Statics bedford 5 chap 10

Problem 10.1 In Active Example 10.1, suppose thatthe distance from point A to point C is increased from14 L to 1

2 L. Draw a sketch of the beam with C in itsnew position. Determine the internal forces amd momentat C.

ABC

L

y

34

L

14

L F

14

Solution: The reactions at A and B are repeated from ActiveExample 10.1.

Passing a plane through the beam at point C and writing the equilib-rium equations for the part of the beam to the left of C, we obtain

Fx : PC D 0,

Fy : 14 F � VC D 0,

MC : MC � � 12 L�� 1

4 F� D 0.

Solving yields

PC D 0, VC D 14 F, MC D 1

8 LF.

Problem 10.2 The magnitude of the triangular distri-buted load is w0 D 2 kN/m. Determine the internalforces and moment at A.

0.4 m0.6 m0.6 m

w0

A

y

x

Solution: The free-body diagram of the beam is shown in Fig. a.From the equilibrium equations

Fx : Bx D 0,

Fy : By C C � 600 N D 0,

MB : C�1.2 m� � �600 N��1 m� D 0,

We obtainBx D 0, By D 100 N, C D 500 N.

Passin a plane through the beam at A and isolating the part of thebeam to the left of A (Fig. b), we obtain

Fx : PA D 0,

Fy : 100 N � VA D 0,

MA : MA � �100 N��0.4 m� D 0.

Solving yields

PA D 0, VA D 100 N, MA D 40 N-m.

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789

Problem 10.3 The C clamp exerts 30-lb forces onthe clamped object. Determine the internal forces andmoment in the clamp at A.

x

y

A

2 in

Solution:

∑Fx : �30 lb C PA D 0

∑Fy : �VA D 0

∑MA : �30 lb��2 in� C MA D 0

Solving: PA D 30 lb, VA D 0, MA D �60 in-lb

MA

PA

VA

30 lb

Problem 10.4 Determine the internal forces andmoment at A.

y

x

900 ft-lbA

400 lb100 lb

3 ft 4 ft 3 ft 4 ft

Solution: Passing a plane through the beam at A and writing theequilibrium equations for the part of the beam to the right of A, weobtain

Fx : �PA D 0,

Fy : VA � 400 lb D 0,

MA : �MA C 900 ft-lb � �400 lb��7 ft� D 0.

Solving yields

PA D 0, VA D 400 lb, MA D �1900 ft-lb.

790


Problem 10.5 The pipe has a fixed support at the leftend. Determine the internal forces and moment at A.

y

x

0.2 m

2 kN

A

0.2 m0.2 m

0.2 m

2 kN

20�

Solution: Use the right section

∑Fx : �PA C 2 kN cos 20° D 0

∑Fy : VA C 2 kN sin 20° C 2 kN D 0

∑MA : �MA � �2 kN cos 20°��0.2 m�

C �2 kN sin 20°��0.2 m� C �2 kN��0.4 m� D 0

Solving:

PA D 1.88 kN, VA D �2.68 kN, MA D 0.561 kN-m

20°

VA

MA

PA

2 kN

2 kN

0.2 m

0.2 m

0.2 m


791

Problem 10.6 Determine the internal forces andmoment at A for each loading.

(b)

2 kN/m

(a)

2 m

1 m

8 kN

4 m

A

1 m

4 m

A

Solution: (a) Denote the reaction at the pinned left end by R, andthe reaction at the roller support by B. The reaction at B:

∑M D �2�8� C B�4� D 0,

from which B D 4 kN. The reaction at R:

∑Fy D Ry � 8 C B D 0,

from which Ry D 4 kN.

∑Fx D Rx D 0.

Make a cut at A: Isolate the left hand part. The sum of moments:

∑M D MA � 4�1� D 0,

from which MA D 4 kN-m VA D 4 kN PA D 0

(b) Determine the reaction at B: The sum of the moments about R:

∑MR D �

∫ 4

02x dx C 4B D 0,

from which

B D(

1

4

) [2

x2

2

]4

0D 16

4D 4 kN.

The reaction at R:

∑Fy D Ry �

∫ 4

02 dx C B D 0,

from which

Ry D 8 � 4 D 4 kN,

∑Fx D Rx D 0.

Make a cut at A: Isolate the left hand part. The sum of moments:

∑M D MA � �1�Ry C

∫ 1

02x dx D 0,

(a)

(b)

Ry = 4 kN

2 kN/m

4 kN

PA

PA

MA

MA

1 m VA

VA

from which MA D Ry � 1 D 3 kN m.

VA D Ry �∫ 1

02 dx D 4 � 2 D 2 kN

PA D 0.

792


Problem 10.7 Model the ladder rung as a simplysupported (pin supported) beam and assume that the750-N load exerted by the person’s shoe is uniformlydistributed. Determine the internal forces and momentat A. A

200 mm 100 mm

375 mm

250 mm

Solution:

∑Fy D B C C � 750 D 0,

∑M�pt. B� D 0.375C � �0.25��750� D 0.

Solving, B D 250 N, C D 500 N.

A

0.2 m 0.1 m 500 N250 N

0.2 m

0.25 m

0.025 m

0.05

0.05 m

250 N

250 N

MA

MA

PA

PA

x

VA

VA

The distributed load is

w D �750 N�/�0.1 m� D 7500 N/m.

From the equilibrium equations

∑Fx D PA D 0,

∑Fy D 250 � VA � 0.05�7500� D 0,

∑M�rightend� D MA � �250��0.25�

C �0.05��7500��0.025� D 0,

we obtain PA D 0, VA D �125 N, MA D 53.1 N-m.

750 N

0.25 m

0.375 m

B C


793

Problem 10.8 In Example 10.2, suppose that thedistance from point A to point B is increased from 3 mto 4 m. Draw a sketch of the beam with B in its newposition. Determine the internal forces and moment at B.

y

x

6 m

60 N/m

3 m 3 m

B CA

Solution: Let us pass a plane through the beam at B. Using similartriangles, the magnitude of the distributed load at B is

4

6�60 N/m� D 40 N/m.

If we represent the distributed load to the left of B by an equivalentforce, its magnitude is

1

2�40 N/m��4 m� D 80 N,

and it acts at the centroid of the distributed load to the left of B. Thedistance from A to the centroid is

2

3�4 m� D 2.67 m.

The equilibrium equations for the part of the beam to the left of B are

Fx : PB D 0,

Fy : 120 N � 80 N � VB D 0,

MB : MB � �120 N��4 m� C �80 N��1.33 m� D 0.

Solving yields

PB D 0, VB D 40 N, MB D 373 N-m.

Problem 10.9 If x D 3 ft, what are the internal forcesand moment at A?

x

y

3 ft 3 ft

600 lb/ft

600 lb/ftx A

Solution: Isolating the part of the beam to the right of A, werepresent the distributed load by an equivalent force. From this free-body diagram, we write the equilibrium equations:

Fx : �PA D 0,

Fy : VA C 900 lb D 0,

MA : �MA C �900 lb��1 ft� D 0.

Solving yields

PA D 0, VA D �900 lb, MA D 900 ft-lb.

794


Problem 10.10 If x D 4 ft, what are the internal forcesand moment at A?

x

y

3 ft 3 ft

600 lb/ft

600 lb/ftx A

Solution: Isolating the part of the beam to the right of A, werepresent the distributed load by an equivalent force.

We can obtain the magnitude of the distributed load by similar trian-gles:

2

3�600 lb/ft� D 400 lb/ft.

If we represent the distributed load to the right of point A by a singleequivalent force, its magnitude is

1

2�400 lb/ft��2 ft� D 400 lb,

and it acts at the centroid of the distributed load to the right of pointA. The distance from A to the centroid is

1

3�2 ft� D 0.667 ft.

From this free-body diagram, we write the equilibrium equations:

Fx : �PA D 0,

Fy : VA C 400 lb D 0,

MA : �MA C �400 lb��0.667 ft� D 0.

Solving yields

PA D 0, VA D �400 lb, MA D 267 ft-lb.


795

Problem 10.11 Determine the internal forces andmoment at A for the loadings (a) and (b).

A B

4 ft3 ft 5 ft

6 ft

A B

2 ft

240 lb 180 lb

3 ft 5 ft4 ft4 ft

60 lb/ft

(a)

(b)

Solution: The external reactions are the same for either loadingcondition∑

MA : ��240 lb��2 ft� � �180 lb��6 ft� C D�10 ft� D 0

) D D 156 lb

∑Fy : C C D � 240 lb � 180 lb D 0 ) C D 264 lb

C D

240 lb180 lb

(a) Use the left section with the distributed loading

∑Fx : PA D 0

∑Fy : C � 180 lb � VA D 0

∑MA : �C�3 ft� C �180 lb��1.5 ft� C MA D 0

Solving PA D 0, VA D 84 lb, MA D 522 ft-lb

C

180 lb

MA

PA

VA

3 ft

(b) Use the left section with the discrete load∑Fx : PA D 0

∑Fy : C � 240 lb � VA D 0

∑MA : �C�3 ft� C �240 lb��1.0 ft� C MA D 0

Solving PA D 0, VA D 24 lb, MA D 552 ft-lb

C

240 lb

MA

PA

VA

3 ft

796


Problem 10.12 For the loadings (a) and (b) shownin Problem 10.11, determine the internal forces andmoment at B.

Solution: The external reactions are the same for either loadingcondition∑

MA : ��240 lb��2 ft� � �180 lb��6 ft� C D�10 ft� D 0

) D D 156 lb

∑Fy : C C D � 240 lb � 180 lb D 0 ) C D 264 lb

240 lb180 lb

C D

(a) Use the right section with the distributed loading

∑Fx : �PB D 0

∑Fy : VB � 125 lb C D D 0

∑MB : �MB � 125 lb

(1

35 ft

)C D�5 ft� D 0

Solving PB D 0, VB D �31 lb, MB D 572 ft-lb

VB

D

PB

MB

125 lb

5 ft

(b) Use the right section with the discrete loads

∑Fx : �PB D 0

∑Fy : VB � 180 lb C D D 0

∑MB : �MB � �180 lb��1 ft� C D�5 ft� D 0

Solving PB D 0, VB D 24 lb, MB D 600 ft-lb

180 lb

4 ft

5 ft

D

VB

PB

MB


797


6 ft

8 ft 4 ft

A

200 lb/ft300 lb/ft

Solution: Use the whole body to find the reactions

∑MC : �B�8 ft� C �1600 lb��4 ft�

C �400 lb��2.67 ft� � �600 lb��1.33 ft� D 0

) B D 833 lb

400 lb

B C

600 lb1600 lb

Now examine the section to the left of the cut∑Fx : PA D 0

∑Fy : B � 1200 lb � 225 lb � VA D 0

∑MA : �B�6 ft� C �1200 lb��3 ft�

C �225 lb��2 ft� C MA D 0

Solving PA D 0, VA D �592 lb, MA D 950 ft-lb

B

6 ft

1200 lb

225 lb

275 lb/ft

200 lb/ft

VA

MA

PA

798


Problem 10.14 Determine the internal forces andmoment at A. A

B

1 m

10 kN

1 m 1 m 1 m 1 m

Solution: The complete structure as a free body : The sum of themoments about the right end:

∑M D 3�10� � 5R D 0,

from which R D 30

5D 6 kN. The sum of forces in the y-Direction:

∑Fy D Ry C Cy � 10 D 0,

from which Cy D 4 kN. The element CA as a free body : The sum ofthe moments about C:

∑MC D �4F1 C 10�3� � F2 D 0.

The sum of the forces:

∑Fy D Cy C F2 � 10 C F1 D 0.

Solve the simultaneous equations: F1 D 8 kN, F2 D �2 kN. Make acut at A: Isolate the left end of CA. The sum of the moments about A:

∑M D MA � 2F1 C 10 D 0,

from which

MA D �10 C 16 D 6 kN-m

VA D 8 � 10 D �2 kN,

PA D 0

RY

CY

CX

10 kN

2 m 3 m

F1 F2 CY

CX10 kN

A

1 m 2 m 1 m

F1

PA

MA

1 m 1 m

10 kN

VA

Problem 10.15 Determine the internal forces andmoment at point B in Problem 10.14.

Solution: Use the solutions to Problem 10.14. Make a cut at pointB : Isolate the left part. The sum of the moments about B:

∑MB D MB C 2F1 � 3Ry D 0,

from which

MB D �16 C 18 D 2 kN-m

VB D Ry � F1 D 6 � 8 D �2 kN

PB D 0

RY

PB

MB

1 m 2 m

F1 = 8

VB6


799


x

0.2m

0.2m

0.6 m

0.4 m

0.4 m

0.4 m 0.4 m

600 Ny

A B

Solution: Use the entire structure to find the reactions∑Fx : Cx C 600 N D 0 ) Cx D �600

∑MC : ��600 N��1.0 m� C D�0.8 m� D 0 ) D D 750 N

∑Fy : Cy C D D 0 ) Cy D �750 N

Cy

Cx

600 N

D

Next examine the vertical bar to find the tension in the cable

∑ME : ��600 N��0.4 m� C 1p

5T�0.8 m� D 0 ) T D 671 N

T

2

1

600 N

Ex

Ey

Finally cut at A and look at the left section

∑Fx : Cx C 1p

5T C PA D 0

∑Fy : Cy C 2p

5T � VA D 0

∑MA : Cx�0.6 m� � Cy�0.2 m� � 2p

5T�0.2 m� C MA D 0

Solving we have

PA D 300 N, VA D �150 N, MA D 330 N-m

T

2

1

PA

VA

Cx

Cy

MA

800


Problem 10.17 Determine the internal forces andmoment at point B of the frame in Problem 10.16.

