civilittee-hu.comcivilittee-hu.com/uploads/1/Static/solution_13th/statics_13th... · 7–3. The...

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7

Three torques act on the shaft. Determine the internal torque at points A, B, C, and D.

Given:

M1 300 N m��

M2 400 N m��

M3 200 N m��

Solution:

Section A:

��x = 0; TA� M1� M2� M3� 0�

TA M1 M2� M3��

TA 100.00 N m��

Section B:

�Mx = 0; TB M3� M2� 0�

TB M3� M2��

TB 200.00 N m��

Section C:

��x = 0; TC� M3� 0�

TC M3�

TC 200.00 N m��

Section D:

��x = 0; TD 0�

2.

Ans.

Ans.

Ans.

Ans.

–

7–3.

The strongback or lifting beam is used for materialshandling. If the suspended load has a weight of 2 kN and acenter of gravity of G, determine the placement d of thepadeyes on the top of the beam so that there is no momentdeveloped within the length AB of the beam. The liftingbridle has two legs that are positioned at 45°, as shown.

SOLUTIONSupport Reactions: From FBD (a),

a

From FBD (b),

Internal Forces: This problem requires Summing moments about point Hof segment EH [FBD (c)], we have

a

Ans.d = 0.200 m

- 1.414 cos 45°10.22 = 0

1.001d + x2 - 1.414 sin 45°1x2+ ©MH = 0;

MH = 0.

FAC = FBC = F = 1.414 kN

2F sin 45° - 1.00 - 1.00 = 0+ c ©Fy = 0;

FAC cos 45° - FBC cos 45° = 0 FAC = FBC = F:+ ©Fx = 0;

FF + 1.00 - 2 = 0 FF = 1.00 kN+ c ©Fy = 0;

FF162 - 2132 = 0 FE = 1.00 kN+ ©ME = 0;

45° 45°

3 m 3 m

0.2 m0.2 m

dd

E

A B

F

G

7

Determine the normal force, shear force, and moment at a section passing through point D of thetwo-member frame.

Units Used:

kN 103 N�

Given:

w 0.75kNm

�

F 4 kN�

a 1.5 m� d 1.5 m�

b 1.5 m� e 3�

c 2.5 m� f 4�

Solution:

� MC = 0;

Bx� c d�( )f

e2 f 2�

�

��

F d� 0�

Bxf d F

e2 f 2� c d�( )� Bx 1.2 kN�

� MA = 0;

w� c d�( )c d�

2�

�� By a b�( )� Bx c d�( )� 0�

By

wc d�( )2

2

��

��

Bx c d�( )�

a b�� By 0.40 kN�

�� Fx = 0; ND� Bx� 0� ND Bx�� ND 1.2� kN�

�� Fy = 0; VD By� 0� VD By�� VD 0.4� kN�

� MD = 0; MD� By b� 0� MD By b� MD 0.6 kN m��

4.

Ans.

Ans.

Ans.

–

7–5.

Determine the internal normal force, shear force, andmoment at points A and B in the column.

SOLUTION

Applying the equation of equilibrium to Fig. a gives

Ans.

Ans.

a

Ans.

and to Fig. b,

Ans.

Ans.

a

Ans.MB = 14.3 kN # m3(1.5) + 8(0.4) + 6 sin 30°(2.9) - 6 cos 30°(0.4) - MB = 0+©MB = 0;

NB = 16.2 kNNB - 3 - 8 - 6 cos 30° = 0+ c©Fy = 0;

VB = 3 kNVB - 6 sin 30° = 0©Fx = 0;:+

MA = 3.82 kN # m8(0.4) + 6 sin 30°(0.9) - 6 cos 30°(0.4) - MA = 0+©MA = 0;

NA = 13.2 kNNA - 6 cos 30° - 8 = 0+ c©Fy = 0;

VA = 3 kNVA - 6 sin 30° = 0©Fx = 0;:+

8 kN6 kN

3 kN

A

B

0.4 m 0.4 m

1.5 m

0.9 m

2 m

30�

7–6.

SOLUTION

a

a

Ans.a =L

3

2PL aL3

- ab = 0

M =2P AL3 - a BL - a

aL3b = 0+©M = 0;

Cy =2P AL3 - a BL - a

-P a2L3

- ab + Cy1L - a2 + Pa = 0+©MA = 0;

Determine the distance a as a fraction of the beam’s lengthL for locating the roller support so that the moment in thebeam at B is zero.

L

AB

C

a L/3

P P

Determine the internal normal force, shear force, and moment at points E and D of thecompound beam.

Given:

M 200 N m�� c 4 m�

F 800 N� d 2 m�

a 2 m� e 2 m�

b 2 m�

Solution:

Segment BC :

M� Cy d e�( )� 0� CyM

d e��

By� Cy� 0� By Cy�

Segment EC :

NE� 0� NE 0N� NE 0.00�

VE Cy� 0� VE Cy�� VE 50.00� N�

ME� M� Cy e� 0� ME Cy e M�� ME 100.00� N m��

Segment DB :

ND� 0� ND 0N� ND 0.00�

VD F� By� 0� VD F By�� VD 750.00 N�

MD� F b� By b c�( )� 0�

MD F� b By b c�( )�� MD 1300� N m�� Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

7–7.

7– . Determine the internal normal force, shear force,and moment at point C.

3 m 2 m

1.5 m

1 m

0.2 m 400 N

AC

B

8

.

.

.

7–9.

Determine the normal force, shear force, and moment at asection passing through point C. Take P = 8 kN.

SOLUTIONa

Ans.

Ans.

a

Ans.MC = 6 kN # m+©MC = 0; -MC + 8(0.75) = 0

VC = - 8 kN

+ c©Fy = 0; VC + 8 = 0

NC = -30 kN

:+ ©Fx = 0; -NC - 30 = 0

+ c©Fy = 0; Ay = 8 kN

:+ ©Fx = 0; Ax = 30 kN

T = 30 kN

+©MA = 0; -T(0.6) + 8(2.25) = 0

0.75 m

C

P

A

B

0.5 m0.1 m

0.75 m 0.75 m

7–10.

SOLUTIONa

Ans.

Ans.

Ans.

a

Ans.MC = 0.400 kN # m+©MC = 0; -MC + 0.533(0.75) = 0

VC = -0.533 kN

+ c©Fy = 0; VC - 0.533 = 0

NC = -2 kN

:+ ©Fx = 0; -NC - 2 = 0

+ c©Fy = 0; Ay = 0.533 kN

:+ ©Fx = 0; Ax = 2 kN

P = 0.533 kN

+©MA = 0; -2(0.6) + P(2.25) = 0

The cable will fail when subjected to a tension of 2 kN.Determine the largest vertical load P the frame will supportand calculate the internal normal force, shear force, andmoment at a section passing through point C for this loading.

0.75 m

C

P

A

B

0.5 m0.1 m

0.75 m 0.75 m

7– . Determine the internal normal force, shear force,and moment at points D and E in the compound beam.Point E is located just to the left of the 10-kN concentratedload.Assume the support at A is fixed and the connection atB is a pin.

10 kN2 kN/m

D

B

EC

A

1.5 m 1.5 m 1.5 m 1.5 m

11

7–12.

Determine the internal normal force, shear force, and themoment at points C and D.

SOLUTIONSupport Reactions: FBD (a).

a

Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)],we have

Ans.

Ans.

a

Ans.

Applying the equations of equilibrium to segment BD [FBD (c)], we have

Ans.

Ans.

a

Ans.MD = 16.5 kN # m8.485132 - 611.52 - MD = 0+ ©MD = 0;

VD + 8.485 - 6.00 = 0 VD = -2.49 kN+ c ©Fy = 0;

ND = 0:+ ©Fx = 0;

MC = 4.97 kN # mMC - 3.515 cos 45°122 = 0+ ©MC = 0;

3.515 sin 45° - NC = 0 NC = 2.49 kNa+©Fy¿ = 0;

3.515 cos 45° - VC = 0 VC = 2.49 kNQ+ ©Fx¿ = 0;

:+ ©Fx = 0 Ax = 0

Ay + 8.485 - 12.0 = 0 Ay = 3.515 kN+ c ©Fy = 0;

By = 8.485 kN

By 16 + 6 cos 45°2 - 12.013 + 6 cos 45°2 = 0+ ©MA = 0;

2 kN/m

3 m 3 m

BD

C

A

6 m

2 m

45˚

7– . Determine the internal normal force, shear force, andmoment at point C in the simply supported beam.

CBA

3 m

4 kN/m

3 m

13

7–14.

Determine the normal force, shear force, and moment at asection passing through point D. Take w = 150 N>m.

SOLUTION

a

Ans.

Ans.

a

Ans.MD = 1200 N # m = 1.20 kN # m

-600142 + 150142122 + MD = 0+ ©MD = 0;

VD = 0

600 - 150142 - VD = 0+ c ©Fy = 0;

ND = -800 N:+ ©Fx = 0;

Ay = 600 N

Ay - 150182 +35110002 = 0+ c ©Fy = 0;

Ax = 800 N

Ax -45110002 = 0:+ ©Fx = 0;

FBC = 1000 N

-150182142 +35FBC182 = 0+ ©MA = 0;

4 m

AD

B

C

4 m

4 m

3 m

w

7–15.

SOLUTIONAssume maximum moment occurs at D;

a

a

(O.K.!)

Ans.w = 100 N/m

FBC = 666.7 N 6 1500 N

+©MA = 0; - 800(4) + FBC (0.6)(8) = 0

w = 100 N/m

800 = 4w(2)

+©MD = 0; MD - 4w(2) = 0

The beam AB will fail if the maximum internal moment atD reaches or the normal force in member BCbecomes 1500 N. Determine the largest load it cansupport.

w800 N # m

4 m

AD

B

C

4 m

4 m

3 m

w

7–16.

Determine the internal normal force, shear force, andmoment at point D in the beam.

SOLUTION

Writing the equations of equilibrium with reference to Fig. a, we have

a

a

Using these results and referring to Fig. b, we have

Ans.

Ans.

a Ans.

The negative sign indicates that ND, VD, and MD act in the opposite sense to thatshown on the free-body diagram.

MD = -300 N # mMD + 600(1)(0.5) = 0+©MD = 0;

VD = -600 N-VD - 600(1) = 0+ c©Fy = 0;

ND = -1350 N = -1.35 kNND + 1350 = 0©Fx = 0;:+

Ax = 1350 NAx - 2250¢25≤ = 0©Fx = 0;:+

Ay = 0600(3)(0.5) - 900 - Ay(2) = 0+©MB = 0;

FBC = 2250 NFBC¢45≤(2) - 600(3)(1.5) - 900 = 0+©MA = 0;

A D B

C

1 m

600 N/m

900 N�m

1 m 1 m

43

5

7–17.

SOLUTION

a

Ans.

Ans.

a

Ans.ME = 2400 N # m = 2.40 kN # m

-5

13120802132 +

121312080212.52 - ME = 0+ ©ME = 0;

VE = 800 N

VE -5

13120802 = 0+ c ©Fy = 0;

NE = -1920 N = -1.92 kN

-NE -1213

120802 = 0:+ ©Fx = 0;

FBC = 2080 N

-1200142 +5

13 FBC162 = 0+ ©MA = 0;

Determine the normal force, shear force, and moment at asection passing through point E of the two-member frame.

3 m

6 m

A DB

C E

2.5 m

400 N/m

7– . Determine the internal normal force, shear force, andmoment at point C in the cantilever beam.

AB

C

w0

L––2

L––2

18

7–19.

Determine the internal normal force, shear force, andmoment at points E and F in the beam.

SOLUTION

With reference to Fig. a,

a

Use these result and referring to Fig. b,

Ans.

Ans.

a

Ans.

Also, by referring to Fig. c,

Ans.

Ans.

a

Ans.

The negative sign indicates that VF acts in the opposite sense to that shown on thefree-body diagram.

MF = 660 N # m664.92(1.5) - 300(1.5)(0.75) - MF = 0+©MF = 0;

VF = -215 N

VF + 664.92 - 300 = 0+ c©Fy = 0;

NF = 0©Fx = 0;:+

ME = 660 N # mME + 300(1.5)(0.75) - 664.92(1.5) = 0+©ME = 0;

VE = 215 N

664.92 - 300(1.5) - VE = 0+ c©Fy = 0;

NE = 470 N

NE - 470.17 = 0©Fx = 0;:+

Ay = 664.92 NAy + 664.92 sin 45° + 664.92 - 300(6) = 0+ c©Fy = 0;

Ax = 470.17 N664.92 cos 45° - Ax = 0©Fx = 0;:+T = 664.92 NT(6) + T sin 45°(3) - 300(6)(3) = 0+©MA = 0;

D BA E F

1.5 m300 N/m

45�

1.5 m 1.5 m 1.5 m

C

7– . Determine the internal normal force, shear force,and moment at points C and D in the simply supportedbeam. Point D is located just to the left of the 5-kN force.

AC D

B

3 kN/m

5 kN

3 m1.5 m 1.5 m

20

7– . Determine the internal normal force, shear force,and moment at points D and E in the overhang beam. PointD is located just to the left of the roller support at B, wherethe couple moment acts.

