Chapter 8

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*8–1. Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. Specify the slope at A and the maximum deflection. EI is constant.

P

P

a

a

A

B x1 x2 L

d2v EI 2 = M(x) dx For M1(x) = Px1 EI EI

d2v1 dx21

= Px1

dv1 Px21 + C1 = dx1 2

(1)

Px21 C x + C1 6 1 1

EIv1 =

(2)

For M1(x) = Pa EI EI

d2v1 dx21

= Pa

dv1 = Pax1 + C1 dx1

EIv1 =

(3)

Pax21 = C3x1 + C4 2

(4)

Boundary conditions: v1 = 0

x = 0

at

From Eq. (2) C2 = 0 Due to symmetry: dv1 = 0 dx1

at

x1 =

L 2

From Eq. (3) 0 = Pa C3 =

L + C3 2

PaL 2

Continuity conditions: v1 = v2

at

x1 = x2 = a

Pa3 Pa3 Pa3L + C1a = + C4 6 2 2 C1a # C4 = dv1 dv2 = dx1 dx2

Pa3 Pa3L 2 2 at

(5)

x1 = x2 = a

267

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8–1.

Continued Pa3 PaL + C1 = Pa3 2 2 C1 =

Pa2 PaL – 2 2

Substitute C1 into Eq. (5) Ca =

Pa3 6

dv1 P = (x2 + a2 - aL) dx1 2EI 1 uA =

Pa(a – L) dv1 2 = dx1 x1 = 0 2EI

Ans.

v1 =

Px1 2 [x + 3a(a – L)] 6EI 1

Ans.

v2 =

Pa + (3x2(x2 – L) + a2) 6EI

Ans.

Vx = 1 = V2 2

= x=

1 2

Pa (4a2 – 3L2) 24EI

Ans.

8–2. The bar is supported by a roller constraint at B, which allows vertical displacement but resists axial load and moment. If the bar is subjected to the loading shown, determine the slope at A and the deflection at C. EI is constant.

P C A

B

L 2

EI EI

d2v1 dx21

= M1 = Px1

dv1 Px21 + C1 = dx2 2

EIv1 = EI

Px21 + C1x1 + C1 6

d2v2 PL = M2 = dx2 2

EI

dv2 PL = x + C3 dx2 2 2

EI v2 =

PL 2 x + C3x3 + C4 4 2

268

L 2

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8–2.

Continued

Boundary conditions: At x1 = 0,

v1 = 0

0 = 0 + 0 + C2; At x2 = 0,

dv2 = 0 dx2

0 + C3 = 0; L , 2

At x1 = Pa

Pa

C2 = 0

C3 = 0 x2 =

L , 2

v1 = v2,

dv1 dv2 = dx1 dx2

L 2 L 2 b PLa b 2 2 L + C1 a b = + C4 6 2 4 L 2 L Pa b b 2 2 + C1 = ; 2 2

C4 = -

3 C1 = - PL3 8

11 3 PL 48

At x1 = 0 dv1 3 PL2 = uA = dx1 8 EI

Ans.

L 2

At x1 =

L 3 b 2 3 L – a PL2 b a b + 0 6EI 8 2

Pa vc =

– PL3 6EI

vc =

8–3.

Ans.

Determine the deflection at B of the bar in Prob. 8–2.

P C

EI EI

d2v1 dx21

A

= M1 = Px1

L 2

dv2 Px21 = + C1 dx1 2

EIv1 =

Px21 + C1x1 + C2 6

EI

d2v2 PL = M2 = dx2 2

EI

dv2 PL x + C3 dx2 2 2

EI v2 =

B

PL 2 x + C3x2 + C4 4 2 269

L 2

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8–3.

Continued

Boundary conditions: v1 = 0

At x1 = 0,

0 = 0 + 0 + C2;

dv2 = 0 dx2

At x2 = 0,

C3 = 0

0 + C3 = 0; At x1 = Pa

Pa

C2 = 0

L , 2

x2 =

L , 2

v1 = v2,

dv1 dv2 = dx1 dx2

L 3 L 2 b PLa b 2 2 L + C1 a b = + C4 6 2 4 L 3 L b Pa b 2 2 + C1 = ; 2 2

C1 = -

3 2 PL 8

11 3 PL 48

C4 =

At x2 = 0, vB = -

11PL3 48EI

Ans.

*8–4. Determine the equations of the elastic curve using the coordinates x1 and x2, specify the slope and deflection at B. EI is constant.

w

C

A

B

x1 a

d2v EI 2 = M(x) dx

For M1(x) = -

EI

d2v1 dx21

EI

= -

x2

x3 L

w 2 wa2 x1 + wax1 2 2 w 2 wa2 x1 + wax1 2 2

dv1 w wa 2 wa2 = - x31 + x1 x + C1 dx1 6 2 2 1

(1)

270

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8–4.

Continued

EIv1 = –

w 4 wa 3 wa2 2 x1 + x – x + C1x1 + C2 24 6 1 4 1

For M2(x) = 0; EI

EI

d2v2 dx32

(2)

= 0

dv2 = C3 dx2

(3)

EI v2 = C3x2 + C4

(4)

Boundary conditions: At x1 = 0,

dv1 = 0 dx1

From Eq. (1),

C1 = 0

At x1 = 0,

v1 = 0

From Eq. (2):

C2 = 0

Continuity conditions: At x1 = a,

x2 = a;

dv1 dv2 = dx1 dx2

From Eqs. (1) and (3), -

wa3 wa3 wa3 + = C3; 6 2 2

C3 = -

wa3 6

From Eqs. (2) and (4), At x1 = a, -

x2 = a

v1 = v2

wa4 wa4 wa4 wa4 + = + C4; 24 6 4 6

C4 =

wa4 24

The slope, from Eq. (3), uB =

dv2 wa3 = dx2 6EI

Ans.