Solution: Use the section to the right of the cut at B

∑Fx : �PB D 0

∑Fy : VB C D D 0

∑MB : �MB C D�0.2 m� D 0

Solving PB D 0, VB D �750 N, MB D 150 N-m

D

VBMB

PB

Problem 10.18 The tension in the rope is 10 kN.Determine the internal forces and moment at point A.

0.6 m

0.6 m

0.8 m

0.8 m 0.8 m

A

y

x

3 kNSolution: Use the whole structure first∑MC : �Bx�1.2 m� � �3 kN��1.6 m� D 0 ) Bx D �4 kN

By

Cy 3 kN

Bx

Cx

Now examine the bent bar (T D 10 kN)

∑MD : �Bx�1.2 m� � By�1.6 m� C 4

5�T��0.6 m� D 0

) By D 6 kN

By

Dy

Bx

Dx

3 kN

T 4

3

Finally cut the bar at A and examine the left section

By

Bx

VA

MA

PA

∑Fx : Bx C PA D 0

∑Fy : By � VA D 0 )

PA D 4 kN

VA D 6 kN

MA D 4.8 kN-m∑MA : �By�0.8 m� C MA D 0


801

Problem 10.19 Determine the internal forces andmoment at point A of the frame.

0.2 m

0.2 m

0.2 m

0.8 m

A

x

y

3 kN

Solution: Use the whole structure first∑MB : ��3 kN��0.4 m� C C�0.8 m� D 0 ) C D 1.5 kN

By

Bx

3 kN

C

Next examine the slanted bar and take advantage of the 2-force member

∑MD : C�0.8 m� � T�0.4 m� D 0 ) T D 3 kN

Dy

Dx

T

C

Now cut the bar at A and look at the lower right section

∑F% : VA � 3

5T C 4

5C D 0

∑F- : PA C 4

5T C 3

5C D 0

∑MA : �T�0.2 m� C C

(2

30.8 m

)� MA D 0

Solving PA D �3.3 kN, VA D 0.6 kN, MA D 0.2 kN-m

4

3

T

C

VA

MA

PA

802



1 m

4 kN/m

1 m 1 m

2 m

B

A x

y

Solution: The free-body diagrams of the horizontal members are

CY

CX

DY

DX

(a)

(b)A

BT

Tθ

θ

1 m

1 m

2 m

2 m RR

1.5 m(3 m)(4 kN/m) = 12 kN

The angle � D arctan�2/1� D 63.4°.

From free-body diagram (a),

∑Fx D Cx � R cos � D 0,

∑Fy D Cy � 12 � R sin � � T D 0,

∑M�pt. C� D ��1.5��12� � 2R sin � � 3T D 0,

and from free-body diagram (b),

∑Fx D Dx C R cos � D 0,

∑Fy D Dy C R sin � C T D 0,

∑M�pt. D� D �1�R sin � C 3T D 0.

Solving, we obtain Cx D �9 kN, Cy D 0, and T D 6 kN.

Cutting member (b) at A,

PA

MA

1 m

6 kNVA

we see that PA D 0, VA D �6 kN, MA D �1��6� D 6 kN-m.

Problem 10.21 Determine the internal forces andmoment at point B of the frame in Problem 10.20.

Solution: See the solution of Problem 10.20. Cutting member (a)at B and including the distributed load acting on the part of the memberto the left of B,

9 kN

9 kN

4 kN

MB

MB

PB

PB

1 m

0.5 m

4 kN/m

VB

VB

we see that PB D 9 kN, VB D �4 kN, MB D ��0.5��4� D �2 kN-m.


803

Problem 10.22 Determine the shear force and bendingmoment as functions of x.

Strategy: Cut the beam at an arbitrary position x anddraw the free-body diagram of the part of the beam tothe left of the plane.

400 lb

3 ft

y

x

Solution: Cut the beam at arbitrary position x and look at sectionto the left of the cut∑

Fy : 400 lb � V D 0

∑Mcut : ��400 lb�x C M D 0

Solving we have

V D 400 lb

M D �400 lb�x

400 lb

V

x

M

P

Problem 10.23 (a) Determine the shear force andbending moment as functions of x.

(b) Draw the shear force and bending moment diagrams.

6 m

48 kN/ my

x

Solution: First determine the reactions∑MB : �A�6 m� C �144 kN��2 m� D 0 ) A D 48 kN

4 m

144 kN

2 m

AB

Now cut the beam at arbitrary x and examine the left section.

R D 1

2x

(x

6 m48

kN

m

)D 4x2 kN

m2

∑Fy : A � R � V D 0

∑Mcut : �Ax C R

( x

3

)C M D 0

2x/3

A

V

R

x P

M

(a) Solving we find

V D �48 � 4x2� kN, M D 4

3�36x � x3�kN-m

(b) The shear and moment diagrams

48 kN

48 kN/m

96 kN

48 kN

�96 kN

150 kN-m

100 kN-m

50 kN-m

0 x

x

x

M

0

V

804


Problem 10.24 (a) Determine the shear force andbending moment as functions of x.

(b) Show that the equations for V and M as functionsof x satisfy the equation V D dM/dx.

Strategy: For part (a), cut the beam at an arbitraryposition x and draw the free-body diagram of the partof the beam to the right of the plane.

12 ft

x

y

60 lb/ft

Solution: Cut the beam at arbitrary x and examine the section tothe right

R D 1

2�12 � x�

(12 � x

12

)�60� D 5

2�12 � x�2

∑Fy : V � R D 0

∑Mcut : �M � R

(12 � x

3

)D 0

(a) Solving V D 5

2�x � 12�2 lb, M D 5

6�x � 12�3 ft-lb

(b)dM

dxD 5

2�x � 12�2 D V

12 − x

VM

R

Problem 10.25 Draw the shear force and bendingmoment diagrams for the beam in Problem 10.24.

Solution: The diagrams

0

360 lb

360 lb

1440 ft-lb

60 ft/lb

0

1440 ft-lb

y

x

x

x

M

V


805

Problem 10.26 Determine the shear force and bendingmoment as functions of x for 0 < x < 2 m.

3600 N-m

x

y

2 m 4 m

Solution: Examine the whole beam first∑MB : �A�6 m� � 3600 N-m D 0 ) A D �600 N

∑Fy : A C B D 0 ) B D 600 N

Now cut at arbitrary x < 2 m and examine the left cut

∑Fy : A � V D 0

∑Mcut : �Ax C M D 0

Solving V D �600 N, M D ��600 N�x for 0 < x < 2 m

3600 N-m

A

2 m 4 m

B

A

M

V

x

Problem 10.27 In Active Example 10.3, suppose thatthe 40 kN/m distributed load extends all the way acrossthe beam from A to C. Draw a sketch of the beamwith its new loading. Determine the shear force V andbending moment M for the beam as functions of x for2 < x < 4 m.

y

x40 kN/m

2 m

A

60 kN

B C

2 m

Solution: Cutting the beam at an arbitrary position x in the range2 m < x < 4 m, we obtain a free-body diagram of the part of the beamto the right of x. We represent the distributed load by an equivalentforce.


Fy : V � �40 kN/m��4 m � x� C 60 kN D 0,

Mleft end : �M C �60 kN��4 m � x�

� [�40 kN/m��4 m � x�][ 12 �4 m � x�] D 0.

We obtain

V�x� D 100 kN � �40 kN/m�x,

M�x� D �80 kN-m C �100 kN�x � �20 kN/m�x2.

806


Problem 10.28

(a) Determine the internal forces and moment as func-tions of x.

(b) Determine the shear force and bending momentdiagrams.

100 lb/ft

x

y

6 ft 6 ft

Solution: The reactions at the left end : The area under the loaddistribution is

F D ( 12

)�6��100� D 300 lb.

The centroidal distance is

d D 6 C ( 23

)6 D 10 ft.

The sum of the moments about the left end:

∑M D MA � dF D 0,

from which MA D 10�300� D 3000 ft lb. The shear and moment asa function of x : Divide the beam into two intervals: 0 � x � 6,and 6 < x � 12. Interval 1 : The shear as a function of x: V1�x� DF D 300 lb. The moment: M1�x� D Fx � MA, from which M1�x� D300x � 3000 ft lb. Interval 2 : The load curve is a straight line, with

intercept �100 lb/ft, and slope100

6. The shear diagram over this

interval is

V2�x� D F �∫ x

6

(�100 C 100

6x

)dx

D F � 100

[�x C 1

12x2

]x

6D 100

(x � x2

12

)

Check : The load curve is continuous at x D 6, hence the shear diagrammust be continuous at x D 6, (since the integral of a continuous func-tion is also continuous) hence

V1�6� D V2�6�

V1�6� D F D 300 D V2�6�

D 100�6� � 100

(36

12

)D 300

check. The force due to the distributed load in the interval 6 � x �12 is

F2�x� D∫ x

6w dx.

Integrate and reduce:

F2�x� D 300 � 100x C 25

3x2.

The centroid distance from x is

d2 D �x � 6�

3.

MA

MAM1(x)

M2(x)

P1(x)

P2(x)MA

100 lb/ft

6 ft 6 ftF

F

F

x

x

V2(x)

V1(x)

–3000

–2500

–2000

–1500

–1000

–500

0

500

0 2 4 6 8 10 12X, ft

Shear & Moment Diagrams

Shear

Moment

The moment about x is M2�x� D Fx � MA � d2F2. Substitute andreduce:

M2�x� D �2400 C 50x2 �(

25

9

)x3 ft lb.

Check : The moment must be zero at x D 12. check. Check : Themoment must be continuous at x D 6, M1�6� D M2�6�, from whichM1�6� D �1200, and M2�6� D �1200 check. The axial forces arezero, P�x� D 0

(b) The graph was drawn with TK Solver Plus.


807

Problem 10.29 The loads F D 200 N and C D800 Nm.

(a) Determine the internal forces and moment as func-tions of x.

(b) Draw the shear force and bending momentdiagrams.

y

x

F

4 m 8 m

C

4 m

Solution: The reactions at the supports: The sum of the momentsabout the left end:

∑M D C C 8B � 16F D 0,

from which

B D ( 18

)��C C 16F� D ( 1

8

)��800 C 16�200�� D 300 N.

The sum of the forces:∑

Fy D Ry C B � F D 0,

from which Ry D �100 N. The intervals as free bodies : Divide theinterval into three parts: 0 � x � 4, 4 < x � 8, and 8 < x � 16.Interval I : The shear is V1�x� D Ry D �100 N. The moment isM1�x� D CRyx D �100x N m. Interval 2 : The shear is V2�x� D Ry D�100 N. The sum of the moments is

∑Mx D M2�x� C C � Ryx D 0,

from which, M2�x� D �100x � 800 Nm. Interval 3 : The shear isV3�x� D RY C B D �100 C 300 D 200 N. The sum of the moments is

∑Mx D M3�x� C C � Ryx � B�x � 8� D 0,

from which

M3�x� D �800 � 100x C 300x � 2400 D 200x � 3200 N m.

The axial forces are zero, P�x� D 0 in all intervals.

(b) The diagrams are shown.

–2000

–1750

–1500

–1250

–1000

–750

–500

–250

0

0 2 4 6 8 10 12 14 16

250

500

X, m

Shear Force

Bending Moment

Shear & Moment Diagram

Problem 10.30 The beam in Problem 10.29 will safelysupport shear forces and bending moments of magni-tudes 2 kN and 6.5 kN-m, respectively. On the basisof this criterion, can it safely be subjected to the loadsF D 1 kN, C D 1.6 kN-m?

Solution: From the solution to Problem 10.29, the shear and themoments in the intervals are Interval 1 : V1 D Ry , M1�x� D Ryx,Interval 2 : V2�x� D Ry , M2�x� D Ryx C C, Interval 3 : V3�x� D Ry CB, M3�x� D �Ry C B�x C C � 8B. The reactions are

B D ( 18

)�16F � C�,

and Ry D F � B.

The maximum shears in each interval have the magnitude rank:

jV1�x�j D jV2�x�j � jV3�x�j,

so that the largest shear for a force F D 1 kN is V3�x� D Ry CB D F � B C B D F D 1 kN, which can be safely supported. Themaximum moment magnitudes in each interval have the rank:jM1�x�j � jM2�x�j � jM3�x�j. The maximum moment magnitudeoccurs in the third interval:

RX

RY

C

B

F

4 m 8 m4 m

M3�x� D �Ry C B�x C C � 8B D Fx C C � �16F � C�

D Fx � 16F C 2C.

