12–6. p 1 x3 ei a - auburn universityhtippur/mech3130/hw-7 solutions-2014.pdf · determine the...

1137

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

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

Slope and Elastic Curve:

For .

(1)

(2)

For ,

(3)

(4)

Boundary Conditions:

at . From Eq. (2),

at . From Eq. (2).

at . From Eq. (4).

(5)

Continuity Condition:

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

From Eq. (5), C4 = -

PL3

4

-

PL2

4+

PL2

12=

PL2

2-

3PL2

2+ C3 C3 =

5PL2

6

dy1

dx1=

dy3

dx3x1 = x3 = L

0 = -

7PL3

12+ C3L + C4

0 =

PL3

6-

3PL3

4+ C3L + C4

x3 = Ly3 = 0

0 = -

PL3

12+ C1L C1 =

PL2

12

x1 = Ly1 = 0

C2 = 0x1 = 0y1 = 0

EI y3 =

P

6 x3

3 -

3PL

4 x3

3 + C3x3 + C4

EI dy3

dx3=

P

2 x2

3 -

3PL

2 x3 + C3

EI d2y3

dx23

= Px3 -

3PL

2

M(x3) = Px3 -

3PL

2

EI y1 = -

P

12 x1

3+ C1x1 + C2

EI dy1

dx1= -

P

4 x1

2+ C1

EI d2y1

dx12 = -

P

2 x1

M(x1) = -

P

2 x1

EI d2y

dx2 = M(x)

12–6. Determine the equations of the elastic curve for thebeam using the and coordinates. Specify the beam’smaximum deflection. EI is constant.

x3x1

L

A

B

P

x1

x3

L2

Hibbeler_12(1130-1148).qxd 3/6/13 1:38 PM 137

The Slope: Substitute the value of C1 into Eq. (1),

The Elastic Curve: Substitute the values of C1, C2, C3, and C4 into Eqs. (2) and (4),

respectively,

Ans.

Ans.

Hence, Ans.ymax = vC =

PL3

8EI T

= -

PL3

8EI

=

P

12EI c2a3

2Lb3

- 9La32

Lb2

+ 10L2a32

Lb - 3L3 d

yC = y3 |x3 =32 L

y3 =

P

12EI A2x3

3 - 9Lx23 + 10L2x3 - 3L3 B

yD = y1 |x1 =L

23

=

P A L

2 3B

12EIa -

L3

3+ L2b =

0.0321PL3

EI

y1 =

Px1

12EI A -x1

2+ L2 B

dy1

dx1= 0 =

P

12EI AL2

- 3x12 B x1 =

L

23

dy1

dx1=

P

12EI AL2

- 3x12 B

12–6. Continued

1138


Ans:

,

vmax =

PL3

8EI T

v3 =

P

12EI (2x3

3 - 9Lx23 + 10L2x3 - 3L3)

v1 =

Px1

12EI (-x1

2+ L2),

Hibbeler_12(1130-1148).qxd 3/6/13 1:38 PM Page 1138

1148


;

Ans.

Ans.ymax = -

5M0 a2

8EI

EI ymax =

12

M0aa2

4b -

M0 a

2 aa

2b -

M0 a2

2

At x2 =

a

2,

UA = -

M0 a

2EI

EI dy1

dx1= -

M0 a

2

At x1 = 0,

Ca =

-M0 a2

2C1 =

-M0 a

2,

C1a = Ca

dy1

dx1=

dy2

dx2y1 = y2,x2 = 0,At x1 = a,

C1 =

-M0 a

2At x2 =

a

2,

dy2

dx2 = 0;

EI y2 =

12

M0x22 + C3x2 + C4

EI dy2

dx2= M0x2 + C2

EI d2y1

dx 22 = M0M2 = M0;

C2 = 0y1 = 0At x1 = 0,

EIy1 = C1x1 + C2

EI dy1

dx1= C1EI

d2y1

dx12 = 0;

M1 = 0

12–13. Determine the maximum deflection of the beamand the slope at A. EI is constant.

