Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 07

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Determine the internal normal force and shear force, and

the bending moment in the beam at points C and D.

Assume the support at B is a roller Point C is located just to

the right of the 8-kip load

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The strongback or lifting beam is used for materials

handling If the suspended load has a weight of 2 kN and a

center of gravity of G, determine the placement d of the

padeyes on the top of the beam so that there is no moment

developed within the length AB of the beam The lifting

bridle has two legs that are positioned at 45°, as shown

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°

0.2 m0.2 m

d d

E

F

G

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The boom DF of the jib crane and the column DE have a

uniform weight of 50 If the hoist and load weigh 300 lb,

determine the normal force, shear force, and moment in the

crane at sections passing through points A, B, and C

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Determine the internal normal force, shear force, and

moment at points A and B in the column.

NB = 16.2 kN

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

NA = 13.2 kN

NA- 6 cos 30° - 8 = 0+ c ©Fy = 0;

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2PL aL3

Determine the distance a as a fraction of the beam’s length

L for locating the roller support so that the moment in the

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Determine the internal normal force, shear force, and

moment at points C and D in the simply-supported beam.

Point D is located just to the left of the 2500-lb force.

+ © MD = 0;

VD = - 125 lb

VD + 2625 - 2500 = 0 + c © Fy = 0;

VC= 1375 lb

2875 - 500(3) - VC = 0+ c ©Fy = 0;

+ ©MB = 0;

By = 2625 lb

By(12) - 500(6)(3) - 2500(9) = 0+ ©MA = 0;

B A

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Determine the normal force, shear force, and moment at a

section passing through point C Assume the support at A

can be approximated by a pin and B as a roller.

VC = 0.5 kip+ c ©Fy = 0; VC- 9.6 + 17.1 - 8 = 0

:+

©Fx = 0; NC = 0

Ay = 20.1 kip+ c ©Fy = 0; Ay - 10 - 19.2 + 17.1 - 8 = 0

:+

©Fx = 0; Ax = 0

By = 17.1 kip+ ©MA = 0; -19.2(12) - 8(30) +By(24) + 10(6) = 0

10 kip

0.8 kip/ft 8 kip

6 ft 12 ft 12 ft 6 ft

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Determine the normal force, shear force, and moment at a

section passing through point C Take P = 8 kN

VC = -8 kN+ c ©Fy = 0; VC + 8 = 0

0.75 m 0.75 m

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VC= -0.533 kN+ c ©Fy = 0; VC - 0.533 = 0

The cable will fail when subjected to a tension of 2 kN

Determine the largest vertical load P the frame will support

and calculate the internal normal force, shear force, and

moment at a section passing through point C for this loading.

0.75 m 0.75 m

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©Fx = 0; ND= 0

MD = 3771 lb#ft = 3.77 kip#ft+ ©MD = 0; -MD + 1886(2) = 0

VC = 2014 lb = 2.01 kip+ c ©Fy = 0; -2500 + 4514 - VC= 0

:+

©Fx = 0; NC= 0

MC = -15000 lb#ft = -15.0 kip#ft+ ©MC = 0; 2500(6) + MC= 0

By = 1886 lb+ c ©Fy = 0; 4514 - 2500 - 900 - 3000 + By = 0

:+

©Fx = 0; Bx = 0

Ay = 4514 lb+ ©MB = 0; -Ay(14) + 2500(20) + 900(8) + 3000(2) = 0

The shaft is supported by a journal bearing at A and a thrust

bearing at B Determine the normal force, shear force, and

moment at a section passing through (a) point C, which is

just to the right of the bearing at A, and (b) point D, which

is just to the left of the 3000-lb force

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Determine the internal normal force, shear force, and the

moment at points C and D.