The maximum magnitude occurs at x D 8, jM3�8�j D 8F D 8 kN mand it exceeds the safe limit by 2.5 kN m. NO

808


Problem 10.31 Model the ladder rung as a simplysupported (pin-supported) beam and assume that the750-N load exerted by the person’s shoe is uniformlydistributed. Draw the shear force and bending momentdiagrams.

200 mm 100 mm

375 mm

x

y

Solution: See the solution of Problem 10.7. The free-bodydiagram of the rung is

250 N 500 N0.2 m 0.1 m

0.375 m

7500 N/m

x

0 < x < 0.2 m

250 N

P

M

x

V

V D 250 N, M D 250x N-m.

0.2 < x < 0.3 m

7500 N/m

250 N 0.2 m

x

M

P

V

250 N

7500 (x–0.2) N

M

P

x

V– (x–0.2)12

V D 250 � 7500�x � 0.2� N,

M D 250x � 12 7500�x � 0.2�2 N-m.

0.3 < x < 0.375 m

P

M

(0.375–x)

500 N

V

V D �500 N, M D 500�0.375 � x� N-m.

500

–500

0

V, N

M, N

–m

0

60

40

20

00 0.1 0.2 0.3 0.4

0.1 0.2 0.3 0.4

X, m

X, m


809

Problem 10.32 What is the maximum bending momentin the ladder rung in Problem 10.31 and where does itoccur?

Solution: See the solution of Problem 10.31. The maximummoment occurs in the interval 0.2 < x < 0.3 m, in which

M D 250x � 3750�x � 0.2�2 N-m.

SettingdM

dxD 250 � 7500�x � 0.2� D 0,

we find that the maximum moment occurs at x D 0.233 m. Substitutingthis value into the expression for M gives M D 54.2 N-m.

Problem 10.33 Assume that the surface the beam restson exerts a uniformly distributed load. Draw the shearforce and bending moment diagrams.

4 kN

6 m

2 m 1 m

2 kN

y

x

Solution: The load density is w D 6

6D 1 kN/m.

The intervals as free bodies: Divide the beam into three intervals:

0 � x < 2 (m),

2 � x < 5 (m),

and 5 � x � 6 (m).

Interval 1 : The shear force is

V1�x� D∫ x

0w dx D x kN.

The force to the left is

F1�x� D∫ x

0w dx D x kN.

The centroid distance from x is d1 D x

2.

The moment is

M1�x� D F1�x�d1 D x2

2kN m.


V2�x� D V1�x� � 4 D x � 4 kN.

The moment is

M2�x� D M1�x� � 4�x � 2� D x2

2� 4x C 8 kN m.


V3�x� D x � 4 � 2 D x � 6 kN.

The bending moment is

M3�x� D M2�x� � 2�x � 5� D x2

2� 6x C 18 kN.

The shear and moment diagrams are shown.

4 kN

2 m5 m

6 m

2 kN

1 kN/m

M1(x)P1(x)

P2(x)

P3(x)

M2(x)

M3(x)

x

x

x

V1(x)

V2(x)

V3(x)

4 kN

4 kN 2 kN

Shear & Moment Diagrams2.5

1.5

.5

–.5

–2–1.5

–2.5

–1

2

0

0 1 2 3 4 5 6

1

X, m

Moment

kN, k

N-m

Shear

810


Problem 10.34 The homogeneous beams AB and CDweigh 600 lb and 500 lb, respectively. Draw the shearforce and bending moment diagrams for beam AB.

200 lb 2 ft

6 ft

B

CD

A

5 ft

Solution: Find reactions∑MB : �200 lb��6 ft� C �600 lb��3 ft� � C�2 ft� D 0

∑Fy : �200 lb � 600 lb C C � B D 0

) C D 1500 lb, B D 700 lb

For 0 < x < 4 ft we have

∑Fy : �200 lb �

(100

lb

ft

)x � V D 0

∑Mcut : �200 lb�x C

(100

lb

ft

)x( x

2

)C M D 0

V D �200 lb �(

100lb

ft

)x

M D ��200 lb�x �(

50lb

ft

)x2

For 4 ft < x < 6 ft we have

100 lb/ft (6 ft�x)

6 ft�x

B

VM

∑Fy : V � B � 100

lb

ft�6 ft � x� D 0

∑Mcut : �M �

(100

lb

ft

)�6 ft � x�2

2� B�6 ft � x� D 0

V D 700 lb � �100 lb/ft��x � 6 ft�,

M D 700 lb�x � 6 ft� � �50 lb/ft��x � 6 ft�2

600 lb

200 lbC

B

(100 lb/ft)x

200 lb

x

M

V

y

200 lb 1500 lb 700 lb

x100 lb/ft

x

900 lb700 lb

0�200 lb

�600 lb

V

0

�1600 ft-lb

M

x


811

Problem 10.35 Draw the shear force and bendingmoment diagrams for beam CD in Problem 10.34.

Solution: Use the reactions from 10.34

100 lb/ft x

V

x

C

M

100 lb/ft x

M

BC

X

V

For 0 < x < 2 ft∑Fy : �C � V � �100 lb/ft�x D 0

∑Mcut : Cx C [�100 lb/ft�x]x/2 C M D 0

V D �1500 lb � �100 lb/ft�x

M D ��1500 lb�x � �50 lb/ft�x2

For 2 ft < x < 5 ft∑Fy : �C C B � �100 lb/ft�x � V D 0

∑Mcut : Cx � B�x � 2 ft� C �100 lb/ft�x2/2 C M D 0

V D �800 lb � �100 lb/ft�x

M D �1400 ft lb � �800 lb�x � �50 lb/ft�x2

y

1500 lb 700 lb 1300 lb

x

x

6650 ft-lb

V

0

�1000 lb

�1300 lb

�1500 lb

�1700 lb

�3200 ft-lb

�6650 ft-lb

M

0 x

100 lb/ft

812


Problem 10.36 Determine the shear force V and bendingmoment M for the beam as functions of x for 0 < x <3 ft.

3 ft

600 lb/ft

3 ft

x

y

600 lb/ft

Solution: From the free-body diagram of the entire beam (Fig. a),we obtain the equilibrium equations

Fx : Ax D 0,

Fy : Ay � 900 lb C 900 lb D 0,

MA : MA � �900 lb��2 ft�

C �900 lb��4 ft� D 0.

We see thatAx D Ay D 0, MA D �1800 ft-lb.

Cutting the beam at an arbitrary position x in the range 0 < x < 3 ft,we obtain a free-body diagram of the part of the beam to the left of x(Fig. b). The distributed load is replaced by the equivalent force

R D 1

2

[(600 lb/ft

3 ft

)x

]x D �100 lb/ft2�x2.

From the equilibrium equations we have

Fy : ��100 lb/ft2�x2 � V�x� D 0,

Mright end : ��1800 lb-ft� C �100 lb/ft2�x2�x/3� C M�x� D 0.

We obtain

V�x� D ��100 lb/ft2�x2, M�x� D ��33.3 lb/ft2�x3 C �1800 lb-ft�.


813

Problem 10.37 Draw the shear force and bendingmoment diagrams for the beam.

3 ft

600 lb/ft

3 ft

x

y

600 lb/ft

Solution: In the solution to Problem 10.36 we found that in therange 0 < x < 3 ft, we have

V�x� D ��100 lb/ft2�x2

M�x� D ��33.3 lb/ft2�x3

C �1800 lb-ft�.

Cutting the beam at an arbitrary position x in the range 3 ft < x < 6 ft,and isolating the right part (Fig. a), we have the free-body diagramshown. The distributed load is replaced by the equivalent force

R D 1

2

[(600

lb

ft

) (6 ft � x

3 ft

)]�6 ft � x� D �100 lb/ft2��6 ft � x�2

From the equilibrium equations we learn

Fy : V�x� C �100 lb/ft2��6 ft � x�2 D 0,

Mleft end : �M�x� C [�100 lb/ft2��6 ft � x�2][ 13 �6 ft � x�] D 0.

Thus

V�x� D ��100 lb/ft2��6 ft � x�2

M�x� D �33.3 lb/ft2��6 ft � x�3.

The resulting diagrams are shown.

3 ft

600 lb/ft

3 ft

x

y

600 lb/ft

814


Problem 10.38 In preliminary design studies, thevertical forces on an airplane’s wing are modeled asshown. The distributed load models aerodynamic forcesand the force exerted by the wing’s weight. The 80-kNforce at x D 4.4 m models the force exerted by theweight of the engine. Draw the shear force and bendingmoment diagrams for the wing for 0 < x < 4.4 m.

y

x

50 kN/m

80 kN

4.4 m 13.0 m

Solution: From the free-body diagram of the entire wing (Fig. a),we obtain the equilibrium equations

Fx : Ax D 0,

Fy : Ay C �220 C 325 � 80� kN D 0,

MA : MA C �220 kN��2.2 m�

� �80 kN��4.4 m�

C �325 kN��17.4 � 8.67� m D 0.

Solving yields

Ax D 0, Ay D �465 kN, MA D �2970 kN-m.

Cutting the wing at an arbitrary position x in the range 0 < x < 4 m,and representing the distributed load by an equivalent force (Fig. b),the equilibrium equations are

Fy : �465 kN C �50 kN/m�x � V�x� D 0,

Mright end : �2970 kN-m C �465 kN�x

� [�50 kN/m�x]�x/2� C M�x� D 0.

Therefore

V�x� D �50 kN/m�x � 465 kN,

M�x� D �25 kN/m�x2 � �465 kN�x C 2970 kN-m.

The resulting diagrams are shown.

x

x

V

�245 kN

2970 kN-m

1408 kN-m

�465 kN

4.4 m

4.4 m

M


815

Problem 10.39 Draw the shear force and bendingmoment diagram for the entire wing. y

x

50 kN/m

80 kN

4.4 m 13.0 m

Solution: The shear force and bending moment diagrams for 0 <x < 4.4 m were obtained in the solution to Problem 10.38. Cut thewing at an arbitrary position x in the range 4.4 m < x < 17.4 m andisolate the right part of the beam (Fig. a). The distributed loading isrepresented by the equivalent force

R D 1

2

[(17.4 m � x

13.0 m

) (50

kN

m

)]�17.4 m � x�

D(

25 kN

13 m2

)�17.4 m � x�2

Using the equilibrium equations we find

V�x� D �(

25 kN

13 m2

)�17.4 m � x�2,

M�x� D R[ 13 �17.4 m � x�] D

(25 kN

39 m2

)�17.4 m � x�3

The diagrams for the entire beam are shown.

816


Problem 10.40* Draw the shear force and bendingmoment diagrams.

x

y

6 m

20 kN-m

6 m 6 m6 kN

4 kN/m

Solution: Start with the reactions∑MA : �6 kN��6 m� C 20 kN-m

� �24 kN��3 m� � �12 kN��8 m� C B�6 m� D 0

∑Fy : A C B � 6 kN � 24 kN � 12 kN D 0

Thus A D 23.3 kN, B D 18.67 kN

20 kN-m

24 kN 12 kN

6 kN A B

The complete diagrams:

V

y

20 kN-m

6kN

20 kN

10 kN

0

�10kN

4kN/m

23.3 kN 18.7 kN

x

x

x

M

0

�20 kN-m

�40 kN-m

�60 kN-m


817

Problem 10.41 Draw the shear force and bendingmoment diagrams.

4 ft

50 lb 50 lb

4 ft

y

x

Solution: In the first region 0 < x < 4 ft

w1 D 0

V1 D 50 lb

M1 D �50 lb�x

In the second region 4 ft < x < 8 ft

w2 D 0

V2 D 0

M2 D 200 ft lb

y

50 lb

50 lb

200 ft-lb

0

0

50 lb200 ft-lb

x

x

x

V

M

818



3600 N-m

x

y

2 m 4 m

Solution: We must first find the reactions∑MB : �A�6 m� � 3600 N-m D 0

∑Fy : A C B D 0

A D �600 N, B D 600 N

In the first region 0 < x < 2 m

w1 D 0

V1 D A D �600 N

M1 D ��600 N�x

In the second region 2 m < x < 6 m

w2 D 0

V2 D �B D �600 N

M2 D ��600 N�x C 3600 N-m

A B

3600 Nm

2 m 4 m

600 N

V

0

�600 N

M

2400 N-m

0

�1200 N-m

x

x

x

600 N

y

3600 N-m


819

Problem 10.43 This arrangement is used to subject asegment of a beam to a uniform bending moment. Drawthe shear force and bending moment diagrams.

50 lb 50 lb

x

y

6 in 6 in12 in

Solution: We first find the reactions∑MB : �A�24 in� C �50 lb��6 in� C �50 lb��18 in� D 0

∑Fy : A C B � 50 lb � 50 lb D 0

A D B D 50 lb

In the first regions 0 < x < 6 in

w1 D 0

V1 D A D 50 lb

M1 D �50 lb�x

In the second region 6 in < x < 18 in

w2 D 0

V2 D 0

M2 D 300 in lb

In the last region 18 in < x < 24 in

w3 D 0

V3 D �B D �50 lb

M3 D ��50 lb�x C 1200 in lb

50 lb 50 lbA B

50 lb 50 lb 50 lb 50 lb

x

x

x

y

V

M

50 lb

0

�50 lb

300 in-lb

0

820


Problem 10.44 Use the procedure described inExample 10.5 to draw the shear force and bendingmoment diagrams for the beam.