B

a a a

A

M0 M0

Ans:

vmax = -

5M0 a2

8EIuA = -

M0a

2EI,


1165


Elastic Curve and Slope:

For

(1)

(2)

For

(3)

(4)


at

From Eq. (2),

Due to symmetry,

at

From Eq. (3),

C3 = -4800

-2880 + 8640 - 960 + C3 = 0

x2 = 6 ftdv2

dx2= 0

C2 = 0

x1 = 0v1 = 0

EIv2 = -3.33x42 + 80x3

2 - 80x22 + C3x2 + C4

EI

dv2

dx2 = -13.33x3

2 + 240x22 - 160x2 + C1

EI

d2v2

dx32

= -40x22 + 480x2 - 160

M2(x) = -40x22 + 480x2 - 160

EIv1 = 53.33x13

+ C1x1 + C2

EI

dv1

dx1= 160x1

2+ C1

EI

d2v1

dx21

= 320x1

M1(x) = 320x1

EI d2v

dx2 = M(x)

12–25. The floor beam of the airplane is subjected to theloading shown. Assuming that the fuselage exerts onlyvertical reactions on the ends of the beam, determine themaximum deflection of the beam. EI is constant.

8 ft2 ft 2 ft

80 lb/ft


1166


Continuity Conditions:

at

From Eqs. (1) and (3),

at

From Eqs. (2) and (4),

occurs at

Ans.

The negative sign indicates downward displacement.

vmax = v2|x2 = 6 =

-18.8 kip # ft3

EI

x2 = 6ft.vmax

v2 =

1EI

(-3.33x42 + 80x3

2 - 80x22 - 4800x2 - 53.33)

C4 = -53.33

426.67 - 9813.33 = -53.33 + 640 - 320 - 9600 + C4

x1 = x2 = 2 ftv1 = v2

C1 = -4906.67

640 + C1 = -106.67 + 960 - 320 - 4800

x1 = x2 = 2 ftdv1

dx1=

dv2

dx2

12–25. Continued

Ans:

vmax =

-18.8 kip # ft3

EI


1189



(1)


at

From Eq. (1),

(2)

at

From Eq. (1),

(3)

Solving Eqs. (2) and (3) yields:

Ans.

8 x - 4.5 93 + 25.1x - 36.4] kN # m3n =

1EI

[-0.25x4+ 0.2088 x - 1.5 93 + 0.258 x - 1.5 94 + 4.625

C2 = -36.42

C1 = 25.12

4.5C1 + C2 = 76.641

0 = -102.516 + 5.625 + 20.25 + 4.5C1 + C2

x = 4.5 mn = 0

1.5C1 + C2 = 1.266

0 = -1.266 + 1.5C1 + C2

x = 1.5 mn = 0

+ 4.6258 x - 4.5 93 + C1x + C2EIn = -0.25x4+ 0.2088 x - 1.5 93 + 0.258 x - 1.5 94

EI dn

dx= -x3

+ 0.625 8 x - 1.5 92 + 8 x - 1.593 + 13.875 8 x - 4.592 + C1

EI d2n

dx2 = M = -3x2+ 1.258 x - 1.5 9 + 38 x - 1.592 + 27.758 x - 4.5 9

M = -3x2+ 1.258 x - 1.5 9 + 38 x - 1.5 92 + 27.758 x - 4.5 9

- (-27.75) 8 x - 4.5 9 8x - 1.5 9 - a- 62b8 x - 1.5 92M = -

628 x - 0 92 - (-1.25)

12–42. The beam is subjected to the load shown.Determine the equation of the elastic curve. EI is constant.

A B

3 m1.5 m

6 kN/m 20 kN

1.5 m

Ans:

+ 25.1x - 36.44 kN # m3

+ 0.258x - 1.594 + 4.6258x - 4.593v =

1EI3-0.25x4

+ 0.2088x - 1.593

Hibbeler_12(1183-1204)_new.qxd 3/6/13 5:23 PM Page 1189

1196


Support Reactions and Elastic Curve: As shown on FBD.

Moment Function: Using the discontinuity function,

Slope and Elastic Curve:

[1]

[2]


at . From Eq. [2],

[3]

at . From Eq. [2],

[4]

Solving Eqs. [3] and [4] yields,

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

Ans.