2 kN/m

5 kN

3 m 1.5 m 34

5

AD B E

C

6 kN � m

1.5 m

21

7–22.

Determine the internal normal force, shear force, andmoment at points E and F in the compound beam. Point F islocated just to the left of the 15-kN force and couple moment.

SOLUTION

With reference to Fig. b, we have

a

a

Using these results and referring to Fig. a, we have

a

With these results and referring to Fig. c,

Ans.

Ans.

a

Ans.

Also, using the result of Dy referring to Fig. d, we have

Ans.

Ans.

a Ans.

The negative sign indicates that VE acts in the opposite sense to that shown in thefree-body diagram.

MF = 2.5 kN # m13.75(2) - 25 - MF = 0+©MF = 0;

VF = 1.25 kNVF - 15 + 13.75 = 0+ c©Fy = 0;

NF = 0©Fx = 0;:+

ME = 4.97 kN # mME + 3(2.25) - 3(2.25)(8.125) = 0+©ME = 0;

VE = -1.17 kN5.583 - 3(2.25) - VE = 0+ c©Fy = 0;

NE = 0©Fx = 0;:+

Ay = 5.583 kN3(6)(1.5) - 1.25(1.5) - Ay(4.5) = 0+©MB = 0;

Ax = 0©Fx = 0;:+

Cy = 1.25 kN15(2) - 25 - Cy(4) = 0+©MD = 0;

Dy = 13.75 kNDy(4) - 15(2) - 25 = 0+©MC = 0;

Cx = 0©Fx = 0;:+

25-kN # m

2 m1.5 m2.25 m2.25 m

2 m

15 kN

25 kN�m

3 kN/m

A

E FCD

B

7–23.

Determine the internal normal force, shear force, andmoment at points D and E in the frame. Point D is locatedjust above the 400-N force.

SOLUTION

With reference to Fig. a, we have

a

Using this result and referring to Fig. b, we have

Ans.

Ans.

a

Ans.

Also, by referring to Fig. c, we can write

Ans.

Ans.

a

Ans.

The negative sign indicates that ND, NE, and VE acts in the opposite sense to thatshown in the free-body diagram.

ME = 190 N # m335.34 cos 30°(1) - 200(1)(0.5) - ME = 0+©ME = 0;

VE = -90.4 NVE + 335.34 cos 30° - 200(1) = 0+ c©Fy = 0;

NE = -168 N-NE - 335.34 sin 30° = 0©Fx = 0;:+

MD = 348 N # m335.34 cos 30°(2) + 335.34 sin 30°(1) - 200(2)(1) - MD = 0+©MD = 0;

ND = -110 N335.34 cos 30° - 200(2) - ND = 0+ c©Fy = 0;

VD = 168 NVD - 335.34 sin 30° = 0©Fx = 0;:+

FB = 335.34 N

FB cos 30°(2) + FB sin 30°(2.5) - 200(2)(1) - 400(1.5) = 0+©MA = 0;

1 m

2 m

1 m

200 N/m

400 N 30�

1.5 m

A

D

BE

C

2.5 m

7–24.

SOLUTIONFree body Diagram: The support reactions at A need not be computed.

Internal Forces: Applying equations of equilibrium to segment BC, we have

Ans.

Ans.

a

Ans.MC = -302 kN # m-24.011.52 - 12.0142 - 40 sin 60°16.32 - MC = 0+ ©MC = 0;

VC = 70.6 kN

VC - 24.0 - 12.0 - 40 sin 60° = 0+ c ©Fy = 0;

-40 cos 60° NC = 0 NC = -20.0 kN:+ ©Fx = 0;

Determine the internal normal force, shear force, andbending moment at point C.

A3 m 3 m

0.3 m

CB

8 kN/m40 kN

3 m

60°

-

7– . Determine the internal normal force, shear force,and moment at point C in the double-overhang beam.

A C B

1.5 m

3 kN/m

1.5 m 1.5 m 1.5 m

25

7–26.

SOLUTION

a

Since ,

Ans.a

b=

14

- a + b =34

b

-16b

(a - b) =18

-16b

(2a + b)(a - b) =14

(2a + b)a 12b

VC = 0

+ c©Fy = 0; -w6b

(2a + b)(a - b) -w4aa +

b

2b - VC = 0

:+ ©Fx = 0; Ax = 0

Ay =w6b

(2a + b)(a - b)

+©MB = 0; -w2

(2a + b) c23

(2a + b) - (a + b) d + Ay (b) = 0

Determine the ratio of for which the shear force will bezero at the midpoint C of the beam.

a>b

BCA

a b/2 b/2

w

a

A BC

7–27.

Determine the normal force, shear force, and moment at asection passing through point D of the two-member frame.

SOLUTION

a

Ans.

Ans.

a

Ans.MD = 1350 N # m = 1.35 kN # m+©MD = 0; -800(3) + 600(1.5) + 150(1) + MD = 0

VD = 50 N

+ c©Fy = 0; 800 - 600 - 150 - VD = 0

:+ ©Fx = 0; ND = 2400 N = 2.40 kN

Ay = 800 N

+ c©Fy = 0; Ay - 1200 - 600 +513

(2600) = 0

:+ ©Fx = 0; Ax =1213

(2600) = 2400 N

FBC = 2600 N

+©MA = 0; -1200(3) - 600(4) +513

FBC (6) = 0

3 m

200 N/m

400 N/m

D

B

A

6 mC

2.5 m

7– . Determine the internal normal force, shear force, andmoment at point C in the simply supported beam. Point C islocated just to the right of the 1500-lb ft couple moment.–

BA

C

500 lb/ft

1500 lb � ft

6 ft

30�

6 ft

FB = 3608.44 kN

2.5 kN · m

2.5 kN · m

2 m 2 m

10 kN/m

2 m 2 m

1 m 1 m

10 (2) kN10 (4) kN

located just to the right of the 2.5 kN · m couple moment.

Writing the moment equation of equilibrium about point A with reference to Fig. a,

�

+ ΣMA = 0; FB cos 30° (4) – 10 (4) (2) – 2.5 = 0 FB = 23.816 kN

Using the result of FB and referring to Fig. b,

+→ ΣFx = 0; –NC – 23.816 sin 30° = 0 NC = –11.908 kN Ans.

+↑ΣFy = 0; VC + 23.816 cos 30° – 10 (2) = 0 VC = –0.625 kN Ans.

�

+ ΣMC = 0; 23.816 cos 30° (2) – 10 (2) (1) – MC = 0 MC = 21.25 kN · m Ans.

The negative sign indicates that NC and VC act in the opposite sense to that shown on the free – body diagram.

82

7–2 . Determine the internal normal force, shear force,and moment at point D of the two-member frame.

2 m1.5 m

250 N/m

300 N/m

4 m

A

C

D

E

B

9

.

.

.

7– . Determine the internal normal force, shear force,and moment at point E of the two-member frame.

2 m1.5 m

250 N/m

300 N/m

4 m

A

C

D

E

B

30

.

.

.

A

45�

75 mm

4 kN

4 kN

7–3 .

The hook supports the 4-kN load. Determine the internalnormal force, shear force, and moment at point A.

SOLUTION


Ans.

Ans.

a

Ans.MA = 0.212 kN # m = 212 N # m4 sin 45°(0.075) - MA = 0+©MA = 0;

NA = 2.83 kNNA - 4 sin 45° = 0+a©Fy¿ = 0;

VA = 2.83 kNVA - 4 cos 45° = 0+Q©Fx¿ = 0;

1

7– . Determine the internal normal force, shear force,and moment at points C and D in the simply supportedbeam. Point D is located just to the left of the 10-kNconcentrated load.

AC D

B

1.5 m

6 kN/m10 kN

1.5 m 1.5 m 1.5 m

32

7– . Determine the distance a in terms of the beam’slength L between the symmetrically placed supports Aand B so that the internal moment at the center of thebeam is zero.

w0 w0

A B

L

a––2

a––2

33

7– . The beam has a weight w per unit length. Determinethe internal normal force, shear force, and moment at pointC due to its weight.

B

A

CL––2

L––2

u

34

7–35.

0.5 m

800 N

1 m

1 m

1 m

1 m

0.5 m

0.5 m

0.25 m

A B

C

D

E

F

G

H

Determine the internal normal force, shear force, andbending moment at points E and F of the frame.

SOLUTIONSupport Reactions: Members HD and HG are two force members. Using method ofjoint [FBD (a)], we have

From FBD (b),

a (1)

From FBD (c),

a (2)

Solving Eqs. (1) and (2) yields,

Internal Forces: Applying the equations of equilibrium to segment DE [FBD (d)],we have

Ans.

Ans.

a Ans.

Applying the equations of equilibrium to segment CF [FBD (e)], we have

Ans.

Ans.

a

Ans.MF = 224 N # mMF + 894.4310.52 - 500 cos 26.57°11.52 = 0+©MF = 0;

NF - 500 sin 26.57° = 0 NF = – 224 N+©Fy¿ = 0;

VF = 447 N

VF + 500 cos 26.57° - 894.43 = 0+Q©Fx¿ = 0;

ME = 0+©ME = 0;

894.43 - NE = 0 NE = 894 N+©Fy¿ = 0;

VE = 0+ ©Fx¿ = 0;

Cy = 0 Cx = 500 N

894.43112 - Cx 12 cos 26.57°2 + Cy 12 sin 26.57°2 = 0+©MA = 0;

Cx 12 cos 26.57°2 + Cy 12 sin 26.57°2 - 894.43112 = 0+©MA = 0;

FHD = FHG = F = 894.43 N

2F sin 26.57° - 800 = 0+ c ©Fy = 0;

FHD = FHG = F

:+ ©Fx = 0 FHG cos 26.57° - FHD cos 26.57° = 0

Q

QQ -

7–3 .

SOLUTIONSupport reactions: FBD(a)

Moments Function:

a

Ans.a =L

3

+©M = 0; 0 +12

(w0)aL

2b a 1

3b aL

2b -

14

w0 Laa

2b = 0

Determine the distance a between the supports in terms ofthe shaft’s length L so that the bending moment in thesymmetric shaft is zero at the shaft’s center. The intensity ofthe distributed load at the center of the shaft is . Thesupports are journal bearings.

w0

L

a

w0

6

7– 7. Determine the internal normal force, shear force,and moment acting at point C. The cooling unit has a totalmass of 225 kg with a center of mass at G.

3 m

F

3 m

30� 30�

0.2 m

G

A B

ED

C

3

.

.

.

7– . Determine the internal normal force, shear force,and moment at points D and E of the frame which supportsthe 100-kg crate. Neglect the size of the smooth peg at C. C

B

E

A

D

38

981

1.35 m

1.2 m

0.6 m

981 N

100 (9.81) N =981 N

100 (9.81) N

1.35 m

1677.5 N·m = 1.68 kN·m

1765.8 N = 1.77 kN

588.6 N = 589 N

1324.35 N·m = 1.32 kN·m

981 N

(0.6) = 0 981 981 (1.35) –

981 981

981

– 981 (1.35) = 0 Ans.

Ans.

1.35 m

1.2 m

0.6 m

0.45 m

0.45 m

7–39.

The semicircular arch is subjected to a uniform distributedload along its axis of per unit length. Determine theinternal normal force, shear force, and moment in the archat u = 45°.

w0

SOLUTIONResultants of distributed load:

At

Ans.

Ans.

a

Ans.M = - 0.0783 r2 w0

+©Mo = 0; -M + r2 w0 ap4 b + (-0.707 r w0)(r) = 0

N = - 0.707 rw0

N = - 0.707 rw0 cos 45° - 0.2929 rw0 sin 45°

+a©Fy = 0; N + FRy cos u + FRx sin u = 0

V = - 0.293 r w0

V = 0.2929 r w0 cos45° - 0.707 r w0 sin 45°

+b©Fx = 0; -V + FRx cos u - FRy sin u = 0

u = 45°

MRo = Lu

0w0 (r du) r = r2 w0u

FRy = Lu

0w0 (r du) cos u = r w0 (sinu) ` u

0= r w0 (sinu)

FRx = Lu

0w0 (r du) sin u = r w0 (-cos u) ` u

0= r w0 (1 - cos u)

O

r

w0

u

7–40.

SOLUTIONResultants of distributed load:

At ,

Ans.

Ans.

a

Ans.M = 1.23 r2w0

+©Mo = 0; -M + r2 w0 (p)a120°180°

b + (-0.866 r w0)r = 0

V = -1.5 rw0

+a©Fy¿ = 0; V + 1.5 r w0 sin 30° + 0.86603 r w0 cos 30° = 0

N = -0.866 r w0

+b©Fx¿ = 0; N + 1.5 r w0 cos 30° - 0.86603 r w0 sin 30° = 0

FRy = r w0 sin 120° = 0.86603 r w0

FRx = r w0 (1 - cos 120°) = 1.5 r w0

u = 120°

MRo = Lu

0w0 (r du) r = r2 w0u

FRy = Lu

0w0 (r du) cos u = r w 0 (sin u) ` u

0= r w0 (sin u)

FRx = Lu

0w0 (r du) sinu = r w0(-cos u) ` u

0= r w0 (1 - cosu)

The semicircular arch is subjected to a uniform distributedload along its axis of per unit length. Determine theinternal normal force, shear force, and moment in the archat .u = 120°

w0

O

r

w0

u

7– . Determine the x, y, z components of internal loadingat a section passing through point C in the pipe assembly.Neglect the weight of the pipe.Take and F2 = 5150i - 300k6 kN.