6 m

4 kN/m

y

x

Solution: We can use the boundary conditions at the right endinstead of calculating the reactions at the left end

w D 4kN

m

V D �(

4kN

m

)x C 24 kN

M D �(

2kN

m

)x2 C �24 kN�x � 72 kN-m

The plots

y

x

x

x

4 kN/m

72 kN-m

24 kN

0

0

�72 kN-m

24 kN

M

V

Problem 10.45 In Active Example 10.4, suppose thatthe 40 kN/m distributed load extends all the way acrossthe beam from A to C. Draw a sketch of the beam withits new loading. Draw the shear force diagram for thebeam.

y

x40 kN/m

2 m

A

60 kN

B C

2 m

Solution: The free-body diagram with the reactions already solvedis shown.Think of the beginning just to the left of the beam, with the initialvalue of the shear force equal to zero. The upward 60-kN reaction atA causes an increase in the shear force of 60-kN magnitude.Between A and B, the distributed load on the beam is constant — thediagram is a straight line. The change in V between A and B can bedetermined from Eq. (10.5),

VB � VA D ��2 m��40 kN/m�

D �80 kN.

Therefore V decreases linearly from 60 kN at A to 60 kN � 80 kN D�20 kN at B.The upward 40-kN reaction at B causes an increase in the shear forceof 40-kN magnitude, so at B the shear force increases from �20 kNto 20 kN.Between B and C, the distributed load on the beam is constant, so theshear diagram between B and C is a straight line. The change in Vbetween B and C can be determined from Eq. (10.5).

VC � VB D ��2 m��40 kN/m� D �80 kN.

Therefore V decreases linearly from 20 kN at B to 20 kN � 80 kN D�60 kN at C.


821


100 lb/ft

x

y

6 ft 6 ft

Solution: Find the reactions first∑MA : �300 lb�10 ft� C B�12 ft� D 0

∑Fy : A C B � 300 lb D 0

A D 50 lb, B D 250 lb

For 0 < x < 6 ft

w D 0

dV

dxD �w D 0 ) V D 50 lb

dM

dxD V D 50 lb ) M D �50 lb�x

For 6 ft < x < 12 ft

w D(

100 lb

6 ft

)�x � 6 ft� D

(50 lb

3 ft

)�x � 6 ft�

V D �(

50 lb

3 ft

)�x � 6 ft�2

2C 50 lb

D �(

25 lb

3 ft

)�x � 6 ft�2 C 50 lb

M D �(

25 lb

3 ft

)�x � 6 ft�3

3C �50 lb�x

D �(

25 lb

9 ft

)�x � 6 ft�3 C �50 lb�x

300 lb

A B

The plots

50 lb

50 lb0

�250 lb

M

V

400 ft-lb

200 ft-lb

0 x

x

250 lb

x

100 lb/ft

y

822


Problem 10.47 Determine the shear force V andbending moment M for the beam as functions of x.

3 ft

600 lb/ft

3 ft

x

y

600 lb/ft

Solution: From the free-body diagram of the entire beam we learnthat Ax D Ay D 0, MA D 1800 ft-lb.From x D 0 to x D 3 ft, the distributed load on the beam is

w D x

(600 lb/ft

3 ft

)D �200 lb/ft2�x.

Using this expression we can integrate Eq. (10.4) to determine V as afunction of x.∫ V

0dV D V D �

∫ x

0wdx D �

∫ x

0�200 lb/ft2�xdx

D ��100 lb/ft2�x2.

The clockwise couple at x D 0 causes an increase in the bendingmoment of 1800 ft-lb. We can integrate Eq. (10.6) to determine Mas a function of x.∫ M

0dM D M D �1800 ft-lb� C

∫ x

0Vdx D �1800 ft-lb� �

∫ x

0�100 lb/ft2�x2dx

D �1800 ft-lb� � �33.3 lb/ft2�x3.

From x D 3 ft to x D 6 ft, the distributed load on the beam canbe expressed as a linear equations w D ax C b. At x D 3ft, w D�600 lb/ft, and at x D 6 ft, w D 0. Using these two conditions todetermine a and b, we find that w is given as a function of x by

w D �200 lb/ft2�x � 1200 lb/ft.

To obtain V as a function of x, let us integrate Eq. (10.4) from anarbitrary position x in the range 3 ft < x < 6 ft to x D 6 ft.

∫ 0

VdV D �

∫ 6 ft

xwdx

V D∫ 6ft

x[�200 lb/ft2�x � 1200 lb/ft]dx

D [�100 lb/ft2�x2 � �1200 lb/ft�x]6 ftx

D �3600 lb C �1200 lb/ft�x � �100 lb/ft2�x2.

Then to obtain M as a function of x, we integrate Eq. (10.6) from anabrbitrary position x in the range 3 ft < x < 6 ft to x D 6 ft.

∫ 0

MdM D �

∫ 6 ft

xVdx

M D �∫ 6 ft

0[��3600 lb� C �1200 lb/ft�x � �100 lb/ft2� x2]x

D [�3600 lb�x � �600 lb/ft�x2 � �33.3 lb/ft2�x3]6 ftx

D �7200 lb-ft� � �3600 lb�x C �600 lb/ft�x2 � �33.3 lb/ft2�x3.

0 < x < 3 ft

V D ��100 lb/ft2�x2, M D �1800 ft-lb� � �33.3 lb/ft2�x3.

3 ft < x < 6 ft

V D ��3600 lb� C �1200 lb/ft�x � �100 lb/ft2�x2,

M D �7200 lb-ft� � �3600 lb�x C �600 lb/ft�x2 � �33.3 lb/ft2�x3.


823

Problem 10.48* Draw the shear force and bendingmoment diagrams.

x

y

6 m

20 kN-m

6 m 6 m6 kN

4 kN/m

Solution: From 10.40 we have A D 23.3 kN, B D 18.67 kN


w1 D 0

V1 D �6 kN

M1 D ��6 kN�x � 20 kN-m


w2 D 4 kN/m

V2 D ��4 kN/m��x � 6 m� C 17.2 kN

M2 D ��2 kN/m��x � 6 m�2 C �17.2 kN��x � 6 m� � 56 kN-m

In the last region 12 m < x < 18 m

w3 D 4 kN/m � 4 kN/m

6 m�x � 12 m�

D(

4kN

m

)�

(2 kN

3 m2

)�x � 12 m�

V3 D �(

4kN

m

)�x � 12 m� C

(1 kN

3 m2

)�x � 12 m�2 C 12 kN

M3 D �(

2kN

m

)�x � 12 m�2 C

(1 kN

9 m2

)�x � 12 m�3

C �12 kN��x � 12 m� � 24 kN-m

BA6 kN

24 kN

20 kNm

12 kN

The plots

V

y

20 kN-m

6kN

20 kN

10 kN

0

�10kN

4kN/m

23.3 kN 18.7 kN

x

x

x

M

0

�20 kN-m

�40 kN-m

�60 kN-m

824


Problem 10.49 Draw the shear force and bendingmoment diagrams for the beam AB.

1 m

400 N/m

1 m 1 m

2 m

A B

y

x

Solution: First we must find the reactions∑MB : �Ax �2 m� � �1200 N��1.5 m�

Using free-body diagram, ACD

∑Fx : Ax � 1p

5TC D 0

∑MA : ��1200 N��1.5 m� �

(2p5

TC

)�2 m� � TD �3 m� D 0

∑Fy : Ay � 1200 N � 2p

5TC � TD D 0

We find

Ax D �900 N, Ay D 0

TC D �2012 N, TD D 600 N

Now we are ready to construct the diagrams.

Ay

By

Bx

Ax

1

1200 N

2

Ay

TC TD

Ax

1200 N

DC


w1 D 400 N/m

V1 D ��400 N/m�x

M1 D ��200 N/m�x2


w2 D 400 N/m

V2 D ��400 N/m��x � 2 m� C 1800 N

M2 D ��200 N/m��x � 2 m�2 C �1800 N��x � 2 m� � 1600 N-m

The plots

400 N/m

600 N900 N900 N

1800 N

1000 N

600 N

0

�800 N

0

�800 N-m

x

x

x

y

BA

V

M


825

Problem 10.50 The cable supports a distributed loadw D 12,000 lb/ft. Using the approach described in ActiveExample 10.6, determine the maximum tension in thecable.

40 ft

w

90 ft

100 ft

Solution: Equation (10.10) must be satisfied for both attachmentpoints:

yL D 40 ft D 1

2axL

2, yR D 90 ft D 1

2axR

2.

Dividing the second equation by the first yieldsxR

2

xL2

D 2.25.

The horizontal span of the bridge is xR � xL D 100 ft.Solving these two equations yields xL D �40 ft and xR D 60 ft.Substituting the coordinates of the right attachment point into Eq.(10.10),

yR D 12 axR

2 ) 90 ft D 1

2a�60 ft�2 ) a D 0.05 ft�1.

Therefore the tension at the lowest point is

T0 D w

aD 12,000 lb/ft

0.05 ft�1 D 240,000 lb.

The maximum tension in the cable occurs at its right end. From Eq.(10.11),

T D T0

√1 C a2x2 D �240 kip�

√1 C �0.05 ft�1�2�60 ft�2 D 759 kip.

759 kip.

Problem 10.51 In Example 10.7, suppose that thetension at the lowest point of one of the main supportingcables of the bridge is two million pounds? What is themaximum tension in the cable?

x

yy � (2.68 � 10–4)x2

xR, yR

Solution:The parameter a D 5.37 ð 10�4 ft�1. The horizontal coordinate of oneof the supporting towers relative to the lowest point of the cable isxR D 735 ft. From Eq. (10.11) the maximum tension in the cable is

T D T0

√1 C a2x2

D �2000 kip�√

1 C �5.37 ð 10�4 ft�1�2�735 ft�2 D 2150 kip.

2.15 million pounds.

826


Problem 10.52 A cable is used to suspend a pipelineabove a river. The towers supporting the cable are 36 mapart. The lowest point of the cable is 1.4 m below thetops of the towers. The mass of the suspended pipe is2700 kg.

(a) What is the maximum tension in the cable?(b) What is the suspending cable’s length?

Solution: The distributed load is

w D �2700 kg��9:81 m/s2�

36 mD 736 N/m.

y

18 m

1.4 m

x

(a) Setting x D 18 m, y D 1.4 m in Eq. (10.10),

1.4 D 12 a�18�2,

we obtain

a D w

T0D 0.00864 m�1.

Therefore the tension at x D 0 is

T0 D w

aD 736

0.00864D 85,100 N.

From Eq. (10.11), the maximum tension is

T D T0

√1 C a2�18�2 D 86,200 N.

(b) Setting x D 18 m in Eq. (10.12), the length of the cable is

2s D 18√

1 C a2�18�2 C 1

aln�18a C

√1 C a2�18�2�

D 36.14 m.

Problem 10.53 In Problem 10.52, let the lowest pointof the cable be a distance h below the tops of the towerssupporting the cable.

(a) If the cable will safely support a tension of 70 kN,what is the minimum safe value of h?

(b) If h has the value determined in part (a), what isthe suspending cable’s length?

Solution: See the solution of Problem 10.52.

(a) The distributed load is w D 736 N/m. Therefore

w D 736 D aT0, (1)

And setting x D 18 m and T D 70,000 N in Eq. (10.11),

70,000 D T0

√1 C �18�2a2. (2)

From Eqs. (1) and (2) we obtain a D 0.0107 m�1, T0 D68,700 N. From Eq. (10.10),

h D 12 �0.0107��18�2

D 1.734 m.

(b) From Eq. (10.12), the length of the cable is

2s D 18√

1 C a2�18�2 C 1

aln�18a C

√1 C a2�18�2�

D 36.22 m.


827

Problem 10.54 The cable supports a uniformlydistributed load w D 750 N/m. The lowest point of thecable is 0.18 m below the attachment points C and D.Determine the axial loads in the truss members ACand BC.

C

0.4 m1.2 m

0.4 m

0.4 m

A B

D

E F

0.4 m

Solution:

y D 12 ax2 :

0.18 D 12 a�0.6�2.

From this equation we obtain a D 1 m�1.

Therefore

T0 D w

aD 750 N

and T D T0

√1 C a2�0.6�2 D 875 N.

From the equation

tan � D ax D �1��0.6�,

we obtain � D 30.96°.

The free-body diagram of joint C is shown.

45° θ

T

PBC

PAC

From the equations

∑Fx D T cos � � PAC cos 45° D 0,

∑Fy D �T sin � � PBC

� PAC sin 45° D 0,

we obtain

PAC D 1061 N,

PBC D �1200 N.

Ty

x

θ0.18 m

0.6 m

828


Problem 10.55 The cable supports a railway bridgebetween two tunnels. The distributed load is w D1 MN/m, and h D 40 m.

