L)

¢ BC = ¢ CB =

6EI(3L)

C -L2 - L2 + (3L)2 D

7By L3
=

18EI

By superposition:

+T

0 =

0 = ¢ B - ¢ BB - ¢ BC
4By L3
7By L3
11wL4
12EI
9EI
18EI

By = Cy =

11wL
10

Ans.

Equilibrium:
a+ ©MD = 0;
Ay =
c + ©Fy = 0;
Dy =
+ ©F = 0;
;
x

3wLa


3L
11wL
11wL
b (L) (2L) - A y (3L) = 0
2
10
10

2wL
5

Ans.

2wL
11wL
11wL
+
+
+ Dy - 3wL = 0
5
10
10
2wL
5

Ans.

Dx = 0

Ans.


1031

C
L

L