F1 = 5350j - 400k6 kN

x

z

y

C

1.5 m

2 m

F1

F2

3 m

41

kN

kN

kN

kN

kN·m

kN·m

MN·m

MN·m

2m

3m

.

.

.

.

.

.

7– . Determine the x, y, z components of internal loading ata section passing through point C in the pipe assembly. Neglectthe weight of the pipe. Take and F2 = 5250i - 150j - 200k6 kN.

F1 = 5-80i + 200j - 300k6 kN

x

z

y

C

1.5 m

2 m

F1

F2

3 m

42

2m

3m

kN

kN

kN

kN

kN·m

kN·m

kN·m

MN·m .

.

.

.

7 4

Determine the x, y, z components of internal loading in the rod at point D.

Units Used:

kN 103 N�

Given:

M 3 kN m��

F7

12�

5�

�

��

kN�

a 0.75 m�

b 0.2 m�

c 0.2 m�

d 0.6 m�

e 1 m�

Solution:

Guesses

Cx 1 N� Cy 1 N� Bx 1 N�

Bz 1 N� Ay 1 N� Az 1 N�

Given

0

Ay

Az

�

��

Bx

0

Bz

�

��

�

Cx

Cy

0

�

��

� F� 0�

e�

b c� d�

0

�

��

0

Ay

Az

�

��

�

0

b c�

0

�

��

Bx

0

Bz

�

��

��

0

0

a

�

��

Cx

Cy

0

�

��

��

0

b c� d�

0

�

��

F��

0

0

M�

�

��

� 0�

3.–

Ay

Az

Bx

Bz

Cx

Cy

�

��

Find Ay Az� Bx� Bz� Cx� Cy��

Ay

Az

Bx

Bz

Cx

Cy

�

��

53.60�

87.00

109.00

82.00�

116.00�

65.60

�

��

kN�

Guesses

VDx 1 N� NDy 1 N�

VDz 1N� MDx 1 N m��

MDy 1 N m�� MDz 1 N m��

Given

Cx

Cy

0

�

��

VDx

NDy

VDz

�

��

� 0�

0

b�

a

�

��

Cx

Cy

0

�

��

�

MDx

MDy

MDz

�

��

�

0

0

M�

�

��

� 0�

VDx

NDy

VDz

MDx

MDy

MDz

�

��

Find VDx NDy� VDz� MDx� MDy� MDz��

VDx

NDy

VDz

�

��

116.00

65.60�

0.00

�

��

kN�

MDx

MDy

MDz

�

��

49.20

87.00

26.20

�

��

kN m��

Ans.

Ans.

Determine the x, y, z components of internal loading in the rod at point E.

Units Used:

kN 103 N�

Given:

M 3 kN m��

F7

12�

5�

�

��

kN�

a 0.75 m�

b 0.4 m�

c 0.6 m�

d 0.5 m�

e 0.5 m�

Solution:

Guesses

Cx 1 N� Cy 1 N� Bx 1 N�

Bz 1 N� Ay 1 N� Az 1 N�

Given

0

Ay

Az

�

��

Bx

0

Bz

�

��

�

Cx

Cy

0

�

��

� F� 0�

d� e�

b c�

0

�

��

0

Ay

Az

�

��

�

0

b0

�

��

Bx

0

Bz

�

��

��

0

0

a

�

��

Cx

Cy

0

�

��

��

0

b c�

0

�

��

F��

0

0

M�

�

��

� 0�

7 44.–

Ay

Az

Bx

Bz

Cx

Cy

�

��

Find Ay Az� Bx� Bz� Cx� Cy��

Ay

Az

Bx

Bz

Cx

Cy

�

��

53.60�

87.00

109.00

82.00�

116.00�

65.60

�

��

kN�

Guesses

NEx 1 N� VEy 1 N�

VEz 1 N� MEx 1 N m��

MEy 1 N m�� MEz 1 N m��

Given

0

Ay

Az

�

��

NEx

VEy

VEz

�

��

� 0�

e�

0

0

�

��

0

Ay

Az

�

��

�

MEx

MEy

MEz

�

��

� 0�

NEx

VEy

VEz

MEx

MEy

MEz

�

��

Find NEx VEy� VEz� MEx� MEy� MEz��

NEx

VEy

VEz

�

��

0.00

53.60

87.00�

�

��

kN�

MEx

MEy

MEz

�

��

0.00

43.50�

26.80�

�

��

kN m��

Ans.

Ans.

b

A

B

C

a

P

7–45.

Draw the shear and moment diagrams for the overhang beam.

SOLUTION

Since the loading discontinues at B, the shear stress and moment equation must bewritten for regions and of the beam. The free-bodydiagram of the beam’s segment sectioned through an arbitrary point in these tworegions are shown in Figs. b and c.

Region

(1)

a (2)

Region

(3)

a (4)

The shear diagram in Fig. d is plotted using Eqs. (1) and (3), while the momentdiagram shown in Fig. e is plotted using Eqs. (2) and (4). The value of moment at Bis evaluated using either Eqs. (2) or (4) by substituting

M ƒ x=b = - Pa

b(b) = -Pa or M ƒ x=b = -P(a + b - b) = -Pa

x = b; i.e.,

M = -P(a + b - x)-M - P(a + b - x) = 0+©M = 0;

V = PV - P = 0©Fy = 0;

b 6 x … a + b, Fig. c

M = - Pa

bxM +

Pa

bx = 0+©M = 0;

V = - Pa

b- Pa

b- V = 0+ c©Fy = 0;

0 … x 6 b, Fig. b

b 6 x … a + b0 … x 6 b

.

.

.

.

7–4 . Draw the shear and moment diagrams for thebeam (a) in terms of the parameters shown; (b) set

L = 12 ft.a = 5 ft,P = 800 lb,

a a

L

P P

1.5 m4 kN

4 kN

4 kN4 kN

4 kN4 kN

V (kN)

M (kN · m) –4

4

6

1.5 m 1.5 m0.6 m

3.6-x

4 kN

4 kN

P = 4 kN, a = 1.5 m, L = 3.6 m.

(a) For 0 � x < a

+↑ΣFy = 0; V = P Ans.

�

+ ΣM = 0; M = Px Ans.

For a < x < L – a

+↑ΣFy = 0; V = 0 Ans.

�

+ ΣM = 0; –Px + P(x – a) + M = 0

M = Pa Ans.

For L – a < x � L

+↑ΣFy = 0; V = –P Ans.

�

+ ΣM = 0; –M + P(L – x) = 0

M = P(L – x) Ans.

(b) Set P = 4 kN, a = 1.5 m, L = 3.6 m

For 0 � x < 1.5 m

+↑ΣFy = 0; V = 4 kN Ans.

�

+ ΣM = 0; M = 4x kN · m Ans.

For 1.5 m < x < 2.1 m

+↑ΣFy = 0; V = 0 Ans.

�

+ ΣM = 0; –4(x) + 4(x – 1.5) + M = 0

M = 6 kN · m Ans.

For 2.1 m < x � 3.6 m

+↑ΣFy = 0; V = –4 kN Ans.

�

+ ΣM = 0; –M + 4(3.6 – x) = 0

M = (14.4 – 4x) kN · m Ans.

7

7–48.

SOLUTION(a)

For

Ans.

a Ans.

For

Ans.

a Ans.

For

Ans.

a Ans.

(b)

Set ,

For

Ans.

c Ans.

For

Ans.

c Ans.

For

Ans.

c Ans.+ ©M = 0; M = 0

+ c ©Fy = 0; V = 0

163

m 6 x … 8 m

+ ©M = 0; M = 500 N # m

+ c ©Fy = 0; V = 0

83

m 6 x 6163

m

+ ©M = 0; M = 0

+ c ©Fy = 0; V = 0

0 … x 683

m

L = 8 mM0 = 500 N # m

+ ©M = 0; M = 0

+ c ©Fy = 0; V = 0

2L

36 x … L

+ ©M = 0; M = M0

+ c ©Fy = 0; V = 0

L

36 x 6

2L

3

+ ©M = 0; M = 0

+ c ©Fy = 0; V = 0

0 … x …L

3

Draw the shear and moment diagrams for the beam (a) interms of the parameters shown; (b) set L = 8 m.

M0 = 500 N # m,

L/3 L/3 L/3

M0 M0

7–49.

If the beam will fail when the maximum shearforce is or the maximum bending moment is

Determine the magnitude of thelargest couple moments it will support.

M0Mmax = 2 kN # m.Vmax = 5 kN

L = 9 m,

SOLUTIONSee solution to Prob. 7–48 a.

Ans.Mmax = M0 = 2 kN # m

L/3 L/3 L/3

M0 M0

7– Draw the shear and moment diagrams for thecantilever beam.

2 kN/m

6 kN � m2 m

A

50.

.

.

.

.

7–52.

SOLUTIONSupport Reactions:

a

Shear and Moment Functions: For [FBD (a)],

Ans.

a Ans.

For 2 m<x ◊ 3 m [FBD (b)],

Ans.

a

Ans.M = -1.2x 2 + 5.2 x - 4.8 kN m#0.4x - 2.41x - 22a x - 2

2b - M = 0+ ©M = 0;

V = 5 - x6 kN

0.4 - 1x - 22 - V = 0+ c ©Fy = 0;

M - 0.4x = 0 M = 10.4x2 kN m#+ ©M = 0;

0.4 - V = 0 V = 0.4 kN+ c ©Fy = 0;

0 ◊ x<2 m

Ay - 2.4 + 2.00 = 0 Ay = 0.40 kN+ c ©Fy = 0;

Cy132 - 2.412.52 = 0 Cy = 2.00 kN+ ©MA = 0;

Draw the shear and moment diagrams for the beam.

CAB

2 m

3 m

0.40

0.40

2.4

0.4

–2

0.8 0.833

2.4 kN/m

2.4

2.405.20

7–53.


SOLUTION

Ans.

a

Ans.

Ans.

c

Ans.M = 20x - 320

+ ©M = 0; M + 20(11 - x) + 100 = 0

V = 20

+ c ©Fy = 0; V - 20 = 0

8 6 x … 11

M = 140x - 20x 2

+ ©M = 0; M + 40xax

2b - 140x = 0

V = 140 - 40x

+ c ©Fy = 0; 140 - 40x - V = 0

0 … x 6 8

40 kN/m20 kN

100 kN m

AB C

8 m 3 m

140

140 kN 200 kN

3.5 m

245

180

–160 7m 100

100 kN·m

7–54. Draw the shear and moment diagrams for thesimply supported beam.

300 N/m

4 m

300 N � m

A B

7–55. Draw the shear and moment diagrams for thesimply supported beam.

300 N/m

4 m

100 N � m

A B

575

575 x 100

575

575 x 100

575

1.917 m

1.917 m

M|x = 1.917 m = –150(1.9172) + 575(1.917) + 100 = 651 N·m

100 N·m

575 N 625 N

100 N·m

575 N

575

1.917

–625

651

1.917

100

7–56.

Draw the shear and bending-moment diagrams for beamABC. Note that there is a pin at B.


a

From FBD (b),

Shear and Moment Functions: For

Ans.

Ans.M =w4A3Lx - 2x2 - L2 B

+ ©M = 0; 3wL

4(x) - wxax

2b -

wL2

4- M =

wL2

4=

0

V =w4

(3L - 4x)

+ c ©Fy = 0;3wL

4- wx - V = 0

0 … x … L [FBD (c)],

+ c ©Fy = 0; AywL

2--

wL

4= 0 Ay =

3wL

4

+ ©MC = 0;

a+ ©MA MA= 0;

wL

2aL

4b - By aL

2b = 0 By =

wL

4

AC

w

B

L2

L2

aL

4bwL

2aL

2bwL

4= 0MA - -

c

7–57. Draw the shear and moment diagrams for the beam. 250 lb/ ft

150 lb � ft150 lb � ft

A B

20 ft6 m 250 N · m250 N · m

5 kN/m

6 m

6 m

6 m

30 kN

15 kN

15 kN

250 N · m

250 N · m250 N · m

250 N · m

5 kN/m

15 kN

–15

0.0167 m

–0.250–0.250

0.0167 m

V (kN)

15

22.25M (kN · m)

250 N · m5x

�

+ ΣMA = 0; –30 (3) + By (6) = 0

By = 15 kN

+→ ΣFx = 0; Ax = 0

+↑ΣFy = 0; Ay – 30 + 15 = 0

Ay = 15 kN

For 0 � x � 3 m

+→ ΣFx = 0; 15 – 5x – V = 0

V = 5 (3 – x) Ans.

�

+ ΣM = 0; –15 (x) + 0.250 + 5x

2

⎛

⎝⎜

⎞

⎠⎟ + M = 0

M = 1

2(30x – 5x2 – 0.5)

M = 15x – 2.5x2 – 0.25 Ans.

7–58.

Draw the shear and moment diagrams for the compoundbeam. The beam is pin-connected at E and F.