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

ND = 0:+

©Fx = 0;

MC = 4.97 kN#m

MC - 3.515 cos 45°122 = 0+ ©MC = 0;

By = 8.485 kN

By16 + 6 cos 45°2 - 12.013 + 6 cos 45°2 = 0+ ©MA = 0;

2 kN/m

B D

C A

6 m

2 m

45˚

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+ c ©Fy = 0; VC + 2000 - 1200 - 800 = 0 VC = 0

:+

©Fx = 0; NC = 0

MD = -1600 lb#ft = - 1.60 kip#ft+ ©MD = 0; -MD- 800(2) = 0

+ c ©Fy = 0; VD - 800 = 0 VD = 800 lb

:+

©Fx = 0; ND = 0

By = 2000 lb+ ©MA = 0; By(8) + 800 (2) - 2400(4) - 800(10) = 0

Determine the internal normal force, shear force, and

moment acting at point C and at point D, which is located

just to the right of the roller support at B.

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Determine the normal force, shear force, and moment at a

section passing through point D Take w = 150 N>m

VD= 0

600 - 150142 - VD = 0+ c ©Fy = 0;

ND = -800 N:+

©Fx = 0;

Ay = 600 N

Ay - 150182 + 3

5110002 = 0+ c ©Fy = 0;

Ax = 800 N

Ax 4

-5110002 = 0:+

©Fx = 0;

FBC= 1000 N

-150182142 + 3

5FBC182 = 0+ ©MA= 0;

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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 at

D reaches or the normal force in member BC

becomes 1500 N Determine the largest load it can

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Determine the internal normal force, shear force, and

moment at point D in the beam.

The negative sign indicates that ND, VD, and MDact in the opposite sense to that

shown on the free-body diagram

MD = -300 N#m

MD + 600(1)(0.5) = 0+ ©MD = 0;

VD = -600 N

-VD - 600(1) = 0+ c ©Fy = 0;

+ ©MB = 0;

FBC = 2250 N

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

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-13120802132 + 12

1312080212.52 - ME = 0+ ©ME = 0;

VE = 800 N

VE 5

-13120802 = 0+ c ©Fy = 0;

NE = -1920 N = - 1.92 kN

-NE 12

-13120802 = 0:+

©Fx = 0;

FBC= 2080 N

-1200142 + 5

13FBC162 = 0+ ©MA = 0;

Determine the normal force, shear force, and moment at a

section passing through point E of the two-member frame.

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VE = -9 kip+ c ©Fy = 0; -VE-3 - 6 = 0

:+

©Fx = 0; ND = 0

By = 6 kip+ c ©Fy = 0; By + 3 - 12(1.5)(12) = 0

:+

©Fx = 0; Bx = 0

Ay = 3 kip+ ©MB = 0; 12(1.5)(12)(4) -Ay(12) = 0

Determine the normal force, shear force, and moment in

the beam at sections passing through points D and E Point

E is just to the right of the 3-kip load.

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Determine the internal normal force, shear force, and

moment at points E and F in the beam.

VF = -215 N

VF + 664.92 - 300 = 0+ c ©Fy = 0;

VE = 215 N

664.92 - 300(1.5) - VE = 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

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Rod AB is fixed to a smooth collar D, which slides freely

along the vertical guide Determine the internal normal

force, shear force, and moment at point C which is located

just to the left of the 60-lb concentrated load

SOLUTION

With reference to Fig a, we obtain

Using this result and referring to Fig b, we have

Ans.

Ans.

a

Ans.

The negative signs indicates that NCand VCact in the opposite sense to that shown

on the free-body diagram

MC = 135 lb#ft

108.25 cos 30°(1.5) - 1

2(15)(1.5)(0.5) - MC = 0+ ©MC = 0;

VC = -22.5 lb

VC - 60 - 1

2(15)(1.5) + 108.25 cos 30° = 0+ c ©Fy = 0;

15 lb/ft

60 lb

B C

A

3 ft 1.5 ft

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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 3000-lb force Assume the

support at A is fixed and the beam segments are connected

together by a short link at B.