The elastic curve: v1 =

w a– x41 + 4ax31 – 6a2x21 b 24EI

Ans.

v2 =

wa3 a– 4x2 + ab 24EI

Ans.

v1 = v2 2

= x3 = L

wa3 a - 4L + a b 24EI

Ans.

271

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8–5. Determine the equations of the elastic curve using the coordinates x1 and x3, and specify the slope and deflection at point B. EI is constant.

w

C

A

a x2

w wa2 For M1(x) = - x21 + wax1 2 2 2 d2v1 wa w EI 2 = - x21 + wax1 2 2 dx1 dv1 w 3 wa 2 wa2 EI = - x1 + x x + C1 dx1 6 2 1 2 1

(1)

w 4 wa 3 wa2 2 x1 + x x + C1x1 + C2 24 6 1 4 1 d2v3 For M2(x) = 0; EI 2 = 0 dx3 dv3 EI = C3 dx3

(2)

(3)

EI v3 = C3x3 + C4

(4)

Boundary conditions: dv1 At x1 = 0, = 0 dx1 From Eq. (1),

At x1 = 0,

C1 = 0

v1 = 0

From Eq. (2), 0 = - 0 - 0 - 0 + 0 + C2;

C2 = 0

Continuity conditions: At x1 = a, -

dv1 dv3 = dx1 dx3

wa3 wa3 -wa3 + = -C3; 6 2 2

At x1 = a, -

x3 = L – a;

x3 = L - a

C3 = +

wa3 6

v1 = v2

wa4 wa4 wa4 wa3 + = (L - a) + C4; 24 6 4 6

C4 =

wa4 wa3L – 24 6

The slope dv3 wa3 = dx3 6EI uB =

x3 L

EIv1 = -

0 = - 0 + 0 - 0 + C1;

B

x1

d2v EI = M(x) dx2

dv3 wa3 2 = dx3 x3 = 0 6EI

Ans.

The elastic curve: wx21 v1 = a – x21 + 4ax1 - 6a2 b 24EI wa3 v3 = a 4x3 + a - 4L b 24EI wa3 V2 = V3 2 = aa - 4Lb 24EI x3 = 0

Ans. Ans. Ans.

272

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8–6. Determine the maximum deflection between the supports A and B. EI is constant. Use the method of integration.

w

B

C

A

Elastic curve and slope:

x1

2

EI

dv = M(x) dx2

EI

d1v1

EI

- wx21 2

=

dx21

L

- wx21 2

For M1(x) =

-wx31 + C1 6

dv1 = dx1

EIv1 = For M2(x) = EI

d2v2 dx23

EI

=

(1)

wx41 + C1x1 + C2 24

(2)

-wLx2 2 - wLx2 2

- wLx22 dv2 = + C3 dx2 4

EIv2 =

-wLx32 412

(3)

+ C3x3 + C4

(4)

Boundary conditions: v2 = 0

at

x2 = 0

From Eq. (4): C4 = 0 v2 = 0

at

x2 = L

From Eq. (4): 0 =

-wL4 + C 3L 12

C3 =

wL3 12

v1 = 0

at

x1 = L

From Eq. (2): 0 = -

x2

wL4 + C1L + C2 24

(5)

273

L

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8–6.

Continued

Continuity conditions: dv1 dv2 = dx1 – dx2

at

x1 = x2 = L

From Eqs. (1) and (3) -

wL3 wL3 wL3 + C1 = - a + b 6 4 12

C1 =

wL3 3

Substitute C1 into Eq. (5) C2 =

7wL4 24

dv1 w = (2L3 – x31) dx1 6EI dv2 w (L3 – 3Lx22) = dx2 12EI uA = v1 =

(6)

dv1 dv2 wL3 2 2 = = dx1 x1 = L dv3 x3 = L 6EI w ( – x41 + 8L3x1 – 7L4) 24EI

(v1)max =

– 7wL4 (x1 = 0) 24EI

The negative sign indicates downward displacement wL (L2x2 - x32) 12EI dv2 (v2)max occurs when = 0 dx2 From Eq. (6) v2 =

(7)

L3 – 3Lx22 = 0 x2 =

L 23

Substitute x2 into Eq. (7), (v2)max =

wL4

Ans.

18 23EI

274

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8–7. Determine the elastic curve for the simply supported beam using the x coordinate 0 … x … L>2. Also, determine the slope at A and the maximum deflection of the beam. EI is constant.

w0

A

B

x L

EI

d2v = M(x) dx2

EI

woL wo 3 d2v = x x 4 3L dx2

EI

woL 2 wo 4 dv = x x + C1 dx 8 12L

EIv =

(1)

wo 5 woL 3 x x + C1x + C2 24 60L

(2)

Boundary conditions: Due to symmetry, at x =

L dv , = 0 2 dx

From Eq. (1), woL L2 wo L4 0 = a b a b + C1; 8 4 12L 16 At x = 0, v = 0

C1 = -

5woL3 192

From Eq. (2), 0 = 0 - 0 + 0 + C2;

C2 = 0

From Eq. (1), wo dv = (24L2x2 - 16x4 - 5L4) dx 192EIL uA =

5woL3 dv 5woL3 2 = = dx x = 0 192EI 192EI

Ans.