(a) What is the maximum tension in the cable?(b) What is the length of the cable?

h

36 m 36 m

Solution: The parameter

a D 2y

x2D 2

40

362D 0.06173.

The tension at the lowest point:

T0 D w

aD 1 ð 106

aD 16200 kN.

The maximum tension: TMAX D T0p

1 C a2x2, which, for x D36 m, TMAX D 39477 kN. The cable length is

s�x� D(

xp

1 C a2x2 C 1

aln

(ax C p

1 C a2x2))

,

which, for x D 36 m, L D 112.66 m.

Problem 10.56 The cable in Problem 10.55 will safelysupport a tension of 40 MN. What is the shortest cablethat can be used, and what is the corresponding valueof h?

Solution: The tension at the lowest point is

T0 D w

a.

The maximum tension is

TMAX D T0p

1 C a2x2.

Square both sides, substitute and reduce algebraically: T2o D T2

MAX �w2x2. The terms on the right are known: T2

MAX D 402�106�, andw2x2 D 362�106�. Solve for the parameter a,

a2 D 106

�402 � 362��106�D 3.29 ð 10�3,

from which a D 0.0574. The height is

h D ( 12

)ax2 D ( 1

2

)�0.0574��362� D 37.165 m.

The length is

s�x� D(

xp

1 C a2x2 C 1

aln

(ax C p

1 C a2x2))

,

which, for x D 36 m, L D 108.26 m


829

Problem 10.57 An oceanographic research ship towsan instrument package from a cable. Hydrodynamic dragsubjects the cable to a uniformly distributed force w D2 lb/ft. The tensions in the cable at 1 and 2 are 800 lband 1300 lb, respectively. Determine the distance h.

h

w1

2

300 ft

Solution: If one assumes that the cable is tangent to the verticalat the point 1, so that the 800 lb is the tension at the lowest point,the data is inconsistent; therefore the point 1 must be at a distance x1

from the lowest point. There are three unknowns in the problem: thedistance x1, the tension at the lowest point T0, and the parameter a.The three equations that define these unknowns are:

(1) 800 D T0

√1 C a2x2

1 ,

(2) T0 D w

aD 2

a,

(3) 1300 D T0

√1 C a2�x1 C 300�2.

These are reduced to two equations in two unknowns:

(1) 800 D(

2

a

) √1 C a2x2

1 ,

(2) 1300 D(

2

a

) √1 C a2�x1 C 300�2 and solved by iteration using

TK Solver Plus. The result: a D 3.596 ð 10�3, x1 D 287.5 ft.Using these values, the distance is

h D ( 12

)a�x1 C 300�2 � ( 1

2

)ax2

1 D 471.94 ft

Problem 10.58 Draw a graph of the shape of the cablein Problem 10.57.

Solution: The following equations are graphed:

(1) IF d > x1 then w D h � ( 12

)ad2 C ( 1

2

)ax2

1 ,

(2) z D 300 C x1 � d, where h D 471.9 ft, a D 3.59573 ð 10�3,x1 D 287.5 ft. The value w is plotted on the abscissa, and z isplotted on the ordinate. The result is a graph of the depth of thecable against the horizontal extension.

0

–50

–100

–150

–200

–300

–250

–3500 100 200 300 400 500

h, ft

Shape of towing cable

dept

h, f

t

830


Problem 10.59 The mass of the rope per unit lengthis 0.10 kg/m. The tension at its lowest point is 4.6 N.Using the approach described in Active Example 10.8,determine

(a) the maximum tension in the rope(b) the rope’s length.

x

y

12 m

Solution: The weight per unit length is

w D �0.1 kg/m��9.81 m/s2� D 0.981 N/m.

(a) a D w

T0D 0.981

4.6D 0.213 m�1.


T D T0 cosh�ax�

D 4.6 cosh[�0.213��6�]

D 8.91 N.

(b) From Eq. (10.22), the length is

2s D 2 sinh[�0.213��6�]

0.213

D 15.55 m.

Problem 10.60 The stationary balloon’s tether ishorizontal at point O where it is attached to the truck.The mass per unit length of the tether is 0.45 kg/m. Thetether exerts a 50-N horizontal force on the truck. Thehorizontal distance from point O to point A wherethe tether is attached to the balloon is 20 m. What isthe height of point A relative to point O?

O

A

Solution:

a D w

T0D �9.81��0.45�

50D 0.0883 m�1.

From equation (10.20),

y D 1

a[cosh�ax� � 1]

h D 1

0.0883fcosh[�0.0883��20�] � 1g D 22.8 m.

y

x

h

A

50 N

20 m


831

Problem 10.61 In Problem 10.60, determine the mag-nitudes of the horizontal and vertical components of theforce exerted on the balloon at A by the tether.

Solution: From the solution to Problems 10.60, a D 0.0883 m�1.The value of the tension at x D 20 m is

T D T0 cosh�ax� D 50 cosh[�0.0883��20�] D 150 N.

The slope at x D 20 m (Equation 10.19) is

� D tan � D sinh�ax� D sinh[�0.0883��20�] D 2.84,

so � D arctan 2.84 D 70.6°. The horizontal and vertical compo-nents are

Tx D �150� cos � D 50 N

Ty D �150� sin � D 142 N.

y

x

20 m50 N

150 N

θ

Problem 10.62 The mass per unit length of lines ABand BC is 2 kg/m. The tension at the lowest point ofcable AB is 1.8 kN. The two lines exert equal horizontalforces at B.

(a) Determine the sags h1 and h2.(b) Determine the maximum tensions in the two lines.

CBA h2h1

40 m60 m

Solution: The lines meet the condition for a catenary. (a) The lineAB. The weight density is

w D 2�9.81� D 19.62 N/m.

The parameter

a1 D w

TABD 19.62

1800D 0.0109.

The sag is

h1 D(

1

a1

)�cosh�30a1� � 1� D 4.949 m.

The line BC. The horizontal component of the tension at B is TAB D1.8 kN. Thus the tension at the lowest point in BC is 1.8 kN, and theparameter a for line BC is equal to a1. The sag is

h2 D(

1

a1

)�cosh�20a1� � 1� D 2.189 m.

(b) The line AB. The maximum tension is

TABMAX D TAB cosh�30a1� D 1897.1 N.

The line BC. The maximum tension is

TBCMAX D TAB cosh�20a1� D 1842.9 N.

832


Problem 10.63 The rope is loaded by 2-kg massessuspended at 1-m intervals along its length. The massof the rope itself is negligible. The tension in the rope atits lowest point is 100 N. Determine h and the maximumtension in the rope.

Strategy: Obtain an approximate answer by modelingthe discrete loads on the rope as a load uniformlydistributed along its length.

10 m

h

Solution: The equivalent distributed load is

w D �2 kg��9.81 m/s2�

1 mD 19.62 N/m.

Therefore a D w

T0D 19.62

100D 0.196 m�1.

From Eq. (10.20),

h D 1

afcosh[a�5�] � 1g D 2.66 m.


T D T0 cosh ax D �100� cosh[a�5�] D 152 N.

Problem 10.64 In Active Example 10.9, what are thetensions in cable segments 1 and 3?

1 m

1 m1

2

3

h2

m1m2

1 m 1 m

Solution: See the solution of Example 10.9. Cutting cablesegment 1, we obtain the free-body diagram

45°

Th

T1

TV

From the equation

∑Fx D �Th C T1 cos 45° D 0,

we obtain T1 D Th

cos 45°D 131

cos 45°D 185 N.

Cutting cable segment 3, we obtain the free-body diagram

m1gm2g

Th

T3

T

β

V

The angle ˇ is

ˇ D arctan

(h2

1

)D arctan�1.25� D 51.3°.

From the equation

∑Fx D �Th C T3 cos ˇ D 0,

we obtain

T3 D Th

cos ˇD 131

cos 51.3°D 209 N.


833

Problem 10.65 Each lamp weighs 12 lb.(a) What is the length of the wire ABCD needed tosuspend the lamps as shown?(b) What is the maximum tension in the wire?

12 in

12 in

18 in 18 in

30 in

A

B

C

D

Solution:

∑MB : �TV�12 in� C TH�12 in� D 0 ) TV D TH

Using the second free-body diagram shown,

∑MC : THh � TV�30 in� C �12 lb��18 in� D 0

Make a cut at the right attachment point and take moments

∑MD : TH�30 in� � TV�48 in� C �12 lb��36 in� C �12 lb��18 in� D 0

Solving together we find

TH D TV D 36 lb, h D 24 in

(a) The length of the cable is then

L D p122 C 122 C

√182 C �24 � 12�2

C√

182 C �30 � 24�2 D 57.6 in

(b) The maximum tension occurs where the angle is thegreatest (AB)

Tmax D TAB D√

362 C 362 D 50.9 lb

12 in

12 in

12 lb

B

TH

TV

TBC

12 in

12 in 18 in

12 lb

12 lb

h

B

C

TBC

TH

TV

834


Problem 10.66 Two weights, W1 D W2 D 50 lb, aresuspended from a cable. The vertical distance h1 D 4 ft.

(a) Determine the vertical distance h2.(b) What is the maximum tension in the cable?

6 ft 10 ft

h1 h2

3 ft

2 ft

W1W2

Solution: The strategy is to make cuts along the string and sumthe moments to the left of each cut. The three simultaneous equationsare then solved for the unknowns. Make a cut at left attachment andto the right of W1. Denote the components of the force exerted by thestring by FH and FV, with sign indicating direction. The sum of themoments about the right end:

∑M D �6FV � h1FH D 0.

Make the cut to the right of W2. The sum of the moments about theright end:

∑M D �h2FH � 16FV C 10W1 D 0.

Make the cut at the right attachment point and sum the moments tothe left:

∑M D �19FV � h2FH C 13W1 C 3W2 D 0.

Solve:

h2 D 4 ft,

FV D 50 lb,

and FH D �75 lb.

From the sum of the forces for the complete string the tension inthe right support string is equal to the tension in the string. Thus themaximum tension is

T D√

F2V C F2

H D 90.14 lb.

6 ft

6 ft

6 ft 3 ft

10 ft

10 ft

Fv

Fv

Fv

FH

FH

T2

T3

T3

FHh1

h2

h2 h2–2 ft

W1

W1 W2

W2W1


835

Problem 10.67 In Problem 10.66, W1 D 50 lb, W2 D100 lb, and the vertical distance h1 D 4 ft.

(a) Determine the vertical distance h2.(b) What is the maximum tension in the cable?

Solution: (a) Cut the cable at the left and just to the right of w1:

∑M�rightend� D h1Th � 6Tv D 0 (1)

Cut the cable at the left and just to the right of w2:

∑M�rightend� D h2Th � 16Tv C 10w1 D 0 (2).

Cut the cable at the left and right:

∑M�rightend� D 2Th � 19Tv C 13w1 C 3w2 D 0. (3)

Knowing

W1 D 50 lb.

W2 D 100 lb

and h1 D 4 ft,

equations (1), (2) and (3) can be solved for Th Tv and h2, obtaining

Th D 89.1 lb,

Tv D 59.4 lb,

h2 D 5.05 ft.

(b) The maximum tension occurs in the segment with the largest sloperelative to the horizontal. In this problem T3 is the largest tension. Theangle between T3 and the horizontal: is

� D arctan

(h2 � 2

3

)D 45.5°.

Summing horizontal forces on the third free body diagram, we obtain�Th C T3 cos � D 0, so

T3 D Th

cos �D 89.1

cos 45.5°D 127 lb.

TV

TV

Th

Th

T3

T2

h1

h1 h2

W1

W1W2

6 ft

6 ft 10 ft

TV

Th

T3

h1h2

W1 W2

6 ft 10 ft 3 ft

2 ft

836


Problem 10.68 Three identical masses m D 10 kg aresuspended from the cable. Determine the verticaldistances h1 and h3 and draw a sketch of theconfiguration of the cable.

2 m

h12 m

m

1

2m

h3

3

4

1 m 3 m 1 m

m

Solution: We make 3 cuts and then draw one diagram of thewhole system

∑MA1 : THh1 � TV �2 m� D 0

∑MA2 : TH �2 m� � TV �3 m� C �98.1 N��1 m� D 0

∑MA3 : THh3 � TV �6 m�

C �98.1 N��4 m� C �98.1 N��3 m� D 0

∑MA4 : �TV �7 m� C �98.1 N��5 m�

C �98.1 N��4 m� C �98.1 N��1 m� D 0

Solving we find

h1 D 1.739 m

h3 D 0.957 m

TH D 161.2 N

TV D 140.1 N

Sketch the configuration

TV

h1

T2

A1

TH

2 m

98.1 N

TV

TH

T3

A2

A1

h1

2 m

1 m2 m

98.1 N98.1 N

TV

T4

A3

A2

A1

TH h1h32 m

1 m2 m 3 m

98.1 N98.1 N

98.1 N

TV

T5

A4

A3

A2

A1

TH h1 h3

2 m

1 m1 m2 m 3 m

98.1 N98.1 N 98.1 N


837

Problem 10.69 In Problem 10.68, what are thetensions in cable segments 1 and 2?