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

uA =

dy

dx`x = 6 ft

=

1EI

e - 43A63 B + 0 + 0 + 590.4 f =

302 kip # ft2

EI

dy

dx=

1EI

e - 43

x3+

1278x - 6 94 + 448x - 6 92 + 590.4 f kip # ft2

C1 = 590.4 C2 = -3110.4

5745.6 = 15C1 + C2

0 = - 13

A154 B +

1135

(15 - 6)5+

443

(15 - 6)3+ C1 (15) + C2

x = 15 fty = 0

432 = 6C1 + C2

0 = - 13

A64 B + 0 + 0 + C1 (6) + C2

x = 6 fty = 0

EI y = - 13

x4+

11358x - 6 95 +

4438x - 6 93 + C1x + C2

EI dy

dx= -

43

x3+

1278x - 6 94 + 448x - 6 92 + C1

EI d2y

dx2 = -4x2+

4278x - 6 93 + 888x - 69

EI d2y

dx2 = M

= -4x2+

4278x - 6 93 + 888x - 69

M = - 12

(8)8x - 0 92 -

16a -

89b8x - 6 93 - (-88)8x - 69

12–49. Determine the displacement C and the slope at Aof the beam. EI is constant.

AB

6 ft 9 ft

x

8 kip/ ft

C

Ans.yC = y |x = 0 =

1EI

{-0 + 0 + 0 + 0 - 3110.4} kip # ft3= -

3110 kip # ft3

EI

+ 590.4x - 3110.4 f kip # ft3 y =

1EI

e - 13

x4+

11358x - 6 95 +

4438x - 6 93

Ans:

vC = -

3110EI

kip # ft3uA =

302EI

kip # ft2,


1197



(1)

(2)


From Eq. (2),

Ans.

Ans.n =

1EI

[0.75x3- 0.125x4

+ 1.758x - 593 + 0.1258x - 594 - 3.125x] kN # m3

dn

dx=

1EI

[2.25x2- 0.5x3

+ 5.258x - 592 + 0.58x - 593 - 3.125] kN # m2

C1 = -3.125

0 = 93.75 - 78.125 + 5C1

n = 0 at x = 5

C2 = 0

n = 0 at x = 0

EIn = 0.75x3- 0.125x4

+ 1.758x - 593 + 0.1258x - 594 + C1x + C2

EIdn

dx= 2.25x2

- 0.5x3+ 5.258x - 592 + 0.58x - 593 + C1

EId2n

dx2 = M = 4.5x - 1.5x2+ 10.58x - 59 + 1.58x - 592

M = 4.5x - 1.5x2+ 10.58x - 59 + 1.58x - 592

M = -(-4.5)8x - 09 -

328x - 092 - (-10.5)8x - 59 - a -3

2b8 x - 5 92

12–50. The beam is subjected to the load shown. Determinethe equations of the slope and elastic curve. EI is constant.

AB

5 m 3 m

xC

3 kN/m

15 kN�m

Ans:

,

+ 0.1258x - 594 - 3.125x4 kN # m3

v =

1EI30.75x3

- 0.125x4+ 1.758x - 593

+ 0.58x - 593 - 3.1254 kN # m2

dv

dx=

1EI32.25x2

- 0.5x3+ 5.258x - 592


1206


Support Reactions and Diagram. As shown in Fig. a,

Moment Area Theorem. Referring to Fig. b,

From the geometry shown in Fig. b,

Ans.uA =

|tB>A|

LAB=

3.375EI

3=

1.125 kN # m2

EI=

1.125(103)

200(109) cp4

(0.054)d= 0.00115 rad

|tB>A| =

153

c12

a 9EIb(1.5)d =

3.375 kN # m3

EI

MEI

12–59. Determine the slope at A of the solid circular shaftof diameter 100 mm. The shaft is made of steel having amodulus elasticity of E = 200 GPa. A B C

1.5 m1.5 m

6 kN 6 kN

1.5 m

Ans:uA = 0.00115 rad


1222


12–75. Determine the slope at B and the deflection at C ofthe beam. E = 200 GPa and I = 65.0(106) mm4.

Support Reactions and Diagram. As shown in Fig. a.