From Eq. (2), v =

wox (40L2x2 - 16x4 - 25L4) 960EIL

vmax = v 2

= x=

L 3

Ans.

woL4 woL4 = 120EI 120EI

Ans.

275

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*8–8. Determine the equations of the elastic curve using the coordinates x1 and x2, and specify the slope at C and displacement at B. EI is constant.

w C A

B

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

Moment Function: As shown on FBD(c) and (c). Slope and Elastic Curve: EI

x1

For M(x1) = wax1 -

3wa2 , 2 EI

d2v1 dx21

EI

= wax1 -

3wa2 2

dv1 wa 2 3wa2 = x1 x + C1 dx1 2 2 1

EIv1 = For M(x2) = -

(1)

wa 3 3wa2 2 x1 x + C1x1 + C2 6 4 1

(2)

w 2 x, 2 2 EI

d2v2 dx22

EI

= -

v1 = 0 at x1 = 0

w 2 x 2 2

dv2 w = - x32 + C3 dx2 6

EIv2 = Boundary Conditions: dv1 = 0 at x1 = 0, dx1

(3)

w 4 x + C3x2 + C4 24 2

From Eq. [1],

C1 = 0

From Eq. [2],

C2 = 0

(4)

Continuity Conditions: dv1 dv2 = From Eqs. [1] and [3], dx1 dx2 wa3 3wa3 7wa3 wa3 = -a+ C3 b C3 = 2 2 6 6 From Eqs. [2] and [4], = a and x2 = a, v1 = v2. wa4 wa4 41wa4 3wa4 5wa4 C4 = = + + C4 6 4 24 6 8

At x1 = a and x2 = a,

At x1

The Slope: Substituting into Eq. [1], dv1 wax1 = (x1 - 3a) dx1 2EI uC =

x2 x3

d2v = M(x) dx2

dv2 wa3 2 = dx2 x1 = a EI

Ans.

The Elastic Curve: Substituting the values of C1, C2, C3, and C4 into Eqs. [2] and [4], respectively wax1 (2x21 - 9ax1) v1 = 12EI w ( -x42 + 28a3x2 - 41a4) v2 = 24EI 41wa4 vB = v2 2 = 24EI x2 = 0 276

Ans. Ans. Ans.

a

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8–9. Determine the equations of the elastic curve using the coordinates x1 and x3, and specify the slope at B and deflection at C. EI is constant.

w C A

B

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

a x1

EI

dv = M(x) dx2

3wa2 , 2 d2v1 3wa2 EI 2 = wax1 2 dx1 dv1 wa 2 3wa2 x x + C1 EI = dx1 2 1 2 1 wa 3 3wa2 2 EIv1 = x1 x + C1x1 + C2 6 4 1

For M(x1) = wax1 -

w 3 x - 2wa2, 2 2 d2v3 EI 2 = 2wax3 dx3 dv3 EI = wax23 dx3 wa 3 x EIv3 = 3 3

(1) (2)

For M(x3) = 2wax3 -

Boundary Conditions: dv1 = 0 at x1 = 0, dx1 v1 = 0 at x1 = 0,

w 2 x - 2wa2 2 3 w 3 x - 2wa2x3 + C3 6 3 w 4 x - wa2x23 + C3x3 + C4 24 3

From Eq. [1],

C1 = 0

From Eq. [2],

C2 = 0

(3) (4)

Continuity Conditions: At x1 = a and x3 = a,

dv1 dv3 = dx1 dx3

From Eqs. [1] and [3],

wa3 3wa3 wa3 = wa3 - 2wa3 + C3 2 2 6 At x1 = a and x3 = a,

v1 = v3,

C3 =

wa3 6

From Eqs.[2] and [4],

wa4 wa4 3wa4 wa4 wa4 = - wa4 + + C4 6 4 3 24 6

C4 = -

wa4 24

The Slope: Substituting the value of C3 into Eq. [3], dv3 w (6ax23 - x33 - 12a2x3 + a3) = dx3 2EI uB =

dv3 7wa3 2 = dx3 x3 = 2a 6EI

Ans.

The Elastic Curve: Substituting the values of C1, C2, C3, and C4 into Eqs. [2] and [4], respectively, wax1 v1 = (2x2 - 9ax1) 12EI 1 vC = v1 2

= x1 = a

v3 =

x2 x3

2

7wa4 12EI

Ans. Ans.

w ( –x43 + 8ax33 - 24a2x23 + 4a3x3 - a4) 24EI 277

Ans.

a

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8–10. Determine the slope at B and the maximum displacement of the beam. Use the moment-area theorems. Take E = 29(103) ksi, I = 500 in4.

15 k A

C B 6 ft

M diagram and the elastic curve shown in Fig. a and b, respectively, EI Theorem 1 and 2 give Using the

uB = ƒ uB>A ƒ =

=

1 90 k # ft a b (6 ft) 2 EI 270 k # ft2 = EI

270 (144) k # in2 k c29(10 ) 2 d (500 in4) in

= 0.00268 rad

Ans.

3

2 1 90 k # ft b(6 ft) d c6 ft + (6 ft) d ¢ max = ¢ C = ƒ tB>A ƒ = c a 2 EI 3 =

2700 k # ft3 EI 2700 (1728) k # in3

=

c29(103)

k d(500 in4) in2

= 0.322 in T

Ans.