A

L

w

B E F CD

L––3

L––3

L––3

L

SOLUTIONSupport Reactions: From FBD (b),

a

From FBD (a),

a

From FBD (c),

a

Shear and Moment Functions: For [FBD (d)],

Ans.

a

Ans.

For [FBD (e)],

Ans.

a

Ans.

For [FBD (f)],

Ans.

Ans.

a7wL

183L - x - w 3L - x

3L - x

2- M = 0+ ©M = 0;

V =w18147L - 18x2

V +7wL

18- w13L - x2 = 0+ c ©Fy = 0;

2L<x ◊ 3L

M =w18127Lx - 20L2 - 9x22

M =w18147Lx - 60L2- 9x2 2

M + wxax

2b -

7wL

18 x -

10wL

91x - L2 = 0+ ©M = 0;

V =w213L - 2x2

7wL

18+

10wL

9- wx - V = 0+ c ©Fy = 0;

L ◊ x<2L

M =w1817Lx - 9x22

M + wxax

2b -

7wL

18x = 0+ ©M = 0;

V =w18

17L - 18x2

7wL

18- wx - V = 0+ c ©Fy = 0;

0 ◊ x<L

By +7wL

18-

4wL

3-

wL

6= 0 By =

10wL

9+ c ©Fy = 0;

4wL

3aL

3b -

wL

6aL

3b - Ay 1L2 = 0 Ay =

7wL

18+ ©MB = 0;

Dy 1L2 +wL

6aL

3b -

4wL

3aL

3b = 0 Dy =

7wL

18+ ©MC = 0;

Ey +wL

6-

wL

3= 0 Ey =

wL

6+ c ©Fy = 0;

FyaL

3b -

wL

3aL

6b = 0 Fy =

wL

6+ ©ME = 0;

7–59.

SOLUTION

a

Mmax = 0.75(1.732) - 0.08333(1.732)3 = 0.866

M = 0.75 x - 0.08333 x3

+ ©M = 0; M + a12b(0.5 x) (x) a1

3xb - 0.75 x = 0

x = 1.732 m

V = 0 = 0.75 - 0.25x2

V = 0.75 - 0.25x2

+ c ©Fy = 0; 0.75 -12

x (0.5x) - V = 0


A B

3 m

1.5 kN/m

7

The beam will fail when the maximum internal moment is Mmax. Determine the position x of theconcentrated force P and its smallest magnitude that will cause failure.

Solution:

For ��< x, M1P� L x�( )

L�

For � > x, M2P x L ��

L�

Note that M1 = M2 when x = �

Mmax M1� M2�P x L x�( )

L�

PL

L x x2��

xL x x2�� d

dL 2x�� x

L2

�

Thus, MmaxPL

L2

��

LL2

��

�P2

L2

��

�

P4Mmax

L�

60.

Ans.

Ans.

–

7–61.


SOLUTION

Ans.Support Reactions: From FBD (a),

a

Shear and Moment Functions: For [FBD (b)],

Ans.

The maximum moment occurs when then

a

Ans.

Thus,

Mmax = 3.0013.4642 -3.4643

12= 6.93 kN # m

M = b3.00x -x3

12r kN # m

M + ¢x2

4≤ ax

3b - 3.00x = 0+ ©M = 0;

0 = 3.00 -x2

4x = 3.464 m

V = 0,

V = b3.00 -x2

4r kN

3.00 -x2

4- V = 0+ c ©Fy = 0;

0 … x … 6 m

9.00122 - Ay162 = 0 Ay = 3.00 kN+ ©MB = 0;

3 m

6 m

A

C

B

3 kN/m

7–62.

SOLUTIONBy similar triangles

Ans.

a Ans.+ ©M = 0; -M-ght

2dx2 ax

3b = 0 M = -

ght

6dx3

+ c ©Fy = 0; V -ght

2dx2 = 0 V =

ght

2dx2

W = gV = ga12

yxtb = g c12ah

dxbxt d =

ght

2dx2

y

x=

h

dy =

h

dx

The cantilevered beam is made of material having a specificweight Determine the shear and moment in the beam asa function of x.

g.

h

dx

t

7–63.


AB

C

4 m 2 m

8 kN/m

SOLUTION:

;

a ;

:

;

a ;

M = - 7.5x - 25

M + 101x - 2.52 - 2.5x = 0+ ©M = 0

V = - 7.5

2.5 - 10 - V = 0+ c ©Fy = 0

5 … x 6 10 m

M = 2.5x - x2

M + 2xa12

xb - 2.5x = 0+ ©M = 0

V = 2.5 - 2x

2.5 - 2x - V = 0+ c ©Fy = 0

0 … x 6 5 m

7–64.


SOLUTION

The free-body diagram of the beam’s segment sectioned through an arbitrary pointshown in Fig. b will be used to write the shear and moment equations. Themagnitude of the resultant force of the parabolic distributed loading and thelocation of its point of application are given in the inside back cover of the book.

Referring to Fig. b, we have

(1)

a (2)

The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1)and (2), respectively. The location at which the shear is equal to zero can beobtained by setting in Eq. (1).

The value of the moment at x = 0.630L is evaluated using Eq. (2).

M ƒ x= 0.630L =w0

12L2 cL3(0.630L) - (0.630L)4 d = 0.0394w0L2

x = 0.630L0 =w0

12L2 aL3 - 4x3b

V = 0

M +13

¢w0

L2x2≤(x)¢x

4≤ -

w0L

12x = 0 M =

w0

12L2aL3x - x4b+ ©M = 0;

V =w0

12L2aL3 - 4x3bw0L

12-

13aw0

L2x2bx - V = 0+ c ©Fy = 0;

A BC

w0

w

x

w0––L2w � x2

L––2

L––2

7–65.


a


Ans.

a

Ans.

For [FBD (c)],

Ans.

a

Ans.M = 5-15017 - x226 N # m

-30017 - x2a7 - x

2b - M = 0+ ©M = 0;

V = 52100 - 300x6 N

V - 30017 - x2 = 0+ c ©Fy = 0;

3 m<x … 7 m

M = 5-650x - 16.7x36 N # m

M + 150.0x22ax

3b + 650x = 0+ ©M = 0;

V = 5-650 - 50.0x26 N

-650 - 50.0x2 - V = 0+ c ©Fy = 0;

0 … x<3 m

Ay 132 + 450112 - 1200122 = 0 Ay = 650 N+ ©MB = 0;

Draw the shear and bending-moment diagrams for the beam.

AB

C

4 m

300 N/m

3 m

7–66.



a


Ans.

The maximum moment occurs when then

a

Ans.

Thus,

Ans.= 0.0940wL2

Mmax =w

12L34L210.5275L2 - 3L10.5275L22 - 10.5275L234

M =w

12L14L2x - 3Lx2 - x32

M +12a w

2Lxbxax

3b +

wx

2ax

2b -

wL

31x2 = 0+ ©M = 0;

0 = 4L2 - 6Lx - 3x2 x = 0.5275L

V = 0,

V =w

12L14L2 - 6Lx - 3x22

wL

3-

w2

x -12a w

2Lxbx - V = 0+ c ©Fy = 0;

0 … x … L

wL

4aL

3b +

wL

2aL

2b - Ay 1L2 = 0 Ay =

wL

3+ ©MB = 0;

w

L

w––2

A B

7–67.

Determine the internal normal force, shear force, andmoment in the curved rod as a function of , where

.

SOLUTION


a

Using this result and referring to Fig. b, we have

Ans.

Ans.

a Ans.M =pr

2 (1 - cos u)

p

2 3r (1 - cos u)4 - M = 0+ ©M = 0;

N =p

2 cos u

p

2 cos u - N = 0©Fy¿ = 0;

V =p

2 sin u

p

2 sin u - V = 0©Fx¿ = 0;

By = p>2By(2r) - p(r) = 0+ ©MA = 0;

0° … u … 90°u

P

r

u

7 6

Determine the normalforce, shear force, andmoment in the curved rodas a function of ��

Given:

c 3�

d 4�

Solution:

For 0 ��

�Fx = 0; Nd

c2 d2�

�

��

P cos �� c

c2 d2�

�

��

P sin �� 0�

NP

c2 d2�

�

��

d cos �� c sin ��

�Fy = 0; Vd

c2 d2�

�

��

P sin �� c

c2 d2�

�

��

P cos �� 0�

VP

c2 d2�

�

��

d sin �� c cos ��

�M = 0;d�

c2 d2�

�

��

P r rcos �� c

c2 d2�

�

��

P rsin �� M� 0�

MP r

c2 d2�

�

��

d d cos �� c sin ��

8.

Ans.

Ans.

Ans.

–

7 6

The quarter circular rod lies in the horizontalplane and supports a vertical force P at itsend. Determine the magnitudes of thecomponents of the internal shear force,moment, and torque acting in the rod as afunction of the angle �.

Solution:

�Fz = 0; V P�

�Mx = 0; M P rcos �� 0�

M P� rcos ��

M P rcos ��

�My = 0; T P r l sin �� 0�

T P� r l sin ��

T P r 1 sin ��

9.

Ans.

Ans.

Ans.

–

7–70.


a

Ay = 6.50 kN

Ay + 9.50 - 2 - 2 - 2 - 2 - 4 - 4 = 0+ c ©Fy = 0;

By = 9.50 kN

-213.252 - 212.252 - 211.252 = 0

By 182 - 417.252 - 416.252 - 214.252+ ©MA = 0;


1.25 m 1 m 1 m 1 m

2 kN 2 kN 2 kN 2 kN4 kN 4 kN

1 m 0.75 m2 m

BA

7–71.


SOLUTION

2 m 2 m

7 kN

A B

12 kN m

4 m

7–72.



a

Ay + 625a35b - 500 - 500 = 0 Ay = 625 N+ c ©Fy = 0;

FCa35b142 - 500122 - 500112 = 0 FC = 625 N+ ©MA = 0;

A B C2 m

250 N/m

500 N

3 m

2 m

7–74.

Draw the shear and moment diagrams for the simply-supported beam.

SOLUTION

w0

2w0

L/2 L/2

A B

7–75.

Draw the shear and moment diagrams for the beam. Thesupport at A offers no resistance to vertical load.

SOLUTIONL

A B

w0

7–76.



a

Ay + 8.90 - 10.0 - 8 = 0 Ay = 9.10 kN+ c ©Fy = 0;

By 1102 - 10.012.52 - 8182 = 0 By = 8.90 kN+ ©MA = 0;

8kN

2 kN/m

5 m 3 m 2 m

A B

8

9.10 8.90

9.10

4.55 –0.90

–8.90

20.7 20.5 17.8

4.55

7–77.

The shaft is supported by a thrust bearing at A and a journalbearing at B. Draw the shear and moment diagrams for theshaft.

SOLUTION

600 N

BA

300 N�m

300 N/m

1.5 m 0.75 m 0.75 m

7–79.


SOLUTION

2 m 1 m 2 m

8 kN

A B

15 kN/m20 kN m

3 m

7–80.

Draw the shear and moment diagrams for the compoundsupported beam.

SOLUTION

3 m 3 m1.5 m 1.5 m

5 kN

3 kN/m

AB C D

7–82.

Draw the shear and moment diagrams for the overhangbeam.

SOLUTION

AB

3 m 3 m 2 m

4 kN/m

6 kN·m

3 kN

7– . Draw the shear and moment diagrams for theshaft. The support at A is a journal bearing and at B it is athrust bearing.

1 ft 4 ft 1 ft

100 lb/ft

A 300 lb � ft

200 lb

B

0.3 m 1.2 m 0.3 m

1000 N 2000 N/m

500 N · m

500 N · m

M (N · m)

V (m)

–1000

2000 N/m

1.2 m 0.3 m

2033.33 N 1366.67 N

0.3 m

1000 N

1033.33

–1366.67

0.517 m

–500–33.1

–300

0.817

x (m)

x (m)

83

7 84.

Draw the shear and moment diagrams forthe beam.

Units Used:

kN 103N �

Given:

w1 20kNm

� w2 10kNm

� a 1.5m �

Solution:

x 0 0.01 a�� a��

V x( ) w2 x� w1xa

��

��

x2

��

��

1kN

� � M x( ) w2 x�x2

� w1xa

��

��

x2

�x3

��

��

1kN m�

� �

0 0.2 0.4 0.6 0.8 1 1.2 1.40

2

4

Distance (m)

Forc

e (k

N)

V x( )

xm

0 0.2 0.4 0.6 0.8 1 1.2 1.40

2

4

Distance (m)

Mom

ent (

kN-m

)

M x( )

xm

–

7–85.


SOLUTION

18 kN�m

3 m

BA C

3 m

2 kN/m 2 kN/m

7–86.


a

Ay =w0L

3

+ c ©Fy = 0; Ay +7w0L

6- w0L -

w0L

2= 0

By =7w0L

6

+ ©MA = 0; By(L) - w0LaL2 b -w0L

2a4L

3b = 0

Draw the shear and moment diagrams for the beam. w0

AB

LL

7–87.


a

Ay - w0L -w0L

2= 0 Ay =

3w0L

2+ c ©Fy = 0;

MA =7w0L

2

1 2

MA - w0L aL4 b -w0L

2a2L

3b = 0+ ©MA = 0;

Draw the shear and moment diagrams for the beam. w0

L––2

L––2

2

2 2

2

12

00

22w L w L

01 (2 )

2 2 2w LLw

12

12

7–88.