Moment Area Theorem. Referring to Fig. b,

From the geometry of the elastic curve, Fig. b,

Ans.

and

Ans.= 0.0138 m = 13.8 mm T

=

180 kN # m3

EI=

180(103)

200(109)[65.0(10- 6)]

=

513EIa3

6b -

76.5EI

¢C = |tA>B|aLBC

Lb - |tC>B|

= 0.00658 rad=

85.5(103)

200(109)[65.0(10- 6)]

uB =

|tA>B|

L=

513>EI

6=

85.5 kN # m2

EI

=

513 kN # m3

EI

|tA>B| = c13

(3) + 3 d c 12

a 51EIb(3) d + c2

3 (3) d c 1

2 a 69

EIb(3) d

=

76.5 kN # m3

EI

|tC>B| = c13

(3) d c 12

a 51EIb(3) d

MEI

A

CB

3 m 3 m

18 kN�m

40 kN

Ans:, ¢C = 13.8 mmTuB = 0.00658 rad


1264


(2)

(3)

Solving Eqs. (2) and (3) yields:

Ans.

c

Ans.MB =

11wL2

192

MB +

3wL

32 (L) -

5wL2

192-

wL

2 aL

4b = 0+ © MB = 0;

MA =

5wL2

192

Ay =

3wL

32

0 =

AyL

6-

MA

2-

wL2

384

tB>A = 0 =

12

aAyL

EIb(L)aL

3b + a -MA

EIb(L)aL

2b +

13

a -wL2

8EIb aL

2b aL

8b

0 =

AyL

2- MA -

wL2

48

uB>A = 0 =

12

aAyL

EIb(L) + a -MA

EIb(L) +

13

a -wL2

8EIb aL

2b

12–111. Determine the moment reactions at the supports A and B. EI is constant.

L–2

A B

w

L–2

Ans:

MA =

5wL2

192, MB =

11wL2

192


1268


Support Reactions: FBD(a).

Ans.

[1]

a [2]

Elastic Curve: As shown.

M/EI Diagrams: M/EI diagrams for P and By acting on a simply supported beam are

drawn separately.

Moment-Area Theorems:

From the elastic curve,

Ans.

Substituting By into Eqs. [1] and [2] yields,

Ans.Cy =

13P

32 Ay =

3P

32

By =

11P

16

7PL3

16EI-

By L3

2EI= 2a5PL3

48EI-

By L3

12EIb

tA>C = 2tB>C

tB>C = (tB>C)1 + (tB>C)2 =

5PL3

48EI-

By L3

12EI

tA>C = (tA>C)1 + (tA>C)2 =

7PL3

16EI-

By L3

2EI

(tB>C)2 =

12

a -

By L

2EIb(L)aL

3b = -

By L3

12EI

=

5PL3

48EI

+ 12

a3PL

8EIb aL

2b aL

2+

L

6b

(tB>C)1 =

12a PL

8EIb aL

2b a2

3b aL

2b + a PL

4EIb aL

2b aL

4b

(tA>C)2 =

12

a -

By L

2EIb(2L)(L) = -

By L3

2EI

=

7PL3

16EI

(tA>C)1 =

12

a 3PL

8EIb a3L

2b a2

3b a3L

2b +

12a3PL

8EIb aL

2b a3L

2+

L

6b

+ ©MA = 0; By (L) + Cy (2L) - Pa3L2b = 0

+ c ©Fy = 0; -Ay + By + Cy - P = 0

:+ ©Fx = 0; Bx = 0

12–115. Determine the vertical reactions at the bearingssupports, then draw the shear and moment diagrams. EI isconstant.

CA B

L

P

L2

L2

Ans:

Ay =

3P

32Cy =

13P

32,By =

11P

16,

Hibbeler_12(1267-1273).qxd 3/6/13 4:48 PM 68

1271


Referring to the FBD of the beam, Fig. a

Ans.

(1)

a

(2)

Referring to Fig. b and the table in Appendix C, the necessary deflections at B arecomputed as follow:

The compatibility condition at support B requires that

Ans.