278

6 ft

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8–11. Solve Prob. 8–10 using the conjugate-beam method.

15 k A

C B 6 ft

The real beam and conjugate beam are shown in Fig. b and c, respectively. Referring to Fig. c, 1 90 k # ft + c a Fy = 0; b(6 ft) = 0 -V¿B - a 2 EI 270 k # ft2 EI 270 (122) k # in2

uB = V¿B = – =

c29(103)

k d (500 in4) in2

Ans.

= 0.00268 rad

Referring to Fig. d, a+ a MC = 0;

M¿C + c

1 90 k # ft 2 a b(6 ft) d c6 ft + (6 ft) d = 0 2 EI 3

2700 k # ft3 EI 2700 (123) k # in3 = = 0.322 in 3 k 4 c29(10 ) 2 d(500 in ) in

¢ max = ¢ C = MC¿ = -

T

279

Ans.

6 ft

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*8–12. Determine the slope and displacement at C. EI is constant. Use the moment-area theorems.

15 k

A

C B 30 ft

M diagram and the elastic curve shown in Fig. a and b, respectively, EI Theorem 1 and 2 give Using the

1 225 k # ft 5062.5 k # ft2 5062.5 k # ft2 ab(45 ft) = = 2 EI EI EI

uC>A =

1 33750 k # ft3 1 225 k # ft ƒ tB>A ƒ = c a b(30 ft) d c (30 ft) d = 2 EI 3 EI 1 225 k # ft 1 1 225 k # ft 2 ƒ tC>A ƒ = c a b(30 ft) d c15 ft + (30 ft) d + c a b(15 ft) d c (15 ft) d 2 EI 3 2 EI 3 =

101250 k # ft3 EI

Then, ¢¿ =

uA =

45 45 33750 k # ft3 50625 k # ft3 (tB>A) = a b = 30 30 EI EI ƒ tB>A ƒ LAB

=

33750 k # ft3>EI 1125 k # ft2 = 30 ft EI

+b uC = uA + uC>A uC =

–1125 k # ft2 5062.5 k # ft2 3937.5 k # ft2 + = EI EI EI

¢ C = ` tC>A ` – ¢ ¿ = =

Ans.

101250 k # ft3 50625 k # ft3 EI EI

50625 k # ft3 T EI

Ans.

280

15 ft

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8–13.

Solve Prob. 8–12 using the conjugate-beam method.

15 k

A

C B 30 ft

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, 1 225 k # ft a+ a MA = 0; B¿y(30 ft) - c a b (30 ft) d (20 ft) 2 EI B¿y =

2250 k # ft2 EI

Referring to Fig. d, + c a Fy = 0;

-V¿C -

uC = V¿C = -

1 225 k # ft 2250 k # ft a b (15 ft) 2 EI EI

3937.5 k # ft2 3937.5 k # ft2 = EI EI

Ans.

1 225 k # ft 2250 k # ft2 a+ a MC = 0; M¿C + c a b (15 ft) d(10 ft) + a b(15 ft) 2 EI EI ¢ C = M¿C =

50625 k # ft3 50625 k # ft3 = EI EI

Ans.

T

281

15 ft

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8–14. Determine the value of a so that the slope at A is equal to zero. EI is constant. Use the moment-area theorems.

P

P C B

A

D a

M diagram and the elastic curve shown in Fig. a and b, EI respectively, Theorem 1 and 2 give

Using the

uA>B =

=

1 PL 1 Pa a b(L) + a b (a + L) 2 4EI 2 EI Pa2 PaL PL2 8EI 2EI 2EI

1 PL L 1 Pa L tD>B = c a b (L) d a b + c a b(L) d a b 2 4EI 2 2 EI 3 =

PaL2 PL3 16EI 6EI

Then uB =

tD>B L

=

PL2 PaL 16EI 6EI

Here, it is required that uB = uA>B PL2 PaL PL2 Pa2 PaL = 16EI 6EI 8EI 2EI 2EI 24a2 + 16La - 3L2 = 0 Choose the position root, a = 0.153 L

Ans.

282

L __ 2

L __ 2

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8–15.

Solve Prob. 8–14 using the conjugate-beam method.

P

P C B

A

D a

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. d, 1 Pa 1 PL 2 1 a+ a MB = 0; D'y(L) + c a b(L) d a Lb - c a b(L) d a b = 0 2 EI 3 2 4EI 2 D¿y =

PL2 PaL 16EI 3EI

It is required that V'A = uA = 0, Referring to Fig. c, c + a Fy = 0;

PL2 PaL Pa2 = 0 16EI 3EI 2EI

24a2 + 16La - 3L2 = 0 Choose the position root, a = 0.153 L

Ans.

283

L __ 2

L __ 2

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*8–16. Determine the value of a so that the displacement at C is equal to zero. EI is constant. Use the moment-area theorems.

P

P C B

A

D a

M diagram and the elastic curve shown in Fig. a and b, EI respectively, Theorem 2 gives Using the

1 PL L 1 Pa L tD>B = c a b (L) d a b + c a – b (L) d a b 2 4EI 2 2 EI 3 =

PL3 PaL2 16EI 6EI

L 1 L 1 Pa L 1 L 1 PL TC>B = c a b a b d c a b d + c a– ba bdc a bd 2 4EI 2 3 2 2 2EI 2 3 2 =

PaL2 PL3 96EI 48EI

It is required that tC>B =

1 t 2 D>B

PL3 PaL2 1 PL3 PaL2 = c d 96EI 48EI 2 16EI 6EI a =

L 3

Ans.