Solution: Use the solution to 10.68

∑Fx : � 2 m√

�2 m�2 C h12

T1 C 1 m√�1 m�2 C �2 m � h1�2

T2 D 0

∑Fx :

h1√�2 m�2 C h1

2T1 � 2 m � h1√

�1 m�2 C �2 m � h1�2T2 � 98.1 N D 0

Solving T1 D 214 N, T2 D 167 N

98.1 N

T1

T2

h1

2 m 1 m2 m � h1

Problem 10.70 Three masses are suspended from thecable, where m D 30 kg, and the vertical distance h1 D400 mm. Determine the vertical distances h2 and h3.

h1 h2h3

m

200 mm

300 mm300 mm700 mm500 mm

1

2m

3

4

2m

Solution: Cutting to the right of the left mass,

TV 0.5 m

2 mg

Th

T2

h1

A1

we obtain

∑M�ptA1� D h1Th � �0.5�Tn D 0. (1)

Cutting to the right of the middle mass,

TV 0.5 m

2 mg

0.7 m

Th

T3

h2

A2

mg

we obtain

∑m�ptA2� D h2Th � �1.2�Tn C �0.7�2 mg D 0. (2)

Cutting to the right of the right mass, we obtain

TV 0.5 m

2 mg

0.7 m 0.3 m

Th

T4

h3

A3

mgmg

∑M�ptA3� D h3Th � �1.5�Tn C �1�2 mg C �0.3� mg D 0. (3)

Finally, cutting at the right attachment point,

TV0.5 m

2 mg

0.7 m 0.3 m 0.3 m

0.2 mTh

T4

A4

mg

mg

we obtain

∑M�ptA4� D �0.2�Th � �1.8�Tn C �1.3�2 mg

C �0.6� mg C �0.3� mg D 0. (4)

Solving Equations (1)–(4), we obtain Th D 831 N, Tn D 665 N, h2 D464 mm, h3 D 385 mm.

838


Problem 10.71 In Problem 10.70, what is the maxi-mum tension in the cable, and where does it occur?

Solution: The tension is greatest in the segment with the greatestslope, which is either segment 1 or segment 4.

Slope of segment 1 (see the solution of Problem 10.71):

h1

500D 400

500D 0.8.

Slope of segment 4:

h3 � 200

300D 385 � 200

300D 0.62.

Cutting segment 1, we obtain

TV

βTh

T1

The angle ˇ is

ˇ D arctan

(h1

500

)D arctan

(400

500

)D 38.7°.

From the equation

∑Fx D �Th C T1 cos ˇ D 0,

we obtain

T1 D Th

cos ˇD 831

cos 38.7°D 1060 N.


839

Problem 10.72 Each suspended object has the sameweight W. Determine the vertical distances h2 and h3.

2 ft 3 ft 4 ft

4 ft

h3

h2

14 ft

5 ft

W

W

W

Solution:

∑MB : �TV�2 ft� C TH�4 ft� D 0

∑MC : �TV�5 ft� C THh2 C W�3 ft� D 0

∑MD : �TV�9 ft� C THh3 C W�7 ft� C W�4 ft� D 0

∑ME : �TV�14 ft� C TH�14 ft� C W�12 ft� C W�9 ft� C W�5 ft� D 0

Solving we find h2 D 8.38 ft, h3 D 12.08 ft

2 ft

4 ft

W

B

TH

TV

TBC

5 ft

B

C

W

W

h2

TV

TCD

TH

9 ft

W

W

W

D

C

B

TV

TH

TDE

h3

840


Problem 10.73 An engineer planning a water systemfor a new community estimates that at maximumexpected usage, the pressure drop between the centralsystem and the farthest planned fire hydrant will be25 psi. Fire fighting personnel indicate that a gagepressure of 40 psi at the fire hydrant is required. Theweight density of the water is � D 62.4 lb/ft3. How tallwould a water tower at the central system have to be toprovide the needed pressure?

Solution: The total pressure will be 65

(lb

in2

). A tower of height h

will produce a pressure of P D �h, where the units are to be consistent.

The tower height is h D(

P

�

).

Let the tower height be in feet:

h ft D(

1

62.4

) (ft3

lb

)�65�

(lb

in2

) (12 in

1 ft

)2

D 150 ft

Problem 10.74 A cube of material is suspended belowthe surface of a liquid of weight density � . By calcu-lating the forces exerted on the faces of the cube bypressure, show that their sum is an upward force ofmagnitude �b3.

b

d

Solution: Neglect the pressures on the supporting wire. The forceon the top surface is FTOP D �db2. The force on the sides is

FS D �

(d C b

2

)b2,

where

(d C b

2

)

is the distance of the centroid of the area from the surface. The forceon the bottom surface is FBOTTOM D ��d C b�b2. The forces on thesides cancel by symmetry. The difference between the downward forceon the top surface and the upward force on the bottom surface is theresultant:

R D ��d C b�b2 � �db2 D �b3.

Problem 10.75 The area shown is subjected to a uniformpressure patm D 1 ð 105 Pa.

(a) What is the total force exerted on the area by thepressure?

(b) What is the moment about the y axis due to thepressure?

x

y

1 m

y � x2

Solution:

(a) F D∫ 1

0

∫ 1

x2105 N

m2dydx D 66,700 N D 66.7 kN

(b) My D∫ 1

0

∫ 1

x2x

(105 N

m2

)dydx D 25,000 Nm D 25 kN-m


841

Problem 10.76 The area shown is subjected to auniform pressure. Determine the coordinates of thecenter of pressure.

Solution:

R D∫ 1

0

∫ 1

x2Pdydx D 2P

3

My D∫ 1

0

∫ 1

x2xPdydx D P

4)

x D My

RD 0.375 m

y D Mx

RD 0.6 m

Mx D∫ 1

0

∫ 1

x2yPdydx D 2P

5

Problem 10.77 The area shown is subjected to auniform pressure patm D 14.7 psi.

(a) What is the total force exerted on the area by thepressure?

(b) What is the moment about the y axis due to thepressure on the area?

x

y

10 in

20 in

Solution: (a) The total force is

F D∫

APatm dA D Patm

∫A

dA D PatmA

D 14.7lb

in2

[1

2��20�2 � 1

2��10�2

]in2 D 6930 lb.

(b) We can represent the pressure by an equivalent force F acting atthe center of pressure. Since the pressure is uniform, the center ofpressure is at the centroid of the plane area. From Appendix b, the xcoordinate of the centroid is

x D

[4�20�

3�

] [1

2��20�2

]�

[4�10�

3�

] [1

2��10�2

]1

2��20�2 � 1

2��10�2

D 9.90 in.

The moment about the y axis is

xF D �9.90��6930� D 68,600 in-lb.

842


Problem 10.78 In Active Example 10.10, suppose thatthe water depth relative to point A is increased from2 ft to 3 ft. Determine the reactions on the gate at thesupports A and B.

A

3 ft

2 ft

B

Solution: The gage pressure pg D �x increses linearly from pg D0 at the surface to pg D ��3 ft� at the bottom of the gate. The centroidof the distribution is located at 2

3 �3 ft� D 2 ft below point B.

The force exerted on the gate by the gage pressure is the “volume” ofthe pressure distribution

F D 1

2�3 ft�

[�62.4 lb/ft3��3 ft�

]�3 ft�

D 842 lb.

The equlibrium equations for the gate are

Fx : Ax C 100 lb D 0,

Fz : Az C B � 842 lb D 0,

My axis : Az�3 ft� � �842 lb��2 ft� D 0.

Solving yields

Ax D �100 lb, Az D 562 lb, B D 281 lb.


843

Problem 10.79 The top of the rectangular plate is2 m below the surface of a lake. Atmospheric pressurepatm D 1 ð 105 Pa and the mass density of water is� D 1000 kg/m3.

(a) What is the maximum pressure exerted on the plateby the water?

(b) Determine the force exerted on a face of the plateby the pressure of the water.

(See Example 10.11.)

2 m

3 m

2 m

Solution:

(a) The maximum pressure occurs at the bottom of the plate:

P D P0 C �x

D 1 ð 105 C �1000��9.81��5�

D 1.49 ð 105 Pa.

(b)

2 m

2 m

3 m

x

x

y

dx

dA = 2 dx

The force is

F D∫

AP dA

D∫ 5

2�P0 C �x�2 dx

D∫ 5

2[1 ð 105 C �1000��9.81�x]2 dx

D 806,000 N.

844


Problem 10.80 In Problem 10.79, how far below thetop of the plate is the center of pressure located?

Solution: See the solution of Problem 10.79. The moment due tothe pressure about the y axis is

M D∫

Axp dA

D∫ 5

2x�P0 C �x�2 dx

D∫ 5

2[1 ð 105x C �1000��9.81�x2]2 dx

D 2,865,000 N-m.

The x-coordinate of the center of pressure is

xP D M

FD 2,865,000

806,000D 3.55 m.

The distance below the top of the plate is xP � 2 m D 1.55 m.

Problem 10.81 The width of the dam (the dimensioninto the page) is 100 m. The mass density of the wateris � D 1000 kg/m3. Determine the force exerted on thedam by the gage pressure of the water (a) by integration;(b) by calculating the “volume” of the pressure distri-bution.

10 m

Solution: For gage pressure, we neglect the pressure of the atmo-sphere. (a) The weight density of the water is

� D 1000�9.81� D 9810 N/m3.

The force on the dam is

F D∫

Ap dA D 100�

∫ 10

0x dx D 5000� D 49.05 ð 106 N

(b) The pressure distribution is a triangle, with base 10� and altitude10 m. The length dimension of this “solid” is 100 m. The volume ofthis triangular “solid” is

V D ( 12

)100�10��9810� D 49.05 ð 106 N

Problem 10.82 In Problem 10.81, how far down fromthe surface of the water is the center of pressure due tothe gage pressure of the water on the dam?

Solution: The center of pressure is the centroid of the pressuredistribution, which is a triangle of altitude 10 m and base 10� . Thecentroid of a triangle is

( 13

)of the altitude, or, from the surface,

d D ( 23

)10 D 6.67 m


845

Problem 10.83 The width of the gate (the dimensioninto the page) is 3 m. Atmospheric pressure patm D1 ð 105 Pa and the mass density of the water is � D1000 kg/m3. Determine the horizontal force and coupleexerted on the gate by its built-in support A.

2 m

A

Solution: The gage pressure at the bottom of the gate is

Pg D �x D �1000��9.81��2�,

so the “volume” of the pressure distribution is

F D 12 �2�Pg�3�

D 12 �2��1000��9.81��2��3�

D 58,860 N.

Ax

Ay

MA

F

13

(2 m)

we see that

Ax D F D 58.9 kN,

MA D 13 �2�F D 39.2 kN-m.

2 m

A

Pg

846


Problem 10.84 The homogenous gate weighs 100 lb,and its width (the dimension into the page) is 3 ft. Theweight density of the water is � D 62.4 lb/ft3, and theatmospheric pressure is patm D 2120 lb/ft2. Determinethe reactions at A and B.

3 ft 2 ft

B

30°

A

Solution: The atmospheric pressure acts on both sides of the gate,so it is ignored. The strategy is use the “volume” of the pressuredistribution to compute the force acting on the face of the gate. Thepressure distribution is a triangle with base 2� . The pressure acts overan area

(6

cos�30�

)D 6.92 ft2.

Thus the “volume” is F D ( 12

)�62.4��2��6.92� D 432.32 lb. This

force acts normally to the surface of the gate, or at an angle of � D 210°

relative to the positive x axis. The centroid of the pressure is

d D(

2

3

) (1

cos 30°

)D 0.7698 ft

along the inner face of the gate from A. The sum of the moments aboutA is

∑MA D dF C W�1.5� sin 30° � 3B D 0,

from which B D 135.933 lb, normal to the inner face of the gate. Thesum of the forces normal to the gate surface is

∑FN D CAN � F C B � W sin 30° D 0,

from which AN D 346.4 lb at an angle ˛ D 30° relative to the positivex axis. The sum of the forces acting parallel to the gate surface is

∑FP D AP � W cos�30°� D 0,

from which AP D 86.6 lb at an angle of 120° . Thus the componentsof the reaction at A are Ax D �346.3 cos�210°� C 86.6 cos�120°� D256.7 lb, to the right, and Ay D �346.3 sin�210°� C 86.6 sin�120°� D248.2 lb upward.

BB

A

y

x

W AN

AP

F

d2ft3ft

30°


847

Problem 10.85 The width of the gate (the dimensioninto the page) is 2 m and there is water of depth d D 1 mon one side. Atmospheric pressure patm D 1 ð 105 Paand the mass density of the water is � D 1000 kg/m3.Determine the horizontal forces exerted on the gate at Aand B.

dA

B

500 mm

Solution: The “volume” of the gage pressure distribution is

F D 12 �gd�2�

D υpd2

D �1000��9.81��1�2

D 9810 N.

Ax

Bx

0.5 mF

13

13

d = (1) m

Applying the equilibrium equations, we find that

Ax D 23 F D 6540 N,

Bx D 13 F D 3270 N.

d

Pg

Problem 10.86 The gate in Problem 10.85 is designedto rotate and release the water when the depth d exceedsa certain value. What is that depth?

