12–3. c - krisada...

884

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, is

a

and is

a

Equations of Slope and Elastic Curve.

For coordinate x1,

(1)

(2)

For coordinate x2

(3)

(4)

Boundary Conditions. At , . Then, Eq. (2) gives

At , . Then, Eq. (2) gives

EI(0) = - 12

W A33 B + C1(3) + 0 C1 = 4.5W

v1 = 0x1 = 3 ft

EI(0) = - 12

W A03 B + C1(0) + C2 C2 = 0

v1 = 0x1 = 0

EIv2 = - 16

Wx2 3 + C3x2 + C4

EI dv2

dx2= -

12

Wx2 2 + C3

EI d2v2

dx2 2 = -Wx2

EIv1 = - 12

Wx1 3 + C1x1 + C2

EI d2v1

dx1= - 3

2 Wx1

2 + C1

EI d2v1

dx1 2 = -3Wx1

EI d2vdx2 = M(x)

+ ©MO = 0; -M Ax2 B - Wx2 = 0 M Ax2 B = -Wx2

M Ax2 B+ ©MO = 0; M Ax1 B + 3Wx1 = 0 M Ax1 B = -3Wx1

M Ax1 B

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

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

3.5 in.

9 ft

A B

C

3 ft 18 in.

2 in.

12 Solutions 46060 6/11/10 11:52 AM Page 884

885



(5)

Continuity Conditions. At and , . Thus, Eqs. (1) and

(3) give

Substituting the value of C3 into Eq. (5),

Substituting the values of C3 and C4 into Eq. (4),

At , . Then,

Ans.W = 112.53 lb = 113 lb

-3.5 =-324W(1728)

1.5 A106 B c 112

(18) A23 B dv2 = -3.5 inx2 = 0

v2 = 1EI

a - 16

Wx2 3 + 49.5Wx2 - 324Wb

C4 = -324W

- 32

W A32 B + 4.5W = - c - 12

W A92 B + C3 d C3 = 49.5W

dv1

dx1= -

dv2

dx2x2 = 9 ftx1 = 3 ft

9C3 + C4 = 121.5W

EI(0) = - 16

W A93 B + C3(9) + C4

v2 = 0x2 = 9 ft

12-3. Continued


902

Elastic curve and slope:

(1)

(2)

(3)

(4)

Boundary Conditions:

at

From Eq. (2),

at

From Eq. (3),

Continuity conditions:

at x1 = x2 = L4

dv1

dx1=

dv2

dx2

C3 = PL2

16

0 = PL2

16+ C3

x2 = L2

dv2

dx2= 0

C2 = 0

x1 = 0v1 = 0

2EIv2 =Px2

3

12+ C3x2 + C4

2EI dv2

dx1=

Px22

4+ C3

2EI d2v2

dx2 2 = P

2 x2

EIv1 =Px1

3

12+ C1x1 + C2

EI dv1

dx1=

Px12

4+ C1

EI d2v1

dx1 2 = P

2 x1

EI d2vdx2 = M(x)

M2(x) = P2

x2

M1 (x) = P2

x1

12–14. The simply supported shaft has a moment of inertiaof 2I for region BC and a moment of inertia I for regionsAB and CD. Determine the maximum deflection of thebeam due to the load P.


CA D

P

–4L –

4L –

4L –

4L

B


903

From Eqs. (1) and (3),

at

From Eqs. (2) and (4)

Ans.vmax = v2 2x2 = L

2

= -3PL3

256EI= 3PL3

256EI T

v2 = P768EI

A32x23 - 24L2 x2 - L3 B

C4 = -PL3

384

PL3

768- 5PL2

128 aL

4b = PL3

1536- 1

2 aPL2

16b aL

4b + 1

2C4

x1 = x2 = L4

v1 = v2

C1 = -5PL2

128

PL2

64+ C1 = PL2

128- 1

2 aPL2

16b

12–14. Continued



914

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

Moment Functions. Referring to the free-body diagrams of the beam’s cutsegments, Fig. b, is

a

and is

a

Equations of Slope and Elastic Curve.