284

L __ 2

L __ 2

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8–17.

Solve Prob. 8–16 using the conjugate-beam method.

P

P C B

A

D a

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a+ a MD = 0;

1 PL L 1 Pa L c a b(L) d a b - c a b(L) d a b - B¿y(L) = 0 2 4EI 2 2 EI 3 –B¿y =

PL2 PaL 16EI 6EI

Here, it is required that M¿C = ¢ C = 0. Referring to Fig. d, a+ a MC = 0;

L 1 L 1 PL c a ba bdc a bd 2 4EI 2 3 2 L 1 L 1 Pa ba bdc a bd –c a 2 2EI 2 3 2 –c

PaL L PL2 da b = 0 16EI 6EI 2

PL3 PaL2 PL3 PaL2 + = 0 96EI 48EI 32EI 12EI a L a L + = 0 96 48 32 12 a =

L 3

Ans.

285

L __ 2

L __ 2

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8–18. Determine the slope and the displacement at C. EI is constant. Use the moment-area theorems.

P

C

A B a

M diagram and the elastic curve shown in Fig. a and b, respectively, EI Theorem 1 and 2 give Using the

1 Pa 2 Pa3 b (a) d a ab = tB>D = c a 2 2EI 3 6EI 1 Pa 5Pa3 2 tC>D = c a b(a) d aa + ab = 2 2EI 3 12EI uC>D =

1 Pa Pa2 a b(a) = 2 2EI 4EI

Then, uC = uC>D =

Pa2 4EI

¢ C = tC>D - tB>D =

Ans. 5Pa3 Pa3 Pa3 = 12EI 6EI 4EI

Ans.

c

286

a

a

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8–19.

Solve Prob. 8–18 using the conjugate-beam method.

P

C

A B a

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a+ a MA = 0;

1 Pa Pa2 c a b(2a) d(a) - B¿y(2a) = 0 B¿y = 2 2EI 4EI

Referring to Fig. d + c a Fy = 0;

Pa2 Pa2 - V¿C = 0 uC = V¿C = 4EI 4EI

Pa2 a+ a MC = 0; M¿C (a) = 0 4EI

¢ C = M¿C =

Pa3 4EI

Ans. Ans.

c

287

a

a

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*8–20. Determine the slope and the displacement at the end C of the beam. E = 200 GPa, I = 70(106) mm4. Use the moment-area theorems.

8 kN 4 kN

A

M Using the diagram and the elastic curve shown in Fig. a and b, EI respectively, Theorem 1 and 2 give uC>A =

3m

1 12 kN # m 1 12 kN # m a b(6 m) + a b(9 m) 2 EI 2 EI

= -

18 kN # m 18 kN # m = EI EI

1 12 kN # m 1 1 12 kN # m tB>A = c a b (6 m) d(3 m) + c a b(6 m) d c (6 m) d 2 EI 2 EI 3 =

36 kN # m3 EI

1 12 kN # m 1 12 kN # m 1 b (6 m) d(6 m) + c a b(6 m) d c3 m + (6 m) d tC>A = c a 2 EI 2 EI 3 1 12 kN # m 2 + c ab(3 m) d c (3 m) d 2 EI 3 = 0 Then uA =

¢¿ =

tB>A LAB

=

36 kN # m3>EI 6 kN # m2 = 6m EI

9 36 kN # m3 54 kN # m3 9 tB>A = a b = 6 6 EI EI

+b uC = uA + uC>A =

=

6 kN # m2 18 kN # m2 + EI EI

24(103) N # m2 24 kN # m2 = 0.00171 rad = 9 EI [200(10 ) N>m2][70(10–6) m4]

¢ C = ¢¿ - tC>A =

=

Ans.

54 kN # m3 - 0 EI 54(103)N # m3 54 kN # m3 = 0.00386 m = EI [200(109) N>m2][70(10–6) m4] = 3.86 mm T Ans.

288

C B

D 3m

3m

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8–21.

Solve Prob. 8–20 using the conjugate-beam method.

8 kN 4 kN

A

C B

D 3m

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c a+ a MA = 0;

1 12 kN # m B¿y (6 m) + c a b (6 m) d(3 m) 2 EI 1 12 kN # m 2 - c a b(6 m) d c (6 m) d = 0 2 EI 3 B¿ y =

6 kN # m2 EI

Referring to Fig. d, a+ a Fy = 0;

-V¿C -

1 12 kN # m 6 kN # m2 a b (3 m) = 0 EI 2 EI

uC = V¿C = -

24(103) N # m2 24 kN # m2 = EI [(200(109) N>m2)] [(70(10 - 6) m4]

= 0.00171 rad a+ a MC = 0;

M¿C + c

Ans.

1 12 kN # m2 2 a b(3 m) d c (3 m) d 2 EI 3 + a

¢ C = M¿C = -

6 kN # m2 b (3 m) = 0 EI 54 (103) N # m3 54 kN # m3 = EI [200(109)N>m2] [70(10 - 6)m4] = 0.00386 m = 3.86 mm T Ans.

289

3m

3m

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8–22. At what distance a should the bearing supports at A and B be placed so that the displacement at the center of the shaft is equal to the deflection at its ends? The bearings exert only vertical reactions on the shaft. EI is constant. Use the moment-area theorems.