Solution: See the solution of Problem 10.85. The gate rotateswhen

13 d > 1

2 m,

d > 32 m.

848


Problem 10.87* The dam has water of depth 4 ft onone side. The width of the dam (the dimension intothe page) is 8 ft. The weight density of the water is� D 62.4 lb/ft3, and the atmospheric pressure patm D2120 lb/ft2. If you neglect the weight of the dam, whatare the reactions at A and B?

A

2 ft

2 ft

B

2 ft

Solution: To simplify the analysis of the pressure forces, we willdraw a free body diagram of the dam and the volume of water shown:The left side and top of the free body diagram are subjected to atmo-spheric pressure, and the right side is subjected to the sum of atmo-spheric pressure and the gage pressure of the water, so we only need toconsider the gage pressure. Let us represent the pressure force on theright side by an equivalent force: We can determine F by calculatingthe “volume” of the pressure distribution:

F D 12 p�4 ft��8 ft� D 1

2 ��4��4��8� D 3990 lb

the complete free body diagram is: The two weights are

w1 D ��2 ft��2 ft��8 ft� D 2000 lb

w2 D 12 ��2 ft��2 ft��8 ft� D 988 lb.


∑Fx D Ax C B � F D 0

∑Fy D Ay � w1 � w2 D 0

∑M�pt.A� D �4B C

(4

3

)F C �1�W1 C 1

3W2 D 0,

we obtain Ax D 2000 lb, Ay D 3000 lb, B D 2000 lb.

F

F

B

W1

W2

Ax Ay

P = r (4 ft)

13

(4) ft

43

ft


849

Problem 10.88* The dam has water of depth 4 ft onone side. The width of the dam (the dimension intothe page) is 8 ft. The weight density of the wateris � D 62.4 lb/ft3, and atmospheric pressure is patm D2120 lb/ft2. If you neglect the weight of the dam, whatare the reactions at A and B?

A

2 ft

2 ft

B

2 ft

Solution: The atmospheric pressure acts on both faces of the dam,so it is ignored. The strategy is to use the “volume” of the pressuredistribution to the determine the reactions. The pressure distribution isa triangle of base 4� and altitude 4 ft. The force on the vertical facesof the dam is

F1 D ( 12

)�4��4��8 D 3993.6 lb.

The moment about A due to the force on the vertical faces is

M1 D ( 43

)F1 D 5324.8 ft lb.

The force on the horizontal face of the dam is

F2 D �2��2��8� D 1996.8 lb.

The moment about A due to the force on the horizontal face is M2 D1F2 D 1996.8 ft lb. The sum of the moments about A:

∑MA D

M1 C M2 � 4B D 0, from which B D 1830.4 lb. The sum of theforces:

∑Fx D Ax � F1 C B D 0, from which Ax D 2163.3 lb to the

right.∑

Fy D Ay � F2 D 0, from which Ay D 1996.8 lb upward.

AX

AY

F1

F2B1 ft

43 ft

850


Problem 10.89 Consider a plane, vertical area A belowthe surface of a liquid. Let p0 be the pressure at thesurface.

(a) Show that the force exerted on the area is F D pA,where p D p0 C �x is the pressure of the liquid atthe centroid of the area.

(b) Show that the x coordinate of the center of pres-sure is

xP D x C �Iy0

pA,

where Iy0 is the moment of interia of the area aboutthe y0 axis through its centroid.

y

x

A

y'

x'

–x

Solution: (a) The definition of the centroid of the area is

xA D∫

Ax dA.

The force on the plate is

F D∫

Ap dA D p0

∫A

dA C �∫

Ax dA D p0A C �xA.

But by definition, the pressure at the centroid of the area is p D p0 C�x, hence F D pA.

(b) The moment about the y axis is

My�axis D∫

Apx dA D p0

∫A

x dA C �∫

Ax2 dA.

The moment of inertia is defined:

IyA D∫

Ax2 dA,

from which the moment is My�axis D p0xA C �Iy . But the moment isalso given by My�axis D FxP, where F is the force on the plate andxP is the distance to the center of pressure. Thus pAxP D p0xA C �Iy ,

from which xP D(

p0

p

)x C �Iy

pA.

From the parallel axis theorem, Iy D x2A C Iy0 . Substitute and reduce:

xP D �p0 C �x�xp

C �Iy0

pAD x C �Iy0

pA,

which demonstrates the required result.


851

Problem 10.90 A circular plate of 1-m radius is belowthe surface of a stationary pool of water. Atmosphericpressure is patm D 105 Pa, and the mass density ofthe water is � D 1000 kg/m3. Determine (a) the forceexerted on the face of the plate by the pressure of thewater; (b) the x coordinate of the center of pressure. (SeeProblem 10.89.)

y

x

1 m

1 m

Solution: (a) From Problem 10.89, the pressure on the face of theplate is F D pA, where p is the pressure at the centroid of the area.

F D �p0 C �2��12� D 375.8 kN.

(b) From Problem 10.99,

xP D x C �Iy0

pAD 2 C

9810

(�14

4

)F

D 2.0205 m

Problem 10.91* A tank consists of a cylinder withhemispherical ends. It is filled with water (� D1000 kg/m3). The pressure of the water at the top ofthe tank is 140 kPa. Determine the magnitude of theforce exerted by the pressure of the water on eachhemispherical end of the tank. (See Example 10.12.)

18 m

6 m

Solution: The free-body diagram is

mg

The force exerted by the pressure distribution on the curved surface(equal and opposite to the force exerted on the tank’s hemisphericalend) can be determined from the fact that the free-body diagram is inequilibrium.

The magnitude of the vertical component is

mg D �1000�1

2

(4

3�R3

)�9.81� D 4.44 ð 106 N.

To determine the horizontal component, see Problem 10.99. Thepressure at the centroid is

p D 140,000 C �1000��9.81��6� D 199,000 Pa,

so the horizontal component is

pA D �199,000��6�2 D 22.5 ð 106 N.

The magnitude of the force is

√�mg�2 C �pA�2 D 22.9 MN.

852


Problem 10.92 An object of volume V and weight Wis suspended below the surface of the stationary liquidof weight density � (Fig. a). Show that the tension in thecord is W � V� . In other words, show that the pressuredistribution on the surface of the object exerts an upwardforce equal to the product of the object’s volume andthe weight density of the water. The result is due toArchimedes (287–212 B.C.).

Strategy: Draw the free-body diagram of a volume ofliquid that has the same shape and position as the object(Fig. b).

V V V

(a) (b)

Solution: The result follows from the free body diagram of thespace occupied by the object and the development of the force exertedby the pressure in terms of the volume of the object. Let A be the areabounding the volume V. The force exerted on a surface A is

F D �∫

Anp dA,

where n is a unit vector normal to the elemental surface dA, posi-tive outward from the surface, and the negative sign comes from theequilibrium condition that the reaction force acts oppositely to the unitvector. Choose an x, y, z coordinate system such that the elementalforces are

np dA D ip dy dz C jp dx dz C kp dx dy.

The force becomes

F D �∫

Aip dy dz �

∫A

jp dx dz �∫

Akp dx dy.

But these integrals can also be written as

F D �i∫

V

∂p

∂xdx dy dz � j

∫V

∂p

∂ydx dy dz � k

∫V

∂p

∂zdx dy dz.

This set of integrals can be collapsed into

F D �∫

Anp dA D �

∫V

rp dV,

where the volume V is bounded by the surface A, rp is a shorthandnotation for

rp D i∂p

∂xC j

∂p

∂yC k

∂p

∂z,

and dV D dx dy dz. The pressure p D p0 C �x, from which rp D i� ,and the integral becomes

F D �∫

Anp dA D �i�

∫V

dV D �i�V.

The weight of the object acts in the positive x direction, so the resultantforce is FR D i�W � �V�

V


853

Problem 10.93 Determine the internal forces andmoment at B (a) if x D 250 mm; (b) if x D 750 mm. A

BC

x

1000 mm

500 mm

20 N-m

Solution: (a) The sum of the moments about A

∑MA D �20 C 1C D 0,

from which C D 20 N. The sum of forces

∑Fy D Ay C C D 0,

from which Ay D �C D �20 N.

∑Fx D Ax D 0

(a) Make a cut at B. Isolate the left hand part. The sum of moments

∑M D MB C 0.25�20� D 0,

from which

MB D �5 N m.

VB D �20 N,

PB D 0.

(b) Make a cut at B : Isolate the left hand part. The sum of moments:

∑M D MB � 20 C 0.75�20� D 0,

from which

MB D 5 N-m.

VB D �20 N,

PB D 0.

Ax

Ay

MB

MB

PB

PB

VB

VB

C

1 m

0.25 m

0.75 m

20 N

20 N

20 N-m

20 N-m

854


Problem 10.94 Determine the internal forces andmoment (a) at B; (b) at C.

x

y

A B C D

80 lb4 ft

12 ft

6 ft 3 ft

Solution: The sum of the moments about A is

∑MA D �4�80� C 12D D 0,

from which D D 80

3lb. The sum of forces:

∑Fy D Ay � 80 C D D 0,

from which Ay D 160

3D 53.33 lb.

∑Fx D Ax D 0

(a) Make a cut at B. Isolate the left hand part. The sum of moments

∑M D MB C 2�80� � 6�53.33� D 0,

from which

MB D 160 ft lb.

VB D 53.33 � 80 D �26.7 lb.

PB D 0

(b) Make a cut at C: The sum of moments

∑M D MC C 80�5� � 9�53.33� D 0,

from which

MC D 80 ft lb.

VC D �80 C 53.33 D �26.7 lb.

PC D 0

Ax

MB

VB

PB

MB

VB

PB

Ay D

80 lb

80 lb

80 lb

53.33 lb

53.33 lb

4 ft12 ft

6 ft

9 ft

(a)

(b)


855

Problem 10.95 (a) Determine the maximum bendingmoment in the beam and the value of x where it occurs.(b) Show that the equations for V and M as functionsof x satisfy the equation V D dM/dx.

x

y

360 lb/ft

3 ft

180 lb/ft

Solution: Find the reactions first∑MA : �A�3 ft� C �540 lb��1.5 ft� C �270 lb��1.0 ft� D 0

∑Fy : A C B � 540 lb � 270 lb D 0

A D 360 lb, B D 450 lb

w D 180 lb/ft C 180 lb/ft

3 ftx D 180 lb/ft C �60 lb/ft2�x

V D ��180 lb/ft�x � �30 lb/ft2�x2 C 360 lb

M D ��90 lb/ft�x2 � �10 lb/ft2�x3 C �360 lb�x

(a) The maximum moment occurs when the shear force is zero

V D 0 ) x D 1.583 ft ) M D 305 ft lb

(b)dM

dxD ��180 lb/ft�x � �30 lb/ft2�x2 C 360 lb D V

540 lb

270 lb

A B


Solution: Plot the solution of 10.95

180 lb/ft

360 lb

360 lb

0

�450 lb

400 ft-lb

200 ft-lb

0

450 lb

360 lb/ft

x

x

x

y

V

M

856


Problem 10.97 Determine the shear force and bendingmoment diagram’s for the beam.

x

12 ft

y

= 10(12x – x2) lb/ft

Solution: Denote the reactions at the left and right ends by A andB, respectively. The total force due to the load is

F D∫ 12

010�12x � x2� dx D 10

[6x2 � x3

3

]12

0D 2880 lb.

The moment about the left end due to the load is

MLOAD D∫ 12

010�12x � x2�x dx D 10

[4x3 � x4

4

]12

0.

MLOAD D 17280 ft lb.

The sum of the moments about the left end:

∑MA D �M C 12B D 0,

from which B D 1440 lb. The sum of the forces:

∑Fy D Ay C B � F D 0,

from which Ay D 1440 lb. Beginning from the left end, the shear is

V�x� D �∫ X

010�12x � x2� C 1440

V�x� D �10

(6x2 � x3

3

)C 1440.

The moment is

M D∫

V�x� dx C C D �10

(2x3 � x4

12

)C 1440x C C.

The moment is zero at x D 0, hence the constant C D 0, and themoment is

M D �10

(2x3 � x4

12

)C 1440x

0–2000

–1000

0

1000

2000

3000

4000

5000

6000

2 4 6 8 10 12

X, ft

Shear Force & Moment Diagram

Bending Moment

Shear Force

Ax

AyB

x

12 ft


857

Problem 10.98 Determine V and M as functions of xfor the beam ABC.

B C

D

4 kN2 m 2 m 2 m

A

1 m

x

y

Solution: First find the reactions∑Mc : �A�6 m� C �4 kN��2 m� D 0

∑Fy : A C C � 4 kN D 0

A D 1.33 kN, C D 2.67 kN


w1 D 0

V1 D 1.33 kN

M1 D �1.33 kN�x


w2 D 0

V2 D �2.67 kN

M2 D ��2.67 kN�x C 16 kN-m

4 kN

A C

858


Problem 10.99 Draw the shear force and bendingmoment diagrams for beam ABC.