VE = 1500 lb = 1.5 kip

3300 - 600(3) - VE = 0+ c ©Fy = 0;

VD = 6 kip

VD - 600(4.5) - 3300 = 0+ c ©Fy = 0;

ND = 0

©Fx = 0;

:+

FB = 3300 lb600(6)(3) + 3000(3) - FB(6) = 0

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Determine the internal normal force, shear force, and

moment at points E and F in the compound beam Point F is

located just to the left of the 15-kN force and

+ ©MF = 0;

VF = 1.25 kN

VF - 15 + 13.75 = 0+ c ©Fy = 0;

VE = -1.17 kN5.583 - 3(2.25) - VE = 0

+ ©MD = 0;

Dy = 13.75 kN

Dy(4) - 15(2) - 25 = 0+ ©MC = 0;

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Determine the internal normal force, shear force, and

moment at points D and E in the frame Point D is located

just above the 400-N force

The negative sign indicates that ND, NE, and VEacts in the opposite sense to that

shown in the free-body diagram

ME = 190 N#m

335.34 cos 30°(1) - 200(1)(0.5) - ME = 0+ ©ME = 0;

VE = -90.4 N

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

ND = -110 N335.34 cos 30° - 200(2) - ND = 0

C

2.5 m

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SOLUTION

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

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

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, and

bending moment at point C.

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Determine the shear force and moment acting at a section

passing through point C in the beam.

MC= 48 kip#ft+ ©MC = 0; -9(6) + 3(2) + MC = 0

:+

©Fx = 0; Ax = 0

Ay = 9 kip+ ©MB = 0; -Ay(18) + 27(6) = 0

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-a + b = 34b

- 16b(a - b) =

18

- 16b(2a + b)(a - b) =

Determine the ratio of for which the shear force will be

zero at the midpoint C of the beam.

a>b

B C A

a b/2 b/2

w

a

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Determine the normal force, shear force, and moment at a

section passing through point D of the two-member frame.

VD = 50 N+ c ©Fy = 0; 800 - 600 - 150 - VD = 0

6 m

C

2.5 m

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VF = -308 lb+ c ©Fy = 0; -307.8 - VF = 0

:+

©Fx = 0; NF = 0

ME = -490 lb#ft+ ©ME = 0; -ME - 245.2(2) = 0

Bx = 250 lb

:+

©Fx = 0; Bx - 500 cos 60° = 0

Cy = 307.8 lb+ ©MB = 0; -120(2) - 500 sin 60°(3) + Cy(5) = 0

Determine the normal force, shear force, and moment at

sections passing through points E and F Member BC is

pinned at B and there is a smooth slot in it at C The pin at C

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Determine the normal force, shear force, and moment

acting at a section passing through point C.

VC = 903 lb+ a ©Fy = 0; 100 sin 30° + 985.1 cos 30° - VC = 0

NC = -406 lb

Q + ©Fx = 0; NC - 100 cos 30° + 985.1 sin 30° = 0

Ay = 985.1 lb+ c ©Fy = 0; Ay - 800 cos 30° - 700 - 600 cos 30° + 927.4 = 0

Ax = 100 lb

:+

©Fx = 0; 800 sin 30° - 600 sin 30° - Ax = 0

By = 927.4 lb+ 600 sin 30°(3 sin 30°) + By(6cos 30° + 6 cos 30°) = 0+ ©MA = 0; -800 (3) - 700(6 cos 30°) - 600 cos 30°(6 cos 30° + 3cos 30°)

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Determine the normal force, shear force, and moment

acting at a section passing through point D.