P

P

A

B

a

a L

M diagram and the elastic curve shown in Fig. a and b, respectively. EI Theorem 2 gives Using the

tB>C = a -

Pa L - 2a L - 2a Pa ba ba b = (L - 2a)2 EI 2 4 8EI

tD>C = a -

Pa L - 2a L - 2a 1 Pa 2 ba b aa + b + ab(a)a a b EI 2 4 2 EI 3

= -c

Pa 2 Pa3 (L - 4a2) + d 8EI 3EI

It is required that tD>C = 2 tB>C Pa 2 Pa3 Pa (L – 4a2) + = 2c - (L - 2a)2 d 8EI 3EI 8EI Pa2L PaL2 7Pa3 + = 0 6EI EI 8EI 56a2 - 48La + 6L2 = 0 Choose a = 0.152 L

Ans.

290

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8–23. Solve Prob. 8–22 using the conjugate-beam method.

P

P

A

B

a

a L

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, B¿y (L - 2a) - c

a+ a MA = 0;

B¿y =

Pa L – 2a (L – 2a) d a b = 0 EI 2

Pa (L – 2a) 2EI

Referring to Fig. d, M¿D +

1 Pa 2 Pa (L - 2a)(a) + c a b(a) d a a b = 0 2EI 2 EI 3

¢ D = M¿D = - c

Pa3 Pa2 (L - 2a) + d 2EI 3EI

Referring to Fig. e, Pa L - 2a Pa L - 2a L - 2a (L - 2a)a b a ba b - M¿C = 0 2EI 2 EI 2 4 ¢ C = M¿C =

Pa (L – 2a)2 8EI

It is required that ƒ¢Dƒ = ¢C Pa3 Pa Pa2 (L – 2a) + = (L – 2a)2 2EI 3EI 8EI 7Pa3 Pa2L PaL2 + = 0 6EI EI 8EI 56a2 - 48La + 6L2 = 0 Choose a = 0.152 L

Ans.

291

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*8–24. Determine the displacement at C and the slope at B. EI is constant. Use the moment-area theorems.

4 kN

4 kN

C A 3m

B 1.5 m

1.5 m

M diagram and elastic curve shown in Fig. a and b, respectively, EI Theorem 1 and 2 give Using the

uB>D = a -

12 kN # m 18 kN # m2 b(1.5 m) = EI EI

tB>D = c a tC>D = c a =

12 kN # m 1 13.5 kN # m3 b (1.5 m) d c (1.5 m) d = EI 2 EI

1 1 2 12 kN # m 12 kN # m b (1.5 m) d c (1.5 m) + 3 m d + c a b(3 m) d c (3 m) d EI 2 2 EI 3

103.5 kN # m3 EI

Then, uB = ƒ uB>D ƒ =

18 kN # m2 EI

¢ C = ƒ tC>D ƒ - ƒ tB>D ƒ =

=

Ans.

103.5 kN # m3 13.5 kN # m2 EI EI 90 kN # m3 EI

Ans.

T

292

3m

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8–25.

Solve Prob. 8–24 using the conjugate-beam method.

4 kN

4 kN

C A 3m

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, B¿y (3 m) - a

a+ a MA = 0;

B¿y = uB =

12 kN # m b(3 m)(1.5 m) = 0 EI

18 kN # m2 EI

Ans.

Referring to Fig.d, a+ a MC = 0; ¢ C = M¿C = -

M¿C + a

18 kN # m2 1 12 kN # m 2 b (3 m) + c a b(3 m) d c (3 m) d = 0 EI 2 EI 3

90 kN # m3 90 kN # m3 = EI EI

Ans.

T

293

B 1.5 m

1.5 m

3m

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8–26. Determine the displacement at C and the slope at B. EI is constant. Use the moment-area theorems.

P

P 2

A

M diagram and elastic curve shown in Fig. a and b, respectively, EI Theorem 1 and 2 give uB>C =

1 Pa Pa 1 Pa 7Pa2 a b(a) + a b (a) + a b(a) = 2 EI EI 2 2EI 4EI

tB>C = c =

1 Pa 2 Pa 1 1 Pa 2 a b (a) d a ab + c (a) d a a + ab + c a b(a) d aa + ab 2 EI 3 EI 2 2 2EI 3

9Pa3 4EI

Then uB = uB>C =

7Pa2 4EI

Ans.

AC = tB>C =

9Pa3 T 4EI

Ans.

294

B

C a

Using the

P 2

a

a

a

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8–27. Determine the displacement at C and the slope at B. EI is constant. Use the conjugate-beam method.

P

P 2

A

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a+ a MA = 0;

c

1 Pa 2 1 Pa 10 a b(a) d a a b + c a b(a) d a a b 2 EI 3 2 2EI 3 + c a

Pa 1 Pa b (2a) + a b(2a) d(2a) EI 2 2EI

-B¿y = (4a) = 0 uB = B¿y =

7Pa2 4EI

Ans.

Referring to Fig. d, a+ a MC = 0;

1 Pa 4 Pa a c a b(a) d a ab + c a b(a) d a b 2 2EI 3 EI 2 1 Pa a 7Pa2 + c a b (a) d a b (2a) 2 2EI 3 4EI -M'C = 0 ¢ C = M'C = -

9Pa3 9Pa3 = 4EI 4EI

Ans.

T

295

P 2

B

C a

a

a

a

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*8–28. Determine the force F at the end of the beam C so that the displacement at C is zero. EI is constant. Use the moment-area theorems.