B

D

600 lb

10 ft

A2 ft

8 ft

C

4 ft

Solution: The structure as a free body: The angle of the cable atD relative to the horizontal is

� D tan�1( 1

4

) D 14°.

Denote the tension in the cable by T. The sum of the moments aboutA is

∑MA D 2T cos � C 8T sin � � 10�600� D 0,

from which T D 1546.2 lb. The sum of the forces:

∑Fy D Ay � 600 C T sin � D 0,

from which Ay D 225 lb. The intervals as free bodies: Divide the beaminto two intervals: 0 � x < 8, and 8 � x � 10. Interval 1: The shearforce is V1�x� D Ay . The internal bending moment is M1�x� D Ayx D225x. Interval 2 : The shear force is V2�x� D Ay C T sin � D 600 lb.The sum of the moments is

∑Mx�x� D M2�x� � Ayx � �T sin ��x � 8� C 2T cos � D 0,

from which M2�x� D 600x � 6000 ft lb.

The diagrams are shown. The discontinuity in the moment at x D 8 isdue to the moment exerted on the beam by the horizontal componentof the cable tension at D. This horizontal component of tension exertsa moment that is independent of the distance x along the beam ABC.In this sense, it behaves like a couple.

Ax

Ax

Ay

Ay

M2(x)

M1(x)

V2(x)

V1(x)

P2(x)

P1(x)

T

T

θ

θ

0–2000–1500–1000–500

0500

100015002000

1 2 3 4 5 6 7 8 9 10X, ft

Internal Forces & Moment

Bending Moment

Shear Force

2 ft

8 ft

8 ft

10 ft600 lb

x

x


859

Problem 10.100 Determine the internal forces andmoments at A.

B

2 m

A

C

1 m 1 m 1 m

1 m

1 m 3 kN/m

Solution: Denote the reactions at the slide support of BC by D,and the reactions at the upper and lower pin supports by E and Grespectively, and the reaction at the pin connection on the beam GBby H. The total force due to the load is

F D∫ 4

03 dx D 12 kN.

The moment about the left end of BC is

M D∫ 4

03x dx D 24 kN m.

The sum of the moments about the point B is

∑MB D �M C 3D D 0,

from which D D 8 kN. The sum of the forces:

∑Fy D B � F C D D 0,

from which B D 4 kN is the vertical reaction at B. The centroiddistance of the load from the left end of BC is

d D M

FD 24

12D 2 m.

The sum of the moments about the upper pin support is

∑M D ��1 C 2�F C 2Gx D 0,

from which Gx D 18 kN. With these known quantities, the followingequations can be written: (See figure to right for assumed positivedirections.)

Sum of moments about H for lower member yields Gy D �8 kN. Thesum of the forces for the lower member yields:

Hy D �16 kN

and Hx D 18 kN.

Divide the lower member into two parts

0 � x < 2 (m)

and 2 � x < 3 (m).

Ey

GYHy

Hy

Hx

Hx

Ex

18 kN

4 kN

8 kN

Part 1: The shear is V1�x� D �8 kN. The moment is

M1�x� D∫

V1�x� dx C C1 D �8x C C1.

The moment at x D 0, M1�0� D 0, from which C1 D 0, andM1�x� D �8x. Part 2: The shear is V2�x� D V1�x� C 16 D 8 kN. Themoment is

M2�x� D∫

V2�x� dx C C2 D 8x C C2.

The moment is continuous at x D 2, M1�2� D M2�2�, from whichC2 D �32, and the moment is M2�x� D 8x � 32 kN m. Thus at pointA, x D 3, the internal forces and bending moment are

V2�3� D 8 kN,

P�3� D 0,

M2�3� D 24 � 32 D �8 kN m

860


Problem 10.101 Draw the shear force and bendingmoment diagrams of beam BC in Problem 10.100.

Solution: From the solution to Problem 10.101, the shear forceand bending moment are:

(1) V1�x� D �8 kN, M1�x� D �8x kN m, �0 � x < 2 m�,(2) V2�x� D 8 kN, M2�x� D 8x � 32 kN m, �2 � x < 4 m�

The shear force and bending moment diagrams are shown.

0–20

–15

–10

–5

0

5

10

.5 1 1.5 2 2.5 3 3.5 4X, ft


Shear Force

BendingMoment

Problem 10.102 Determine the internal forces andmoments at B (a) if x D 250 mm; (b) if x D 750 mm.

AB

x

1000 mm

500 mm

20 N-m

40 NSolution: The complete beam: The sum of the moments about A is

∑MA D MA � 20 C 1�40� D 0,

from which MA D �20 N m. The sum of the forces:

∑Fy D Ay C 40 D 0,

from which Ay D �40 N.

∑Fx D Ax D 0

The internal forces at x D 250 mm. The shear is V1�x� D �40 N. Themoment is

M1�x� D∫

V1�x� dx C C1 D �40x C C1

At x D 0, M1�0� D �MA D 20 N m, from which C1 D 20 kN. Thusthe moment is M�0.25� D �40�0.25� C 20 D 10 Nm. The internalforces at x D 750 mm. Divide the beam into two segments:

0 � x < 0.5 (m),

and 0.5 � x < 0.75 (m).

The shear in the second segment is V2�x� D �40 N. The moment is

M2�x� D∫

V2�x� dx C C2 D �40x C C2.

A known discontinuity exists in the moment at x D 0.5 m, M1�0.5� �M2�0.5� D �20 N m, from which C2 D 40 and the moment isM2�x� D �40x C 40 Nm. At x D 0.75, M2�0.75� D �40�0.75� C40 D 10 N m. The axial forces P�x� D 0 everywhere.


861


A

1000 mm

500 mm

20 N-m

40 N

x

y

Solution: From the solution to Problem 10.102, the shear andbending moment are

V1�x� D �40 N,

M1�x� D �40x C 20 Nm �0 � x < 0.5 m�,

and V2�x� D �40 N,

M2�x� D �40x C 40 Nm, �0.5 � x < 1 m�


0–50–40–30–20–10

01020304050

.1 .2 .3 .4 .5 .6 .7 .8 .9 1

X, m


Bending Moment

Shear Force

Problem 10.104 The homogenous beam weighs1000 lb. What are the internal forces and bendingmoments at its midpoint?

10 ft

2 ft 3 ft

Solution: Denote the support reactions by A and B. The load distri-bution is w D 1000/10 D 100 lb/ft. The moment about the left end dueto the load of the beam is

ML D∫ 10

0wx dx D 100

[x2

2

]10

0D 5000 ft lb.

The sum of the moments about the left end is

∑M D �ML C 2A C 7B D 0

The sum of the forces:

∑Fy D A C B � 1000 D 0.

Solve the two simultaneous equations to obtain: A D 400 lb, B D600 lb.

∑Fx D 0.

Divide the left half of the beam into two parts: 0 � x < 2 (ft) and2 � x < 5 (ft).

Part 1: The shear is V1�x� D �100x. The moment is

M1�x� D∫

V1�x� dx C C1 D �50x2 C C1.

At x D 0, M1�0� D 0, thus C1 D 0, and the moment is M1�x� D�50x2.

Part 2: The shear is V2�x� D V1�x� C 400 lb. The moment is

M2�x� D∫

V2�x� dx C C2 D �50x2 C 400x C C2.

The moment is continuous at x D 2, M1�2� D M2�2�, from whichC2 D �800, and the moment is M2�x� D �50x2 C 400x � 800. At themidpoint, x D 5 ft, the shear is V2�5� D �100�5� C 400 D �100 lband the moment is M2�5� D �50�25� C 400�5� � 800 D �50 ft lb.The axial force is zero, P�5� D 0.

862



Solution: From the solution for Problem 10.104, the shear forceand bending moment are:

(1) V1�x� D �100x lb, M1�x� D �50x2 ft lb, �0 � x < 2 ft�.(2) V2�x� D �100x C 400 lb, M2�x� D �50x2 C 400x � 800 ft lb,

�2 � x < 7 ft�.(3) V3�x� � 100x C 1000 lb, M3�x� D �50x2 C 1000x � 5000 ft lb,

�7 � x < 10 ft�.


0–500

–400

–300

–200

–100

0

100

200

300

1 2 3 4 5 6 7 8 9 10X, ft


Bending Moment

Shear Force

Problem 10.106 At A the main cable of the suspensionbridge is horizontal and its tension is 1 ð 108 lb.

(a) Determine the distributed load acting on the cable.(b) What is the tension at B?

x

y

A

B

900 ft

300 ft

Solution: (a) The parameter

a D 2y

x2D 2

300

9002D 7.4074 ð 10�4.

The distributed load is

w D T0a D �1 ð 108��7.4 ð 10�4� D 7.4074 ð 104 lb/ft

(b) The tension at B is

TB D T0

√1 C a2�900�2 D 1.2 ð 108 lb


863

Problem 10.107 The power line has a mass of1.4 kg/m. If the line will safely support a tension of5 kN, determine whether it will safely support an iceaccumulation of 4 kg/m.

12°

40 m

Solution: The power line meets the conditions for a catenary. Theweight density with an ice load is

� D �1.4 C 4��9.81� D 52.974 N/m.

The angle at the attachment point is related to the length and theparameter a by sa D tan �. But sa D sinh�ax�, and x is known. Thusthe parameter can be found from

a D sinh�1�tan ��

xD sinh�1�0.2126�

20D 0.01055.

The tension at the lowest point is

T0 D w

aD 5021.53 D 5.02 kN.

The maximum tension is

T D T0 cosh�ax� D 5133.7 D 5.133 kN.

Thus the line will not sustain the load.

Problem 10.108 The water depth at the center ofthe elliptical aquarium window is 20 ft. Determine themagnitude of the net force exerted on the window bythe pressure of the seawater (� D 64 lb/ft3) and theatmospheric pressure of the air on the opposite side.(See Problem 10.89.)

6 ft

L

3 ft 6 in

Solution: The force on the plate is

F D∫

Ap dA.

The pressure is p D p0 C �x � p0 D �x, where the atmosphericpressure cancels, since it appears on both sides. The force:

F D �∫

Ax dA D �xA,

hence the force on the window is F D 1280A lb. The area of the ellipseis A D �ab D 21�, and the force is F D 84445 lb

3 ft 6 in

6 ft

L

864


Problem 10.109 In Problem 10.108, determine themagnitude of the net moment exerted on the windowabout the horizontal axis L by the pressure of theseawater (� D 64 lb/ft3) and the atmospheric pressureof the air on the opposite side. (See Problem 10.89.)

Solution: The moment is

M D∫

px dA D �∫

x2 dA D �Iy.

The area moment of inertia for an ellipse is

Iy D �ab3

4D 202 ft4,

and the moment is M D �Iy D 12930.8 ft lb.

Problem 10.110* The gate has water of 2-m depth onone side. The width of the gate (the dimension intothe paper) is 4 m, and its mass is 160 kg. The massdensity of the water is � D 1000 kg/m3 and atmosphericpressure is patm D 105 Pa. Determine the reactions onthe gate at A and B. (The support B exerts only ahorizontal reaction on the gate.)

A

B

2 m

Solution: Consider the free-body diagram shown. The weight ofthe gate is

W D �160��9.81� D 1570 N,

and x2 D R � 2R

�D 0.727 m.

The pressure force

F D 12 ��R�R�4�

D 12 �1000��9.81��2�2�4�

D 78,480 N,

and y D 13 R D 0.667 m.

The area below the gate is

Ab D 14 �R2 D 3.142 m2,

and the centroid of Ab is at

xb D R � 4R

3�D 1.151 m.

The area above the gate is

Aa D R2 � Ab D 0.858 m2.

The centroid of Aa [ Ab is at x D R/2, so

R

2D x1Aa C xbAb

Aa C Ab,

from which we obtain x1 D 0.447 m.

The weight of the water is Q D �Aa�4� D 33,700 N.

y

F

Q

x

X2

X1

Y

AX

AY

W

B

R = 2 m


∑Fx D F C Ax � B D 0,

∑Fy D Ay � W � Q D 0,

∑M�ptA� D RB � yF � x1Q � x2W D 0,

we obtain

Ax D �44.2 kN,

Ay D 35.3 kN,

B D 34.3 kN.


865

Problem 10.111 A spherical tank of 400-mm innerradius is full of water (� D 1000 kg/m3). The pressureof the water at the top of the tank is 4 ð 105 Pa.

(a) What is the pressure of the water at the bottom ofthe tank?

(b) What is the total force exerted on the inner surfaceof the tank by the pressure of the water?

Strategy: For (b), draw a free-body diagram of thesphere of water in the tank.

400 mm

Solution: The weight density is � D �g D 9810 N/m3. ((a) Thepressure distribution is P�x� D p0 C �x.) The pressure at the bottomof the tank is P�0.8� D p0 C � �0.5� D 4 ð 105 C �9810��0.8� D4.0785 ð 105 Pa.

(b) From the free body diagram of the sphere of water, the unbalancedforce is the weight of water, acting downward:

W D(

4

3

)�R3� D 2629.9 N

866


Statics bedford 5 chap 10

Engineering

Transcript of Statics bedford 5 chap 10