For coordinate x1,

(1)

(2)

For coordinate x2,

(3)

(4)

Boundary Conditions. At , . Then, Eq. (2) gives



(5)L2

C3 + C4 = wL4

384

EI(0) = - w24

aL2b4

+ C3aL2b + C4

v2 = 0x2 = L2

EI(0) = - wL48AL3 B + C1L + 0 C1 = wL3

48

v1 = 0x1 = L

EI(0) = - wL48A03 B + C1(0) + C2 C2 = 0

v1 = 0x1 = 0

EIv2 = - w24

x2 4 + C3x2 + C4

EI dv2

dx2= - w

6 x2

3 + C3

EI d2v2

dx2 2 = - w

2 x2

2

EIv1 = - wL48

x1 3 + C1x1 + C2

EI dv1

dx1= - wL

16 x1

2 + C1

EI d2v1

dx1 2 = - wL

8 x1

EI d2vdx2 = M(x)

+ ©MO = 0; -M(x2) - wx2ax2

2b = 0 M(x2) = - w

2 x2

2

M(x2)

+ ©MO = 0; M(x1) + wL8

x1 = 0 M(x1) = - wL8

x1

M(x1)

•12–21. Determine the elastic curve in terms of the andcoordinates and the deflection of end C of the overhangbeam. EI is constant.


w

L L2

C

B

A

x1

x2


915

Continuity Conditions. At Land , . Thus, Eqs. (1) and

(3) give

Substituting the value of C3 into Eq. (5),

Substituting the values of C3 and C4 into Eq. (4),

At C, . Thus,

Ans.vC = v2! x2 = 0 = - 11wL4

384EI= 11wL4

384EI T

x2 = 0

v2 = w384EI

A -16x2 4 + 24L3x2 - 11L4 B

C4 = - 11wL4

384

- wL16AL2 B + wL3

48= - C - w

6aL

2b3

+ C3S C3 = wL3

16

dv1

dx1= -

dv2

dx2x2 = L

2x1 = L

•12–21. Continued



920

For

(1)

(2)

For

(3)

(4)

Boundary conditions:

At .

From Eq. (1),

At .

From Eq. (2):

Continuity conditions:

At



At

- wa4

24+ wa4

6- wa4

4= - wa4

6+ C4 ; C4 = wa4

24

x1 = a, x2 = a y1 = y2

- wa3

6+ wa3

2- wa3

2= C3; C3 = - wa3

6

x1 = a, x2 = a ; dy1

dx1=

dy2

dx2

C2 = 0

y1 = 0x1 = 0

C1 = 0

dy1

dx1= 0x1 = 0

EI y2 = C3x2 + C4

EI dy2

dx2= C3

M2(x) = 0 ; EI d2y2

dx2 2 = 0

EI y1 = - w24

x14 + wa

6 x1

3 - wa2

4x1

2 + C1x1 + C2

EI dy1

dx1= - w

6 x1

2 + wa2

x12 - wa2

2 x1 + C1

EI d2y1

dx12 = - w

2 x1

2 + wax1 - wa2

2

M1(x) = - w2

x12 + wax1 - wa2

2

EI d2y

dx2 = M(x)

12–26. Determine the equations of the elastic curve usingthe coordinates and and specify the slope and deflectionat B. EI is constant.

x2 ,x1


L

A

B

a

w

x1

x2

C


921

The slope, from Eq. (3).

Ans.

The elastic curve:

Ans.

Ans.

Ans.yB = y2 2x3 = L

= wa3

24EI a -4L + ab

y2 = wa3

24EI a -4x2 + ab

y1 = w24EI

a -x14 + 4ax1

3 - 6a2 x12b

uB =dy2

dx2= - wa3

6EI

12–26. Continued



12–3. c - krisada...

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