F P D A

C B a

M diagram and elastic curve shown in Fig. a and b, respectively, EI Theorem 2 gives Using the

1 Pa 1 Fa 1 Pa3 2Fa3 tB/A = c a b (2a) d(a) + c a b (2a) d c (2a) d = 2 2EI 2 EI 3 2EI 3EI 1 Pa 1 Fa 1 tC/A = c a b (2a) d (2a) + c a b(2a) d c (2a) + a d 2 2EI 2 EI 3 1 Fa 2 + c ab(a) d c (a) d 2 EI 3 =

2Fa3 Pa3 2EI 3EI

It is required that tC>A =

3 t 2 B>A

Pa3 2Fa3 3 Pa3 2Fa3 = c d EI EI 2 2EI 3EI F =

P 4

Ans.

296

a

a

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8–29. Determine the force F at the end of the beam C so that the displacement at C is zero. EI is constant. Use the conjugate-beam method.

F P D A

C B a

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a+ a MA = 0;

c

1 Pa 1 Fa 2 a b(2a) d(a) - c a b (2a) d c (2a) d - B¿y (2a) = 0 2 2EI 2 EI 3

B¿y =

Pa2 2Fa2 4EI 3EI

Here, it is required that ¢ C = M¿C = 0. Referring to Fig. d, a+ a MC = 0;

c

1 Fa 2 Pa2 2Fa2 a b(a) d c (a) d - a b (a) = 0 2 EI 3 4EI 3EI

F =

P 4

Ans.

297

a

a

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8–30. Determine the slope at B and the displacement at C. EI is constant. Use the moment-area theorems.

P

P

A

C

B a

M diagram and elastic curve shown in Fig. a and b, EI Theorem 1 and 2 give

Using the

uB>A =

Pa2 1 Pa Pa2 ab (a) = = 2 EI 2EI 2EI

tB>A = c

Pa 1 Pa3 1 ab(a) d c (a) d = 2 EI 3 6EI

tC>A = c

1 Pa Pa3 ab(2a) d(a) = 2 EI EI

Then uA =

¢¿ =

ƒ tB>A ƒ LAB

=

Pa3>6EI Pa2 = 2a 12EI

3 Pa3 3 Pa3 ƒ tB>A ƒ = a b = 2 2 6EI 4EI

uB = uA + uB>A a+ uB = -

Pa2 Pa2 5Pa2 + = 12EI 2EI 12EI

¢ C = ƒ tC>A ƒ - ¢ ¿ =

Pa3 3Pa3 Pa3 = EI 4EI 4EI

Ans.

T

Ans.

298

a

a

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8–31. Determine the slope at B and the displacement at C. EI is constant. Use the conjugate-beam method.

P

P

A

C

B a

The real beam and conjugate beam are shown in Fig. c and d, respectively. Referring to Fig. d, 1 Pa 2 c a b (a) d a a + ab - B¿y (2a) = 0 2 EI 3

a+ a MA = 0;

uB = B¿y =

5Pa2 12EI

Ans.

Referring to Fig. c, a+ a MC = 0;

-M¿ C - c

1 Pa 2 5Pa2 a b(a) d a ab - a b(a) = 0 2 EI 3 12EI

¢ C = M¿C = -

3Pa3 3Pa3 = 4EI 4EI

Ans.

T

299

a

a

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*8–32. Determine the maximum displacement and the slope at A. EI is constant. Use the moment-area theorems.

M0 C

A L __ 2

M diagram and the elastic curve shown in Fig. a and b, respectively, EI Theorem 1 and 2 give Using the

uD>A =

M0 2 1 M0 a xb(x) = x 2 EIL 2EIL

M0L2 1 M0 L 1 L tB>A = c a ba bdc a bd = 2 2EI 2 3 2 48EI Then, uA =

ƒ tB>A ƒ LAB

=

M0L2>48EI M0L = L>2 24EI

Ans.

Here uD = 0. Thus, uD = uA + uD>A a+

0 = -

M0L M0 2 + x 24EI 2EIL

x =

L 212

= 0.2887L

1 1 0.2113M0 ¢ max = ¢ D = tB>D = c a b(0.2113L) d c (0.2113L) d 2 EI 3 + ca =

0.2887M0 1 b (0.2113L) d c (0.2113L) d EI 2

0.00802M0L2 T EI

Ans.

300

B

L __ 2

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8–33. Determine the maximum displacement at B and the slope at A. EI is constant. Use the conjugate-beam method.

M0 C

A L __ 2

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c 1 M0 L L a+ a MB = 0; A¿y(L) - c a ba bd a b = 0 2 2EI 2 3 A¿y = uA =

M0L 24EI

Ans.

Here it is required that uD = V¿D = 0. Referring to Fig. d, c a Fy = 0; x =

M0L 1 M0 a xb (x) = 0 2 EIL 24EI L 212

M0L L a+ a MD = 0; M¿D + a ba b 24EI 212 -

1 M0 L L 1 L a ba ba bc a bd = 0 2 EIL 3 212 212 212

¢ max = ¢ D = M¿D = -

=

0.00802M0L2 EI

0.00802M0L2 EI Ans.

T

301

B

L __ 2

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8–34. Determine the slope and displacement at C. EI is constant. Use the moment-area theorems.

P M0  Pa A

C a

M diagram and the elastic curve shown in Fig. a and b, respectively, EI Theorem 1 and 2 give Using the

uC = ƒ uC>A ƒ =

1 Pa 2Pa 1 Pa 3Pa2 a b(a) + a b(a) + a b (a) = 2 EI EI 2 EI EI

¢ C = ƒ tC>A ƒ = c

1 Pa 2 a 2Pa a b (a) d aa + ab + c a b (a) d aa + b 2 EI 3 EI 2 + c

=

Ans.