SOLUTION


L/2

w

L/2

w

7–89.

The shaft is supported by a smooth thrust bearing at A and asmooth journal bearing at B. Draw the shear and momentdiagrams for the shaft.

SOLUTION

300 N

300 N�mBA

100 N/m

1.5 m 1.5 m 3 m

6 kN3 kN/m

6 m1.5 m

AB

7–90.


SOLUTION

The maximum span moment occurs at the position where shear is equal to zerowithin the region of the beam. This location can be obtained using themethod of sections. By setting Fig. b, we have

Ans.

Using this result,

M ƒ x=1.76 m = 3.73 kN # m

M ƒ x=1.76 m +12

(6 - 1.76)(1.76)a1.762b +

14

(1.76)(1.76) c23

(1.76) d - 4.5(1.76) = 0+©M = 0;

x = 1.76 m4.5 -12

a12

xbx -12

(6 - x)(x) = 0 + c©Fy = 0;

V = 0,0 … x 6 6 m

7

The beam consists of three segmentspin connected at B and E. Draw theshear and moment diagrams for thebeam.

Units Used: kN 103 N�

Given: MA 8 kN m�� F 15 kN� w 3kNm

� a 3 m� b 2 m�

c 2 m� d 2 m� e 2 m� f 4 m�

Guesses Ay 1 N� By 1 N� Cy 1 N�

Dy 1 N� Ey 1 N� Fy 1 N�Given

Ay Cy� Dy� Fy� F� w f� 0� F b MA� Ay a b�( )� 0�

w� ff2

��

Fy f� 0� MA F a� By a b�( )� 0�

By Cy� Dy� Ey� 0� By� c Dy d� Ey d e�( )� 0�

Ay

By

Cy

Dy

Ey

Fy

�

��

Find Ay By� Cy� Dy� Ey� Fy��

Ay

By

Cy

Dy

Ey

Fy

�

��

4.40

10.60�

15.20

1.40

6.00�

6.00

�

��

kN�

x1 0 0.01a� a��

V1 x( ) Ay1

kN� M1 x( ) Ay x MA�� 1

kN m��

x2 a 1.01a� a b��

V2 x( ) Ay F�� 1kN

� M2 x( ) Ay x MA� F x a�( )�� 1

kN m��

x3 a b� 1.01 a b�( )� a b� c��

V3 x( ) By1

kN� M3 x( ) By x a� b�( )

1kN m�

�

x4 a b� c� 1.01 a b� c�( )� a b� c� d��

V4 x( ) By Cy�� 1kN

� M4 x( ) By x a� b�( ) Cy x a� b� c�( )�� 1

kN m��

92.–

x5 a b� c� d� 1.01 a b� c� d�( )� a b� c� d� e��

V5 x( ) Ey�1

kN� M5 x( ) Ey a b� c� d� e� x�( )

1kN m�

�

x6 a b� c� d� e� 1.01 a b� c� d� e�( )� a b� c� d� e� f��

V6 x( ) Fy� w a b� c� d� e� f� x�( )�� 1

kN�

M6 x( ) Fy a b� c� d� e� f� x�( ) wa b� c� d� e� f� x�( )2

2�

��

��

1kN m�

�

0 2 4 6 8 10 12 1415

10

5

0

5

10

Distance (m)

Forc

e (k

N)

V1 x1� �V2 x2� �V3 x3� �V4 x4� �V5 x5� �V6 x6� �

x1 x2� x3� x4� x5� x6�

0 2 4 6 8 10 12 1430

20

10

0

10

20

30

Distance (m)

Mom

ent (

kN-m

)

M1 x1� �M2 x2� �M3 x3� �M4 x4� �M5 x5� �M6 x6� �

x1 x2� x3� x4� x5� x6�

7 3.

Draw the shear and momentdiagrams for the beam.

Units Used:

kN 103 N�

Given:

w 3kNm

� a 3 m� b 3 m�

Solution:

Ay� a b�( )wb2

��

2b3

��

�wa2

��

ba3

��

� 0� Ay

wb2

3wa2

��

ba3

��

�

a b��

Ay By�w2

a b�( )� 0� Byw2

a b�( ) Ay��

x1 0 0.01a� a��

V1 x( ) Ay12

wxa

�

��

x��

��

1kN

� M1 x( ) Ay x12

wxa

�

��

xx3

��

��

��

1kN m�

�

x2 a 1.01a� a b��

V2 x( ) By�12

wa b� x�

b�

��

a b� x�( )��

��

1kN

�

M2 x( ) By a b� x�( )12

wa b� x�

b�

��

a b� x�( )a b� x�

3�

��

��

1kN m�

�

0 1 2 3 4 5 65

0

5

Distance (m)

Forc

e (k

N)

V1 x1� �V2 x2� �

x1 x2�

9–

0 1 2 3 4 5 65

0

5

10

Distance (m)

Mom

ent (

kN-m

)

M1 x1� �M2 x2� �

x1 x2�

7–9 Determine the tension in each segment of thecable and the cable’s total length. Set .P = 80 lb

P

A

B

C

D

2 ft

3 ft

50 lb

5 ft

4 ft3 ft

0.6 m

1.5 m

0.9 m1.2 m0.9 m

250 N

0.9 m 1.2 m

0.9 m

250 N

250 N

Ax = 210.112 kN

Ay = 314.763 kN

TBD = 390.935 kN

400 N

cable and the cable’s total length. Set P = 400 N.

From FBD (a)

�

+ ΣMA = 0; TBD cos 59.04° (0.9) + TBD sin 59.04° (2.1) – 250 (2.1) – 400 (0.9) = 0

TBD = 390.935 N = 390.9 N Ans.

+→ ΣFx = 0; 390.935 cos 59.04° – Ax = 0

Ax = 210.112 N

+↑ΣFy = 0; Ay + 390.935 sin 59.04° – 400 – 250 = 0

Ay = 314.763 N

Joint A :

+→ ΣFx = 0; TAC cos � – 210.112 = 0 (1)

+↑ΣFy = 0; –TAC sin � + 314.763 = 0 (2)

Solving Eqs. (1) and (2) yields :

� = 56.28°

TAC = 378.4 N Ans.

Joint D :

+→ ΣFx = 0; 390.935 cos 59.04° – TCD cos � = 0 (3)

+↑ΣFy = 0; 390.935 sin 59.04° – TCD sin � – 250 = 0 (4)

Solving Eqs. (3) and (4) yields :

� = 22.97°

TCD = 218.4 N Ans.

Total length of the cable:

lT = 1.5

sin 59.04° +

1.2

cos 22.97° +

0.9

cos 56.28° = 4.674 m Ans.

4.

7–9 If each cable segment can support a maximum tensionof 75 lb, determine the largest load P that can be applied.

P

A

B

C

D

2 ft

3 ft

50 lb

5 ft

4 ft3 ft

1.5 m

0.6 m

0.9 m1.2 m0.9 m

250 N

0.6 m

0.9 m

0.9 m 1.2 m

250 N

Ay = 288.394 N

Ax = 192.915 N

TAC = 346.969 N

� = 56.22°

359.966 N

of 375 N, determine the largest load P that can be applied.

�

+ ΣMA = 0; –TBD (cos 59.04°) (0.6) + TBD (sin 59.04°) (3) – 250 (2.1) – P (0.9) = 0

TBD = 0.39754P + 231.899

+→ ΣFx = 0; –Ax + TBD cos 59.04° = 0

+↑ΣFy = 0; Ay – P – 250 + TBD sin 59.04° = 0

Assume maximum tension is in cable BD,

TBD = 375 N

P = 359.966 N

Ax = 192.915 N

Ay = 288.394 N

Pin A :

TAC = (192.915) + (288.394)2 2 = 346.969 N < 375 N OK

� = tan–1 288.394

192.915

⎛⎝⎜

⎞⎠⎟

= 56.22°

Joint C :

+→ ΣFx = 0; TCD cos � – 346.969 cos 56.22° = 0

+↑ΣFy = 0; TCD sin � + 346.969 sin 56.22° – 359.966 = 0

TCD = 205.766 N < 375 N OK

� = 20.355°

Thus, P = 360 N Ans.

5.

7 9

Cable ABCD supports the lamp of mass M1 and the lamp of mass M2. Determine the maximumtension in the cable and the sag of point B.

Given:

M1 10 kg�

M2 15 kg�

a 1 m�

b 3 m�

c 0.5 m�

d 2 m�

Solution:

Guesses yB 1 m� TAB 1 N� TBC 1 N� TCD 1 N�

Givena�

a2 yB2�

�

��

TABb

b2 yB d�� 2�

��

��

TBC� 0�

yB

a2 yB2�

�

��

TAByB d�

b2 yB d�� 2�

��

��

TBC� M1 g� 0�

b�

b2 yB d�� 2�

��

��

TBCc

c2 d2�

�

��

TCD� 0�

yB d��

b2 yB d�� 2�

��

��

TBCd

c2 d2�

�

��

TCD� M2 g� 0�

yB

TAB

TBC

TCD

�

��

Find yB TAB� TBC� TCD��

TAB

TBC

TCD

�

��

100.163

38.524

157.243

�

��

N�

Tmax max TAB TBC� TCD�� Tmax 157.2 N� yB 2.43 m�

6.

Ans.

–

7–9 . The cable supports the three loads shown. Determinethe sags and of points B and D. Take P2 = 250 lb.

P1 = 400 lb,yDyB 4 ft

12 ft 20 ft 15 ft 12 ft

AE

B

C

D

yB yD14 ft

P2 P2

P1

1.2 m

4.2 m

3.6 m 6 m 4.5 m 3.6 m

the sags yB and yD of points B and D. Take P1 = 2000 N, P2 = 1250 N.

1250 N

63.6

4.2 – yD4.2 – yB

yD + 1.24.2 – yD

2000 N

1250 N

3.6

4.5

4.53.6

yB4.2 – yB

At B

+→ ΣFx = 0;

6

(4.2 – ) + 362yB

TBC – 3.6

+ 12.962yB

TAB = 0

+↑ΣFy = 0; –4.2 –

(4.2 – ) + 362

y

y

B

B

TBC + y

y

B

B2 + 12.96

TAB – 1250 = 0

9.6 – 15.12

(4.2 – ) + 362

y

y

B

B

TBC = 4500 (1)

At C

+→ ΣFx = 0;

4.5

(4.2 – ) + 20.252yD

TCD – 6

(4.2 – ) + 362yB

TBC = 0

+↑ΣFy = 0; 4.2 –

(4.2 – ) + 20.252

y

y

D

D

TCD + 4.2 –

(4.2 – ) + 362

y

y

B

B

TBC – 2000 = 0

–6 + 44.1 – 4.5

(4.2 – ) + 362

y y

y

D B

B

TBC = 9000 (2)

–6 + 44.1 – 4.5

(4.2 – ) + 20.252

y y

y

D B

D

TCD = 12000 (3)

At D

+→ ΣFx = 0;

3.6

(1.2 + ) + 12.962yD

TDg – 4.5

(4.2 – ) + 20.252yD

TCD = 0

+↑ΣFy = 0; 1.2 +

(1.2 + ) + 12.962

y

y

D

D

TDg – 4.2 –

(4.2 – ) + 20.252

y

y

D

D

TCD – 1250 = 0

–9.72 + 8.1

(4.2 – ) + 20.252

y

y

D

D

TCD = 4500 (4)

Combining Eqs. (1) & (2)

23.7yB + 6yD = 74.34

Combining Eqs. (3) & (4)

13.5yB + 82.8yD = 210.06

yB = 2.602 m Ans.

yD = 2.113 m Ans.

7

7–9 The cable supports the three loads shown.Determine the magnitude of if and

Also find the sag yD.yB = 8 ft.P2 = 300 lbP1 4 ft

12 ft 20 ft 15 ft 12 ft

AE

B

C

D

yB yD14 ft

P2 P2

P1

1.2 m

4.2 m

3.6 m 6 m 4.5 m 3.6 m

Determine the magnitude of P1 if P2 = 1500 N and yB = 2.4 m. Also find the sag yD.

1500 N

4.2 – yD

61.8

2.43.6

4.51.8

6

4271.04 N

yD + 1.24.2 – yD

1500 N4.5

3.6

At B

+→ ΣFx = 0;

6

39.24TBC –

3.6

18.72TAB = 0

+↑ΣFy = 0; –1.8

39.24TBC +

2.4

18.72TAB – 1500 = 0

TAB = 4916.66 N

TBC = 4271.03 N

At C

+→ ΣFx = 0;

–6

39.24(4271.04) +

4.5

(4.2 – ) + 20.252yD

TCD = 0 (1)

+↑ΣFy = 0; 1.8

39.24(4271.04) +

4.2 –

(4.2 – ) + 20.252

y

y

D

D

TCD – P1 = 0 (2)

At D

+→ ΣFx = 0;

3.6

(1.2 + ) + 12.962yD

TDg – 4.5

(4.2 – ) + 20.252yD

TCD = 0

+↑ΣFy = 0; 1.2 +

(1.2 + ) + 12.962

y

y

D

D

TDg – 4.2 –

(4.2 – ) + 20.252

y

y

D

D

TCD – 1500 = 0

TCD = 5400 20.25 + (4.2 – )

8.1 – 9.72

2y

yD

D

Substitute into Eq. (1) :

yD = 1.933 m Ans.