Substitute this result into Eq (1) and (2)

Ans.Ay =

3P

4 MA =

PL

4

By =

7P

4

0 =

7PL3

12EI+ a -By L3

3EIb

(+ T) 0 = vP + vBy

yBy=

PL3AB

3EI=

By L3

3EI c

=

7PL3

12EI T

=

P(L2)

6EI c3a3

2 Lb - L d

yP =

Px2

6EI (3LAC - x)

MA = By L -

32

PL

+ ©MA = 0; -MA + By L - Pa32

Lb = 0

Ay = By - P

+ c ©Fy = 0; By - P - Ay = 0

:+ ©Fx = 0; Ax = 0

12–118. Determine the reactions at the supports A and B.EI is constant.

A BL2

P

L

Ans:

, , , MA =

PL

4Ay =

3P

4By =

7P

4Ax = 0


1272


Support Reaction: FBD(b).

Ans.

[1]

a [2]

Method of Superposition: Using the table in Appendix C, the requireddisplacements are

The compatibility condition requires

Ans.

Substituting By into Eqs. [1] and [2] yields,

Ans.Ay = 2.625 kip Cy = 14.6 kip

By = 30.75 kip

0 =

6480EI

+

2376EI

+ a -

288By

EIb

(+ T) 0 = yB¿ + yB– + yB– ¿

yB– ¿ =

PL3

48EI=

By A243 B48EI

=

288By ft3

EI c

=

12(6)(12)

6EI(24) A242

- 62- 122 B =

2376 kip # ft3

EI T

yB– =

Pbx

6EIL AL2

- b2- x2 B

yB¿ =

5wL4

768EI=

5(3) A244 B768EI

=

6480 kip # ft3

EI T

+ ©MA = 0; By (12) + Cy (24) - 12(6) - 36.0(18) = 0

+ c ©Fy = 0; Ay + By + Cy - 12 - 36.0 = 0

:+ ©Fx = 0; Cx = 0

12–119. Determine the reactions at the supports A, B, andC, then draw the shear and moment diagrams. EI is constant.

6 ft 12 ft

3 kip/ ft

A BC

6 ft

12 kip

Ans:

Cy = 14.6 kipAy = 2.625 kip,By = 30.75 kip,Cx = 0,


1275


Referring to the FBD of the beam, Fig. a,

Ans.

(1)

a

(2)

Referring to Fig. b and the table in the appendix, the necessary deflections are:

Compatibility condition at roller support B requires

Ans.

Substitute this result into Eq. (1) and (2)

Ans.Ay =

3M0

2L MA =

M0

2

By =

3M0

2L

0 =

M0L2

2EI+ a -

ByL3

3EIb

(+ T) 0 = vM0

+ (vB)y

vBy=

PL3

3EI=

ByL3

3EI c

vM0=

M0L2

2EI T

MA = ByL - M0

+ ©MA = 0; By(L) - M0 - MA = 0

+ c ©Fy = 0; By - Ay = 0

:+ ©Fx = 0; Ax = 0

12–122. Determine the reactions at the supports A and B.EI is constant.

L

A

M0

B

Ans:

, , , MA =

M0

2Ay =

3M0

2LBy =

3M0

2LAx = 0


1276


Support Reactions: FBD (a).

Ans.

a [1]

Method of Superposition: Using the table in Appendix C, the requireddisplacements are

The compatibility condition requires

Substituting By into Eq. [1] yields,

Ans.Cy =

P

3

By =

2P

3

By L3

48EI=

PL3

24EI+ a -

By L3

24EIb

(+ T) yB = yB¿ + yB–

yB– =

PLBD3

3EI=

By L3

24EI c

yB¿ =

PLBD3

3EI=

P AL2 B33EI

=

PL3

24EI T

yB =

PL3

48EI=

By L3

48EI T

+ ©MA = 0; Cy(L) - By aL

2b = 0

:+ ©Fx = 0; Cx = 0

12–123. Determine the reactions at support C. EI is thesame for both beams.

A C

D

P

B

L2

L2

Ans:

, Cy =

P

3 Cx = 0


12–6. p 1 x3 ei a - auburn universityhtippur/mech3130/hw-7 solutions-2014.pdf · determine the...

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