2 1 Pa a b(a) d a ab = 0 2 EI 3

25Pa3 6EI

Ans.

T

302

B

a

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8–35. Determine the slope and displacement at C. EI is constant. Use the conjugate-beam method.

P M0  Pa A

C a

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, + c a Fy = 0;

-V¿C -

1 Pa 2Pa 1 Pa a b (a) - a b(a) - a b(a) = 0 2 EI EI 2 EI

uC = V¿C = -

3Pa2 3Pa2 = EI EI

Ans.

1 Pa 2 2Pa a a+ a MC = 0; M¿C + c a b (a) d a ab + c a b(a) d aa + b 2 EI 3 EI 2 2 1 Pa + c a b(a) d a a + ab = 0 2 EI 3 ¢ C = M¿C = -

25Pa3 25Pa3 = 6EI 6EI

Ans.

T

303

B

a

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*8–36. Determine the displacement at C. Assume A is a fixed support, B is a pin, and D is a roller. EI is constant. Use the moment-area theorems.

25 kN

C

3m

M diagram and the elastic curve shown in Fig. a and b, respectively, EI Theorem 1 and 2 give Using the

1 37.5 kN # m 2 112.5 kN # m3 ¢ B = ƒ tB>A ƒ = c a b (3 m) d c (3 m) d = 2 EI 3 EI

T

1 37.5 kN # m 1 56.25 kN # m3 tC>D = c a b(3 m) d c (3 m) d = 2 EI 3 EI 337.5 kN # m3 1 37.5 kN # m b (6 m) d(3 m) = tB>D = c a 2 EI EI Then uD =

¢ B + tB>D LB>D

=

¢ C + tC>D =

112.5 kN # m3>EI + 337.5 kN # m3>EI 75 kN # m2 = 6m EI

Ans.

1 (¢ + tB>D) 2 B

¢C +

56.25 kN # m3 1 112.5 kN # m3 337.5 kN # m3 = a + b EI 2 EI EI

¢C =

169 kN # m3 T EI

Ans.

304

D

B

A

3m

3m

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8–37. Determine the displacement at C. Assume A is a fixed support, B is a pin, and D is a roller. EI is constant. Use the conjugate-beam method.

25 kN

C

3m

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, a+ a MB = 0;

1 37.5 kN # m c a b (6 m) d(3 m) 2 EI 1 37.5 kN # m + c a b(3 m) d(2 m) - D¿y (6 m) = 0 2 EI

uD = Dy¿ =

75 kN # m2 EI

Ans.

Referring to Fig. d, a+ a MC = 0;

1 37.5 kN # m c a b (3 m) d(1 m) 2 EI -a

75 kN # m2 b(3 m) – M¿C = 0 EI

¢C =

-

168.75 kN # m3 168.75 kN # m3 = T EI EI

305

D

B

A

Ans.

3m

3m

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8–38. Determine the displacement at D and the slope at D. Assume A is a fixed support, B is a pin, and C is a roller. Use the moment-area theorems.

6k

B

A 12 ft

M diagram and elastic curve shown in Fig. a and b, respectively, EI Theorem 1 and 2 give Using the

1 72 k # ft 2 3456 k # ft3 c ¢ B = tB>A = c a b(12 ft) d c (12 ft) d = 2 EI 3 EI uD>B =

72 k # ft 864 k # ft2 864 k # ft2 1 ab (24 ft) = = 2 EI EI EI

1 72 k # ft 1 1728 k # ft3 tC>B = c a b (12 ft) d c (12 ft) d = 2 EI 3 EI 1 72 k # ft 10368 k # ft3 b(24 ft) d(12 ft) = tD>B = c a 2 EI EI Then, ¢¿ = 2(¢ B – ƒ tC>B ƒ = 2 a uBR =

3456 k # ft3 1728 k # ft3 3456 k # ft3 – b = EI EI EI

3456 k # ft3>EI 144 k # ft2 ¢¿ = = LBD 24 ft EI

uD = uBR + uD>B +b uD =

144 k # ft2 864 k # ft2 1008 k # ft2 + = EI EI EI

Ans.

¢ D = ƒ tD>B ƒ + ¢¿ – ¢ B =

10368 k # ft3 3456 k # ft3 3456 k # ft3 + EI EI EI

=

10,368 k # ft3 T EI

Ans.

306

D

C 12 ft

12 ft

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8–39. Determine the displacement at D and the slope at D. Assume A is a fixed support, B is a pin, and C is a roller. Use the conjugate-beam method.

6k

B

A 12 ft

The real beam and conjugate beam are shown in Fig. a and b, respectively. Referring to Fig. c, 1 72 k # ft C¿y(12 ft) – c a b(12 ft) d(16 ft) 2 EI

a+ a MB = 0;

= 0 C¿y =

576 k # ft2 EI

Referring to Fig. d, + c a Fy = 0;

1 72 k # ft 576 k # ft2 -V¿D - a b(12 ft) = 0 2 EI EI

uD = V¿D =

-

1008 k # ft2 1008 k # ft2 = EI EI

Ans.

1 72 k # ft 576 k # ft2 a+ a MC = 0; M¿D + c a b(12 ft) d(8 ft) + a b(12 ft) = 0 2 EI EI M¿D = ¢ D =

-

10,368 k # ft3 10368 k # ft3 = T EI EI

Ans.

307

D

C 12 ft

12 ft