TCD = 4580.6 N

P1 = 3288.1 N Ans.

8.

7–99.

If cylinders E and F have a mass of 20 kg and 40 kg,respectively, determine the tension developed in each cableand the sag yC.

SOLUTION

First, TAB will be obtained by considering the equilibrium of the free-body diagramshown in Fig. a. Subsequently, the result of TAB will be used to analyze theequilibrium of joint B followed by joint C. Referring to Fig. a, we have

a

Ans.

Using the free-body diagram shown in Fig. b, we have

Solving,

Ans.

Using the result of and the geometry of the cable, yC is given by

Ans.

Using the results of yC, , and TBC and analyzing the equilibrium of joint C, Fig. c,we have

Ans.

(Check!)281.85 sin 28.44° + 357.86 sin 46.17° - 40(9.81) = 0+ c©Fy = 0;

TCD = 357.86 N = 358 N

TCD cos 46.17° - 281.85 cos 28.44° = 0©Fx = 0;:+

u

yC = 3.083 m = 3.08 m

yC - 2

2= tan u = 28.44°

u

TBC = 281.85 N = 282 N

u = 28.44°

413.05a45b - 20(9.81) - TBC sin u = 0+ c©Fy = 0;

TBC cos u - 413.05a35b = 0©Fx = 0;:+

TAB = 413.05 N = 413 N

40(9.81)(2) + 20(9.81)(4) - TABa35b(1) - TABa4

5b(4) = 0+©MD = 0;

2 m

1 m

1.5 m 2 m 2 m

B

E

F

C

D

A

yC

7–100.

If cylinder E has a mass of 20 kg and each cable segment cansustain a maximum tension of 400 N, determine the largest massof cylinder F that can be supported. Also, what is the sag yC?

SOLUTION

We will assume that cable AB is subjected to the greatest tension, i.e., TAB = 400 N.Based on this assumption, MF can be obtained by considering the equilibrium of thefree-body diagram shown in Fig. a. We have

a+

Ans.

Analyzing the equilibrium of joint B and referring to the free-body diagram shownin Fig. b, we have

Solving,

Using these results and analyzing the equilibrium of joint C,

Solving,

By comparing the above results, we realize that cable AB is indeed subjected to thegreatest tension. Thus,

Ans.

Using the result of either or , the geometry of the cable gives

Ans.

or

Ans.yC = 3.03 m

yC - 1

2= tan f = tan 45.45°

yC = 3.03 m

yC - 2

2= tan u = tan 27.29°

fu

MF = 37.5 kg

TCD = 342.11 Nf = 45.45°

TCD sin f + 270.05 sin 27.29° - 37.47(9.81) = 0+ c©Fy = 0;

TCD cos f - 270.05 cos 27.29° = 0©Fx = 0;:+

TBC = 270.05 N

u = 27.29°

400a45b - 20(9.81) - TBC sin u = 0+ c©Fy = 0;

TBC cos u - 400a35b = 0©Fx = 0;:+

MF = 37.47 kg

MF(9.81)(2) + 20(9.81)(4) - 400a35b(1) - 400a4

5b(4) = 0©MD = 0;

2 m

1 m

1.5 m 2 m 2 m

B

E

F

C

D

A

yC

7–10 . If h = 5 m, determine the maximum tensiondeveloped in the chain and its length. The chain has a massper unit length of .8 kg>m A B

50 m

h � 5 m

1

7–10 . The cable supports the uniform distributed loadof . Determine the tension in the cable ateach support A and B.

w0 = 600 lb>ftA

w0

B

25 ft

10 ft

15 ft

7.5 m

3 m

4.5 m

4.5 m

7.5 – x

3 m

of w0 = 12 kN/m. Determine the tension in the cable at

Use the equations of Example 7.12.

y = w

FH

0

2x2

4.5 = 12

2 FH

x2

3 = 12

2 FH

(7.5 – x)2

12

2 (4.5)x2 =

12

2 (3)(7.5 – x)2

x2 = 1.5 (56.26 – 15x + x2)

0.5x2 – 22.5x + 84.375 = 0

Choose root < 7.5 m

x = 4.1288 m

FH = w

y0

2x2 =

12

2 (4.5) (4.1288)2 = 22.729 kN

At B :

y = w

FH

0

2x2 =

12

2 (22.729)x2

dy

dx = tan �B = 0.52796 x |x = 4.1288 = 2.180

�B = 65.36°

TB = FH

Bcos � =

22.729

cos 65.36° = 54.52 kN Ans.

At A :

y = w

FH

0

2x2 =

12

2 (22.729)x2

dy

dx = tan �A = 0.52796 x |x = (7.5 – 4.1288) = 1.780

�A = 60.67°

TA = FH

Acos � =

22.729

cos 60.67° = 46.40 kN Ans.

2

7–103. Determine the maximum uniform distributedload the cable can support if the maximum tension thecable can sustain is 4000 lb.

w0

A

w0

B

25 ft

10 ft

15 ft

7.5 m

3 m

4.5 m

4.5 m

7.5 – x

3 m

cable can sustain is 20 kN.

Use the equations of Example 7.12.

y = w

FH

0

2x2

4.5 = w

FH

0

2x2

3 = w

FH

0

2(7.5 – x)2

x2

4.5 =

1

3(7.5 – x)2

x2 = 1.5 (56.26 – 15x + x2)

0.5x2 – 22.5x + 84.375 = 0

Choose root < 7.5 m

x = 4.1288 m

FH = w

y0

2x2 =

w0

2 (4.5) (4.1288)2 = 1.89411 w0

Maximum tension occurs at B since the slope y of the cable is greatest there.

y = w

FH

0

2x2

dy

dx|x = 4.1288

= tan �max = w x

FH

0 = w0 (4.1288)

1.89411

�max = 65.36°

Tmax = FH

maxcos �

20 = 1.89411

cos 65.36° w0 = 4.40 kN/m Ans.

7–104.

The cable AB is subjected to a uniform loading of 200 N/m.If the weight of the cable is neglected and the slope anglesat points A and B are 30° and 60°, respectively, determinethe curve that defines the cable shape and the maximumtension developed in the cable.

SOLUTION

Ans.

Ans.Tmax = 5.20 kN

Tmax =FH

cos umax=

2598cos 60°

= 5196 N

umax = 60°

y = 138.5x2 + 577x2110-32 m

FH = 2598 Ndy

dx= tan 60°;At x = 15 m,

y =1FH1100x2 + FH tan 30°x2

C1 = FH tan 30°dy

dx= tan 30°;At x = 0,

C2 = 0y = 0;At x = 0,

dy

dx=

1FH1200x + C12

y =1FH1100x2 + C1x + C22

y =1FHL ¢L200 dx≤ dx

15 m200 N/m

y

xA

B

60°

30°

7–10 . The bridge deck has a weight per unit length of. It is supported on each side by a cable. Determine

the tension in each cable at the piers A and B.80 kN>m A

B

1000 m

150 m75 m

5

7–10 . If each of the two side cables that support thebridge deck can sustain a maximum tension of 50 MN,determine the allowable uniform distributed load causedby the weight of the bridge deck.

w0

A

B

1000 m

150 m75 m

6

A

B

CD

12 m

3 m

h

7–107.

Cylinders C and D are attached to the end of the cable. If Dhas a mass of 600 kg, determine the required mass of C, themaximum sag h of the cable, and the length of the cablebetween the pulleys A and B. The beam has a mass per unitlength of 50 kg m.

SOLUTION

From the free-body diagram shown in Fig. a, we can write

a

Solving,

Ans.

Ans.

Thus, . As shown in Fig. a, the origin of the coordinate system is set at the lowest point of the cable. Using Eq. (1) of Example 7–12,

Using Eq. (4) and applying two other boundary conditions at x = x0and y = h at we have

Solving these equations yields

Ans.

The differential length of the cable is

Thus, the total length of the cable is

Ans. = 13.2 m

= 0.1140e 12

cx276.89 + x2 + 76.89 ln ax + 276.89 + x2b d f `-3.808 m

8.192 m

= 0.1140L8.192 m

-3.808 m 276.89 + x2 dx

L = Lds = L 8.192 m

-3.808 m 21 + 0.01301 x2

ds = 2dx2 + dy2 = A1 + adydxb2

dx = 21 + 0.01301x2 dx

x0 = 8.192 m

h = 0.8268 m = 0.827 m

h = 0.05702[-(12 - x0)]2

h + 3 = 0.05702x02

x = -(12 - x0),y = (h + 3) m

y = 0.05702x2

y =w0

2FHx2 =

50(9.81)

2(4301.00)x2

x - yFH = TB cos uB = 4301.00 N

uA = 23.47°

mC = 477.99 kg = 478 kg

mC(9.81) sin uA + 600(9.81) sin 43.05° - 50(12)(9.81) = 0+ c©Fy = 0;

600(9.81) cos 43.05° - mC(9.81) cos uA = 0©Fx = 0;:+uB = 43.05°

600(9.81) sin uB(12) - 600(9.81) cos uB(3) - 50(12)(9.81)(6) = 0+©MA = 0;

>

.

7–109.

If the pipe has a mass per unit length of 1500 kg m,determine the maximum tension developed in the cable.

SOLUTION

As shown in Fig. a, the origin of the x, y coordinate system is set at the lowest pointof the cable. Here, . Using Eq. 7–12,we can write

Applying the boundary condition at x = 0, results in c1= 0.

Applying the boundary condition y = 0 at x = 0 results in c2 = 0. Thus,

Applying the boundary condition y = 3 m at x = 15 m, we have

Substituting this result into Eq. (1), we have

The maximum tension occurs at either points at A or B where the cable has thegreatest angle with the horizontal. Here,

Thus,

Tmax =FH

cos umax=

551.8(103)

cos 21.80°= 594.32(103) N = 594 kN

umax = tan-1 adydx`15 mb = tan-1 [0.02667(15)] = 21.80°

dy

dx= 0.02667x

FH = 551.81(103) N3 =7.3575(103)

FH(15)2

y =7.3575(103)

FHx2

dy

dx= 0

=1FHa14.715(103)

2x2 + c1x + c2b

y =1FHL aLw0dxbdx

w(x) = w0 = 1500(9.81) = 14.715(103) N>m

> 30 m

3 mA B

Ans.

7–110.

If the pipe has a mass per unit length of 1500 kg m,determine the minimum tension developed in the cable.

SOLUTION

As shown in Fig. a, the origin of the x, y coordinate system is set at the lowest pointof the cable. Here, . Using Eq. 7–12,we can write

Applying the boundary condition at x = 0, results in c1= 0.

Applying the boundary condition y = 0 at x = 0 results in c2 = 0. Thus,

Applying the boundary condition y = 3 m at x = 15 m, we have

Substituting this result into Eq. (1), we have

The minimum tension occurs at the lowest point of the cable, where . Thus,

Tmin = FH = 551.81(103) N = 552 kN

u = 0°

dy

dx= 0.02667x

FH = 551.81(103) N3 =7.3575(103)

FH(15)2

y =7.3575(103)

FHx2

dy

dx= 0

=1FHa14.715(103)

2x2 + c1x + c2b

y =1FHL aLw0dxbdx

w(x) = w0 = 1500(9.81) = 14.715(103) N>m

> 30 m

3 mA B

Ans.

7–111.

If the slope of the cable at support A is zero, determine thedeflection curve y = f(x) of the cable and the maximumtension developed in the cable.

SOLUTION

Using Eq. 7–12,

Applying the boundary condition at x = 0 results in C1 = 0.

Applying the boundary condition y = 0 at x = 0, we have

Thus,

Applying the boundary condition y = 4.5 m at x = 12 m, we have

Substituting this result into Eqs. (1) and (2), we obtain

and

Ans.

The maximum tension occurs at point B where the cable makes the greatest anglewith the horizontal. Here,

Thus,

Ans.Tmax =FH

cos umax=

51.876(103)

cos 30.50°= 60.207(103) N = 60.2 kN

umax = tan-1adydx`x = 12 m

b = tan-1 c0.5890 sina p24

(12)b d = 30.50°

= 4.5a1 - cos p

24xb m

y =2304(103)

p 2(51.876)(103)

c1 - cos p

24x d

= 0.5890 sin p

24x

dy

dx=

96(103)

p(51.876)(103) sin p

24x

FH = 51.876(103) N

4.5 =2304(103)

p2FHc1 - cos

p

24 (12) d

y =2304(103)

p2FHc1 - cos

p

24x d

C2 =2304(103)

p2FH

0 = -24pc96(103)

pFH cos 0° d + C2

dy

dx= 0

y = - 24p

c96(103)

pFH cos

p

24x d + C1x + C2

y =1FHL

24pc4(103) d sin

p

24x + C1

y =1FHL aL4 cos

p

24* dxbdx

y =1FHL aLw(x)dxbdx

12 m

4.5 m

4 kN/m

y

x

B

A

––24w � 4 cos xp

7–112.