VD = -203 lb

Q + ©Fy = 0; VD - 600 + 927.4 cos 30° = 0

ND = -464 lb+ a ©Fx = 0; ND - 927.4 sin 30° = 0

Ay = 985.1 lb+ c ©Fy = 0; Ay - 800 cos 30° - 700 - 600 cos 30° + 927.4 = 0

Ax = 100 lb

:+

©Fx = 0; 800 sin 30° - 600 sin 30° - Ax = 0

By = 927.4 lb+ 600 sin 30°(3 sin 30°) + By(6 cos 30° + 6 cos 30°) = 0+ ©MA = 0; -800(3) - 700(6 cos 30°) - 600 cos 30°(6 cos 30° + 3 cos 30°)

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Determine the distance a between the supports in terms of

the shaft’s length L so that the bending moment in the

symmetric shaft is zero at the shaft’s center The intensity of

the distributed load at the center of the shaft is The

supports are journal bearings

w0

L a

w0

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If the engine weighs 800 lb, determine the internal normal

force, shear force, and moment at points F and H in the

VF = 1155 lb3695.04 sin 30° - 800 cos 30° - VF = 0

+ Q ©Fy¿ = 0;

NF = 2800 lb3695.04 cos 30° - 800 sin 30° - NF = 0

+ a ©Fx¿ = 0;

FAC = 3695.04 lb

800 cos 30°(4) - FAC sin 30°(1.5) = 0+ ©MB = 0;

B F C D

A

60⬚

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The jib crane supports a load of 750 lb from the trolley

which rides on the top of the jib Determine the internal

normal force, shear force, and moment in the jib at point C

when the trolley is at the position shown The crane

members are pinned together at B, E and F and supported

C B H D

VC= - 844 lb+ c ©Fy = 0; -843.75 - VC = 0

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The jib crane supports a load of 750 lb from the trolley

which rides on the top of the jib Determine the internal

normal force, shear force, and moment in the column at

point D when the trolley is at the position shown The crane

members are pinned together at B, E and F and supported

C B H D

Ax = 1062.5 lb

:+

©Fx = 0; Ax - 687.5 - 375 = 0

TB= 687.5 lb+ ©MA = 0; TB(6) - 750 (9) + 375(7) = 0

FEF = 2656.25 lb+ ©MB = 0; FEFa35b (4) - 750 (9) + 375(1) = 0

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Determine the internal normal force, shear force, and

bending moment at points E and F of the frame.

SOLUTION

Support Reactions: Members HD and HG are two force members Using method of

joint [FBD (a)], we have

From FBD (b),

From FBD (c),

Solving Eqs (1) and (2) yields,

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

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 - Cx12 cos 26.57°2 + Cy12 sin 26.57°2 = 0+ ©MA = 0;

Cx12 cos 26.57°2 + Cy12 sin 26.57°2 - 894.43112 = 0+ ©MA = 0;

FHD = FHG = F = 894.43 N

2F sin 26.57° - 800 = 0+ c ©Fy = 0;

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The hook supports the 4-kN load Determine the internal

normal force, shear force, and moment at point A.

NA = 2.83 kN

NA - 4 sin 45° = 0+ a ©Fy¿ = 0;

VA= 2.83 kN

VA - 4 cos 45° = 0+ Q ©Fx¿ = 0;

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Determine the normal force, shear force, and moment acting

at sections passing through point B on the curved rod.

MB = -480 lb#ft

MB + 400(2sin30°) + 300(2 - 2cos30°) = 0+ ©MB = 0;

VB = -496 lb

VB + 400cos30° + 300sin30° = 0+ R ©Fy = 0;

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Determine the normal force, shear force, and moment acting

at sections passing through point C on the curved rod.

MC = -1590 lb#ft = - 1.59 kip#ft

-MC - 1200 - 400(2sin45°) + 300(2 - 2cos45°) = 0+ ©MC = 0;

MA = 1200 lb#ft

MA - 300(4) = 0+ ©MA = 0;

Ay = 300 lb+ c ©Fy = 0;

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The semicircular arch is subjected to a uniform distributed

load along its axis of per unit length Determine the

internal normal force, shear force, and moment in the arch

w0

u

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The semicircular arch is subjected to a uniform distributed

load along its axis of per unit length Determine the

internal normal force, shear force, and moment in the arch

at u = 120°

w0

O r

w0

u

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