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12 Solutions 46060 6/11/10 11:52 AM Page 883 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in wri...

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12 Solutions 46060

6/11/10

11:52 AM

Page 883

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

•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: 1 M = 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 ⫽ 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 b C 131 A 109 B D = 582 MPa E = a 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, -3.5 =

-324W(1728) 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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Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. EI is constant.

•12–5.

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 L 2 P 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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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).

B

x1

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

L

Slope and Elastic Curve: d2v = M(x) dx2

EI For M(x1) = -

x3

P x. 2 1 EI

d2y1 dx21

P x 2 1

dy1 P = - x21 + C1 dx1 4

EI

EI y1 = For M(x3) = Px3 -

= -

[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 = -

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-

L3 0.0321PL3 + L2 b = 3 EI

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

Ans.

yC = y3 |x3 = 23 L =

2 P 3 3 3 3 c2 a Lb - 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

Px21 dy1 = + 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), PI 2 1 PI 2 PL2 1 c+ C1 d = c+ 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 Pl3 Pl2 PL2 Pl2 1 e+ c a+ b + dl + C2 f EIBC 6 IAB 2 2 2 =

Pl3 PL2l PL3 1 c+ d EIAB 6 2 3

C2 =

IBC Pl3 IBC PL3 Pl3 IAB 3 IAB 3 3

Therefore, y1 =

Px1 3 IBC Pl2 PL2 Pl2 1 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 Pl3 P I e f = e l3 - L3 - a bl f EIBC IAB 3 IAB 3 3 3EIAB IBC IAB 3 P e a1 bl - 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

d 2v = 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) =

P 2 PL (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

P L 3 PL L 2 PL2 L 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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Determine the equations of the elastic curve using the x1 and x2 coordinates. EI is constant.

•12–9.

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 = Pa a1 EI

EI

d2y2 dx2 2

x2 b L

= Pa a1 -

x2 b L

x22 dy2 = 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 a2 a3 Pb 3 a + C1a = Pa a b + C3a + C4 6L 2 6L

(6)

dy2 dy1 2 2 = dx1 x1 = a dx2 x2 = a a2 Pb 2 a + C1 = Pa aa 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 x 3 - A L2 - b 2 B x 1 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...