SOLUTIONThe Equation of The Cable:

(1)

(2)

Boundary Conditions:

at then from Eq. (1)


Thus, (3)


and

at and the maximum tension occurs when

The maximum tension in the cable is

Ans.Tmax =FH

cos umax=

1 267 265.47cos 12.61°

= 1 298 579.01 N = 1.30 MN

umax = 12.61°

tan umax =dy

dx `x=100 m

= 0.4735110-3211002 + tan 10°

u = umax .x = 100 mu = umax

= 0.4735110-32x + tan 10°

=600

1 267 265.47x + tan 10°

dy

dx=

w0

FHx + tan 10°

20 =6002FH110022 + tan 10°11002 FH = 1 267 265.47 N

x = 100 m,y = 20 m

y =w0

2FHx2 + tan 10°x

tan 10° =1FH1C12 C1 = FH tan 10°x = 0,

dy

dx= tan 10°

0 =1FH1C22 C2 = 0x = 0,y = 0

dy

dx=

1FH1w0x + C12

=1FH¢w0

2x2 + C1x + C2≤

y =1FHL11w1x2dx2 dx

Determine the maximum tension developed in the cable ifit is subjected to a uniform load of 600 N/m.

x20 m

100 m

600 N/m

y

A

B10°

7– . Determine the maximum uniform distributedloading N/m that the cable can support if it is capable ofsustaining a maximum tension of 60 kN.

w0

60 m

7 m

w0

113

.

7 114.

A cable has a weight density � and is supported at points that are a distance d apart and at thesame elevation. If it has a length L, determine the sag.

Given:

� 30N

cm�� d 25m�� L 26m��

Solution:

Guess FH 1000N��

GivenL2

FH�

sinh�

FH

d2

��

�

��

��

��

��

� 0= FH Find FH� �� FH 770 N�

hFH�

cosh12

�

FHd��

��

�

1��

�

�� h 3.104 m� Ans.

–

7 115.

A wire has a weight denisty �. If it can span a distance L and has a sag h determine the length ofthe cable. The ends of the cable are supported from the same elevation.

Given :

� 2Nm

�� d 10m�� h 1.2m��

Solution :

From Eq. (5) of Example 7-15 :

hFH�

�2

d FH

��

�

2

2

� �

��

�= FH18

�d2

h�� FH 20.83 N�

From Eq. (3) of Example 7-15 :

L2

FH�

sinh�

FH

d 2

��

�

��

��

�= L 2FH�

sinh12

�d

FH��

��

�

�� L 10.39 m� Ans.

–

7–116.

The cable is suspended between the supports and . If the cable can sustain a maximum tension of

and the maximum sag is , determine themaximum distance between the supports

SOLUTION

The origin of the x, y coordinate system is set at the lowest point of the cable. HereUsing Eq. (4) of Example 7–13,

Applying the boundary equation m at we have

The maximum tension occurs at either points A or B where the cable makes thegreatest angle with the horizontal. From Eq. (1),

By referring to the geometry shown in Fig. b, we have

Thus,

(3)

Solving Eqs. (2) and (3) yields

Ans.

FH = 1205.7 N

L = 16.8 m

1500 = FH cosh¢49.05LFH

≤Tmax =

FHcos umax

cos umax =1

A1 + sinh2 ¢ 49.05LFH

≤ =1

cosh ¢49.05LFH

≤ tan umax = sinh ¢49.05L

FH≤

3 =FH

98.1 B cosh ¢49.05L

FH≤ - 1R

x =L

2,y = 3

y =FH

98.1 B cosh ¢98.1x

FH≤ - 1R

y =FHw0

B cosh ¢ w0

FHx≤ - 1Rw0 = 10(9.81) N>m = 98.1 N>m.

L3 m1.5 kN

BA10 kg>m

A B

L

3 m

7–117.

Show that the deflection curve of the cable discussed inExample 7–13 reduces to Eq. 4 in Example 7–12 when thehyperbolic cosine function is expanded in terms of a seriesand only the first two terms are retained. (The answerindicates that the catenary may be replaced by a parabola inthe analysis of problems in which the sag is small. In thiscase, the cable weight is assumed to be uniformlydistributed along the horizontal.)

SOLUTION

Substituting into

Using Eq. (3) in Example 7–12,

We get QEDy =4hL2x

2

FH = w0L

2

8h

= w0x

2

2FH

=FHw0

B 1 +w2

0x2

2F2H

+ Á - 1R y =

FHw0

B cosh ¢ w0

FHx≤ - 1R

cosh x = 1 +x2

21+ Á

7–11 . If the horizontal towing force is T = 20 kN and thechain has a mass per unit length of , determine themaximum sag h. Neglect the buoyancy effect of the wateron the chain. The boats are stationary.

15 kg>m 40 m

hT T

8

7–119.

The cable has a mass of 0.5 and is 25 m long.Determine the vertical and horizontal components of forceit exerts on the top of the tower.

kg>m,

SOLUTION

Performing the integration yields:

(1)

rom Eq. 7-13

At ; . Hence

(2)

Applying boundary conditions at ; to Eq.(1) and using the resultyields . Hence

(3)

At ; . From Eq.(3)

By trial and error

At point A, From Eq.(2)

Ans.

Ans.(FH)A = FH = 73.9 N

(Fv )A = FH tan uA = 73.94 tan 65.90° = 165 N

tan uA =dy

dx2s=25 m

=4.905(25)

73.94+ tan 30° uA = 65.90°

s = 25 m

FH = 73.94 N

15 =FH

4.905b sin h-1 c 1

FH(4.905(25) + FH tan 30°)R - sin h-1(tan 30°) rs = 25 mx = 15 m

x =FH

4.905b sin h-1 c 1

F H(4.905s+FH tan 30°) d - sin h-1(tan 30°)rC2 = - sin h-1(tan 30°)C1 = FH tan 30°

s = 0x = 0

dy

dx=

4.905sFH

+ tan 30°

C1 = FH tan 30°dy

dx= tan 30°s = 0

dy

dx=

1FH

(4.905s + C1)

dy

dx=

1FHLw0ds

x =FH

4.905 b sin h-1 c 1

FH(4.905s + C1) d + C2 r

x =

Ldsb 1 +

1F2H

(w0ds)2 r 1

2

30B

A

15 m

7 1

A cable of length L is suspended between two points a distance d apart and at the same elevation.If the minimum tension in the cable is Tmin, determine the total weight of the cable and themaximum tension developed in the cable.

Units Used: MN 103kN �

Given: L 50 m� � d 15 m� � Tmin 200 kN� �

Tmin FH= FH Tmin � FH 200 kN�Solution:

From Example 7-15: sFHw0

sinhw0 x�

FH

�

��

�=

Guess w0 1kNm

�

GivenL2

FHw0

sinhw0FH

d2

��

��

�= w0 Find w0� � � w0 79.93kNm

�

Totalweight w0 L� � Totalweight 4.00 MN�

tan �max� �w0FH

L2

�= �max atanw0

L2

�

��

�

FH

��

��

� �max 84.284 deg�

Then,

TmaxFH

cos �max� � � Tmax 2.01 MN�

20.

Ans.

Ans.

–

7 121.

The chain of length L is fixed at its ends and hoisted at its midpoint B using a crane. If the chainhas a weight density w, determine the minimum height h of the hook in order to lift the chaincompletely off the ground. What is the horizontal force at pin A or C when the chain is in thisposition? Hint: When h is a minimum, the slope at A and C is zero.

Given:

L 80m �

d 60m �

w 0.5kNm

�

Solution:

Guesses FH 10kN � h 1m �

Given hFHw

coshw

FH

d2

��

��

1��

��

�=L2

FHw

sinhw

FH

d2

��

��

�=

h

FH

�

��

Find h FH�� FA FH � FC FH �FA

FC

�

��

11.1

11.1�

��

kN�

h 23.5 m�

Ans.

Ans.

–

7 122.

A uniform cord is suspended between two points havingthe same elevation. Determine the sag-to-span ratio sothat the maximum tension in the cord equals the cord’stotal weight.

SOLUTION

From Example 7–15.

At

when

Ans.h

L= 0.141

0.1547 L2h

= 0.5493

h =FHw0

e 1

C1 - tanh2a w0L

2FHb

- 1 u = 0.1547¢FHw0≤

h =FHw0

Bcosh ¢ w0

FH x≤ - 1R

y = hx =L

2,

w0L

2FH= tanh-110.52 = 0.5493

tanh¢w0L

2FH≤ =

12

2FH sinh¢w0L

2FH≤ = FH cosh ¢w0L

2FH≤

w012s2 = FH cosh ¢w0L

2FH≤

Tmax =FH

cos umax

cos umax =1

coshaw0L

2FHb

dy

dx`max

= tan umax = sinh¢w0L

2FH≤

x =L

2,

y =FHw0

Bcosh ¢ w0

FH x≤ - 1R

s =FHw0

sinh ¢ w0

FH x≤

–

*7–108. A cable having a weight per unit length of is suspended between supports A and B. Determine theequation of the catenary curve of the cable and the cable’slength.

5 lb>ftA B

150 ft

30� 30�

50 m7–12 . A cable having a weight per unit length of 0.1 kN/m is suspended between supports A and B. Determine the equation of the catenary curve of the cable and the cable’s length.

As shown in Fig. a, the origin of the x, y coordinate system is set at the lowest point of the cable. Here, w(s) = 0.1 kN/m.

d y

dx

2

2 =

0.1

FH

= 1 +2

dy

dx

⎛⎝⎜

⎞⎠⎟

If we set u = dy

dx, then

du

dx =

d y

dx

2

2. Substituting these two values into the equation,

du

u1 + 2 =

0.1

FH

dx

Integrating,

ln (u + 1 + 2u ) = 0.1

FH

x + C1

Applying the boundary condition u = dy

dx = 0 at x = 0 results in C1 = 0. Thus,

ln(u + 1 + 2u ) = 0.1

FH

x

u + 1 + 2u = e FHx0.1

dy

dx = u =

e eF FH Hx x0.1 0.1

–

2

–

Since sinh x = e ex x–

2

–

, then

dy

dx = sinh

0.1

FH

x (1)

Applying the boundary equation dy

dx = tan 30° at x = 25 m,

tan 30° = sinh0.1

(25)FH

⎡

⎣⎢⎢

⎤

⎦⎥⎥

FH = 4.5512 kN

Substituting this result into Eq. (1),

dy

dx = sinh[(0.0219722)x] (2)

3

25 m 25 m

FH = 4.5512 kN

Integrating,

y = 45.512 cosh[(0.0219722)x] + C2

Applying the boundary equation y = 0 at x = 0 results in C2 = –45.512. Thus,

y = 45.512 {cosh[(0.0219722)x] – 1} m Ans.

If we write the force equation of equilibrium along the x and y axes by referring to the free – body diagram shown in Fig. b, we have

+→ ΣFx = 0; T cos � – 4.5512 = 0

+↑ΣFy = 0; T sin � – 0.1s = 0

Eliminating T,

dy

dx = tan � = 0.0219722 (3)

Equating Eqs. (2) and (3),

(0.0219722)s = sinh[(0.0219722)x]

s = 45.512 [(0.0219722)x] m

Thus, the length of the cable is

L = 2{45.512 sinh[(0.0219722) (25)]}= 52.553 m Ans.

7 1 4.

A fiber optic cable is suspended over the poles so that the angle at the supports is �.Determine the minimum tension in the cable and the sag. The cable has a mass density " andthe supports are at the same elevation.

Given:

� 22 deg�

" 0.9kgm

�

a 30 m�

g 9.81m

s2�

Solution:

�max ��

w0 "g�

xyd

dtan �� sinh w0

a2

FH

�

��

�

FH

w0a2

��

asinh tan �� FH 336 N�

TmaxFH

cos �� Tmax 363 N�

hFHw0

coshw0

a2

FH

�

��

1�

�

��

� h 2.99 m�

2

Ans.

Ans.

–

7 125.

The beam is supported by a pin at C and a rod AB. Determine the internal normal force, shearforce, and moment at point D.

Units Used:

kN 103N �

Given:

F 4kN �

a 6m �

b 5m �

c 3m �

d 6m �

� 60deg �

Solution:

Guesses

FAB 1N � ND 1N � VD 1N � MD 1N m� �

Given F� sin �( )� b c� d�( )�a

a2 b c�( )2�

FAB� b c�( )�� 0=

ND�b c�

a2 b c�( )2�

FAB�� F cos �( )�� 0=

VDa

a2 b c�( )2�

FAB� F sin �( )�� 0=

a

a2 b c�( )2�

FAB� c� F sin �( )� c d�( )�� MD� 0=

FAB

ND

VD

MD

�

��

Find FAB ND� VD� MD�� ND

VD

�

��

6.08�

2.6��

��

kN� MD 12.99� kN m�� Ans.

–

7–126.

SOLUTION

Ans.

a

Ans.M = 10x - x2 - 30

+©M = 0; M + 30 - 10x + 2x ax2b = 0

V = 10 - 2x

+ c©Fy = 0; -V + 10 - 2x = 0


5 m

2 kN/m

5 kN m

BA

7–127.


a

Free body Diagram: The FBD for segment AC sectioned through point C is drawn.

Internal Forces: This problem requires . Summing moments about pointC [FBD (b)], we have

a

Ans.a = 0.366L

2a2 + 2aL - L2 = 0

+©MC = 0;wa2aa

4b +

w4

(L - a) c16

(2a + L) d -w4

(L + a)aa2b = 0

MC = 0

+©MC = 0;w2

(L + a) aa2b - By (a) = 0 By =

w4

(L + a)

Determine the distance a between the supports in terms ofthe beam’s length L so that the moment in the symmetricbeam is zero at the beam’s center.

L

a

w

7– 2 . The stacker crane supports a 1.5-Mg boat with thecenter of mass at G. Determine the internal normal force,shear force, and moment at point D in the girder.The trolleyis free to roll along the girder rail and is located at theposition shown. Only vertical reactions occur at A and B.

3.5 m

D

G

C

BA

5 m

7.5 m

1 m1 m2 m

2 m

1 8

7–129.

The yacht is anchored with a chain that has a total length of 40m and a mass per unit length of and the tension inthe chain at A is 7 kN. Determine the length of chain whichis lying at the bottom of the sea. What is the distance d?Assume that buoyancy effects of the water on the chain arenegligible. Hint: Establish the origin of the coordinate systemat B as shown in order to find the chain length BA.

ld

18 kg/m,

SOLUTIONComponent of force at A is

From Eq. (1) of Example 7 - 13

Since , , then

Also , , so that and the above equation becomes

(1)

or,

(2)

From Example 7 - 13

(3)

Substituting Eq. (2) into Eq. (3). Integrating.

Since , , then

Thus,

(4)

Slope of the cable at point A is

Using Eq. (3),

Length of chain on the ground is thus

Ans.

From Eq. (1), with

Using Eq. (4),

Ans.d = y = 19.8 m

y = 19.82 acos ha26.1019.82

b - 1b

x = 19.82 asin h-1a34.3319.82

b b = 26.10 m

s = 34.33 m

ld = 40 - 34.33 = 5.67 m

sAB = 19.82 (1.732) = 34.33 m

dy

dx= tan 60° = 1.732

y = 19.82 acos ha x19.82

b - 1b

C3 = - 19.82y = 0x = 0

dy

dx= sin h a x

19.82b y = 19.82 cos ha x

19.82b + C3

dy

dx=

w0 s

FH=

18 (9.81)

3500s =

s

19.82

s = 19.82 asin ha x19.82

b b

x = 19.82 asin h-1a s

19.82b b

C2 = 0s = 0x = 0

dy

dx=

1FH

(w0 s + C1); C1 = 0

s = 0dy

dx= 0

x =3500

18 (9.81) asin h-1 c 1

3500(18)(9.81)s + C1 d + C2b

FH = T cos u = 7000 cos 60° = 3500 N

d

A

Bs

x

y

60

dl

7–130.

SOLUTIONSupport Reactions: The 6 kN load can be replacde by an equivalent force andcouple moment at B as shown on FBD (a).

a


Ans.

a Ans.

For [FBD (c)],

Ans.

a

Ans.M = 527.0 - 4.50x6 kN # m6.364 sin 45°16 - x2 - M = 0+ ©M = 0;

V + 6.364 sin 45° = 0 V = -4.50 kN+ c ©Fy = 0;

3 m<x ◊ 6 m

M - 1.50x = 0 M = 51.50x6 kN # m+ ©M = 0;

1.50 - V = 0 V = 1.50 kN+ c ©Fy = 0;

0 ◊ x<3 m

Ay + 6.364 sin 45° - 6 = 0 Ay = 1.50 kN+ c ©Fy = 0;

FCD sin 45°162 - 6132 - 9.00 = 0 FCD = 6.364 kN+ ©MA = 0;

Draw the shear and moment diagrams for the beam ABC.

A C

D

B

3 m 1.5 m

1.5 m

1.5 m

6 kN

45°

7– . The shaft is supported by a thrust bearing at A anda journal bearing at B. Determine the x, y, z components ofinternal loading at point C.

1 m

1 m

0.5 m0.2 m

0.2 m

1 m

750 N

750 N

600 N

z

C

y

x900 N

A

B

131

7–132.

SOLUTION

Performing the integration yields:

(1)

From Eq. 7-14

At

(2)

Applying boundary conditions at ; to Eq. (1) and using the result yields . Hence

(3)

Substituting Eq. (3) into (2) yields:

(4)

Performing the integration

Applying boundary conditions at ; yields . Therefore

At

By trial and error

At . From Eq. (4)

Ans.Tmax =FH

cos umax=

75.25cos 11.346°

= 76.7 kN tan umax =

dy

dx 2x=30 m

= sin ha0.5(30)

75.25b umax = 11.346°

x = 30 m; u = umax

FH = 75.25 kN

x = 30 m; y = 3 m; 3 =FH0.5

ccos ha0.5FH

(30)b - 1 d

y =FH0.5ccos ha0.5

FHxb - 1 d

C3 = -

FH0.5

y = 0x = 0

y =FH0.5

cos ha0.5FH

xb + C3

dy

dx= sin ha0.5x

FHb

s =FH0.5

sin ha0.5FH

xbC2 = 0

C1 = 0s = 0x = 0

dy

dx= tan u =

0.5sFH

s = 0;dy

dx= 0 hence C1 = 0

dy

dx=

1FH

(0.5s + C1)

dy

dx=

1FHLw0ds

x =FH0.5b sin h-1B 1

FH(0.5s + C1)R + C2 r

x = Lds

e1 +1F2HL(w0ds)

2 f12

A chain is suspended between points at the same elevationand spaced a distance of 60 m apart. If it has a weight perunit length of and the sag is 3 m, determine themaximum tension in the chain.

0.5 kN>m

30 m 30 m

7–133.


SOLUTION

5 m 5 m

2 kN/m

A

50 kN m

BC

7–134.

Determine the normal force, shear force, and moment atpoints B and C of the beam.

5 m5 m 3 m

2 kN/m1 kN/m

7.5 kN

40 kN m

6 kN

1 m

A BC

SOLUTIONFree body Diagram: The support reactions need not be computed for this case .

Internal Forces: Applying the equations of equilibrium to segment DC [FBD (a)],we have

Ans.

Ans.

a

Ans.

Applying the equations of equilibrium to segment DB [FBD (b)], we have

Ans.

Ans.

a

Ans.MB = -184.5 kN # m- 4.00(7) - 6(9) - 40 = 0

+©MB = 0; -MB - 10.0(2.5) - 7.5(5)

VB = 27.5 kN

+ c©Fy = 0; VB - 10.0 - 7.5 - 4.00 - 6 = 0

:+ ©Fx = 0; NB = 0

MC = -62.5 kN # m+©MC = 0; -MC - 3.00(1.5) - 6(3) - 40 = 0

+ c©Fy = 0; VC - 3.00 - 6 = 0 VC = 9.00 kN

:+ ©Fx = 0; NC = 0

7 135.

Draw the shear and momentdiagrams for the beam.

Units Used:

kN 103 N�

Given:

a 2 m� b 4 m� w 1.5kNm

�

Solution:

w b a�( )b a�

2�

�� Ay b� 0� Ay

w b a�( )2

2b� Ay 0.75 kN�

x1 0 0.01a� a��

Ay V1 x( )� 0� V1 x( ) Ay1

kN� Ans.

–

Ay� x M1 x( )� 0� M1 x( ) Ay x1

kN m��

x2 b a� 1.01 b a�( )� b��

Ay w x a�( )� V2 x( )� 0� V2 x( ) Ay w x a�( )�� 1

kN�

Ay� x w x a�( )x a�

2�

�� M2 x( )� 0� M2 x( ) Ay x w

x a�( )2

2�

��

��

1kN m�

�

0 0.5 1 1.5 2 2.5 3 3.5 44

2

0

2

Distance (m)

Forc

e (k

N)

V1 x1� �V2 x2� �

x1 x2�

0 0.5 1 1.5 2 2.5 3 3.5 41

0

1

2

Distance (m)

Mom

ent (

kN-m

)

M1 x1� �M2 x2� �

x1 x2�

Ans.

Ans.

Ans.

7–136.

If the 45-m-long cable has a mass per unit length of 5 kg m,determine the equation of the catenary curve of the cableand the maximum tension developed in the cable.

SOLUTION

As shown in Fig. a, the orgin of the x, y coordinate system is set at the lowest pointof the cable. Here,

Set

Integrating,

ln

Applying the boundary condition results in . Thus,

ln

Since , then

Integrating,

Applying the boundary equation at results in . Thus,

If we write the force equation of equilibrium along the x and y axes by referring tothe free-body diagram shown in Fig. b,

T sin u - 5(9.81)s = 0+ c©Fy = 0;

T cos u - FH = 0©Fx = 0;:+

y =FH

49.05 ccosh a49.05

FHxb - 1 d m

C2 = -

FH49.05

x = 0y = 0

y =FH

49.05 cosh a49.05

FHxb + C2

dy

dx= sinh

49.05FHx

sinh x =ex - e-x

2

dy

dx= u = e

49.05FH

x

- e- 49.05FH

x

2

u + 11 + u2 = e 49.05FHx

au + 11 + u2b =49.05FHx

C1 = 0u =dy

dx= 0 at x = 0

au + 11 + u2b =49.05FHx + C1

du

11 + u2=

49.05FH

dx

u =dy

dx, then

du

dx=d2y

dx2, then

d2y

dx2 =49.05FH

A1 + adydx b2

w(s) = 5(9.81) N>m = 49.05 N>m.

>A B

40 m

7–136. (continued)

Eliminating T,

(3)

Equating Eqs. (1) and (3) yields

Thus, the length of the cable is

Solving by trial and error,

Substituting this result into Eq. (2),

Ans.

The maximum tension occurs at either points A or B where the cable makes thegreatest angle with the horizontal. Here

Thus,

Ans.Tmax =FH

cos umax=

1153.41cos 43.74°

= 1596.36 N = 1.60 kN

umax = tan-1adydx`x=20m

b = tan-1 e sinha49.05FH

(20)b f = 43.74°

y = 23.5 [cosh 0.0425x - 1] m

FH = 1153.41 N

L = 45 = 2e FH49.05

sinh a49.05FH

(20)b f

s =FH

49.05= sinh a49.05

FHb

49.05sFH

= sinh a49.05FHxb

dy

dx= tan u =

49.05sFH

7–137.

The traveling crane consists of a 5-m-long beam having auniform mass per unit length of 20 kg/m. The chain hoistand its supported load exert a force of 8 kN on the beamwhen . Draw the shear and moment diagrams forthe beam. The guide wheels at the ends A and B exert onlyvertical reactions on the beam. Neglect the size of thetrolley at C.

x = 2 m


a

Shear and Moment Functions: For

Ans.

a

Ans.

For

Ans.

a

Ans.M = {16.0 - 2.71x - 0.0981x2} kN # m

+©M = 0; 3.6905(5 - x) -20(9.81)

1000 (5 - x)a5 - x

2b - M = 0

V = {-0.196x - 2.71} kN

+ c©Fy = 0; V + 3.6905 -20(9.81)

1000 (5 - x) = 0

2 m 6 x … 5 m [FBD (c)],

M = {5.29x - 0.0981x2} kN # m

+©M = 0; M + 0.1962xax2b - 5.2905x = 0

V = {5.29 - 0.196x} kN

+ c©Fy = 0; 5.2905 - 0.1962x - V = 0

0 … x 6 2 m [FBD (b)],

+ c©Fy = 0; Ay + 3.6905 - 8 - 0.981 = 0 Ay = 5.2905 kN

+©MA = 0; By (5) - 8(2) - 0.981 (2.5) = 0 By = 3.6905 kN

x 2 m

A

C

5 m

8 kN

B

7 138.

Draw the shear and moment diagramsfor the beam.

Units Used: kN 103 N�

Given:

w 3kNm

� F 10 kN�

L 6 m�

Solution:

V x( ) w L x�( )� F� 0�

V x( ) w L x�( ) F�[ ]1

kN�

M x( )� w L x�( )L x�

2�

�� F L x�( )� 0�

M x( ) w�L x�( )2

2F L x�( )�

��

��

1kN m�

�

0 1 2 3 4 5 60

20

40

Distance (m)

Forc

e (k

N)

V x( )

x

0 1 2 3 4 5 6150

100

50

0

Distance (m)

Mom

ent (

kN-m

)

M x( )

x

Ans.

Ans.

–

300-kg

0.45 m 0.45 m

0.45 m:

–150 (9.81) – V = 0

V = –1471.5 N = –1.47 kN Ans.

1.4715 Ans.

For 0.45 m < x ≤ 0.9 m:

2943 – 1471.5 – V = 0

V = 1471.5 N = 1.47 kN

= 1.4715 x – 1.32435 + 1.4715 x – 2.943 (x – 0. 45) = 0

Ans.M

M

M + 1.4715 x = 0

=M

150 (9.81) N

0.45 m

300 (9.81) N = 2943 N

150 (9.81) N = 1471.5 N

(kN) 1.47

(m)

(m)

(kN·m)

–1.47

–0.662

x–

.

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