VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Chapter 7.1, Problem 7.15P

(a)

To determine

The internal forces exerted at the point C in a frame.

(a)

Expert Solution
Check Mark

Answer to Problem 7.15P

The internal forces of shearing force is V=576N acting in the upward direction, axial force F=480N acting in the leftside of the frame, and the bending moment acting at that point is M=0.

Explanation of Solution

Sketch the free body diagram for the internal forces acting on the frame and pulley system as shown in the Figure 1.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 7.1, Problem 7.15P , additional homework tip  1

Write the equation of the axial force exerted at the axial point A of the frame from x direction.

Fx=0Ex=0 (I)

Here, the force exerted on the frame at the point E from x direction in equilibrium condition is Fx.

Write the equation of the moment of couple formed in the bending moment of the frame and pulley system supported at the point E rotated in counterclockwise moment (Refer fig 1).

ME=0FB(r+l)FB(l1r)Ad=0 (II)

Here, the axial force exerted on the pulley at point B is FB, the radius of the each pulley is r, distance between the point A on the frame to the point D is l, distance between the point A on the frame to the point B is l1, and total distance of the frame is d.

Write the equation of the axial force exerted at the axial point of the frame from y direction (Refer fig 1).

Fy=0Ey+AFB+FD=0 (III)

Here, the axial force exerted on the pulley at point D is FD.

Sketch the free body diagram for the cable as shown in the Figure 2.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 7.1, Problem 7.15P , additional homework tip  2

The slope of the cable (Refer fig 2):

BC=CD=200mmBG=DH=120mm

The angle formed in the slope of the cable:

sinα=120mm200mm=35cosα=45

Rewrite the above relation to find the angle.

α=cos1(45)=cos1(0.8)=36.87°

Write the equation of the axial force exerted at the axial point AC of the given free body diagram from x direction.

Fx=0F+FBcosα=0 (IV)

Here, the angle between the pulley B to the frame is α and the axial force exerted on the frame and pulley system is F.

Sketch the free body diagram for the cable for the point AC as shown in the Figure 3.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 7.1, Problem 7.15P , additional homework tip  3

Write the equation of the axial force exerted at the point A of the frame from y direction.

Fy=0V+ABxcosαF=0 (V)

Here, shearing force acting on the semicircular rod is V

At the pulley B it is resolved into two components Bx and By.

Write the equation of the moment of couple formed in the bending moment supported at the point C rotated in counterclockwise moment.

MC=0Mx1A+(Bxcosα+F)x2=0 (VI)

Here, the moment of couple exerted at the point C in the frame is represented as MC and moment of couple acting in the bending beam is M.

Conclusion:

Substitute 600N for FB, 120mm for r, 700mm for l, 1000mm for d, and 300mm for l1 in equation (II) to solve for A.

(600N)(120mm+700mm)(600N)(300mm120mm)A(1000mm)=0(600N)(820mm)(0.001m1mm)(600N)(180mm)(0.001m1mm)A(1000mm)(0.001m1mm)=0(600N)(0.820m)(600N)(0.180m)A(1m)=0

Solve the above equation for A.

(492Nm)(108Nm)A(1m)=0(384Nm)=A(1m)A=384N

Substitute 600N for FB, 600N for FD, and 384N for A in equation (III) to solve for Ey.

Ey+384N600N+600N=0Ey=384N

Substitute 600N for FB and 36.87° for α in equation (IV) to solve for F.

F+(600N)cos36.87°=0F+480N=0F=480N

Substitute 384N for A, 600N for Bx, 480N for F, and 36.87° for α in equation (V) to solve for V

V+384N(600N)cos36.87°480N=0V+384N480N480N=0V576N=0V=576N

Substitute 384N for A, 600N for Bx, 480N for F, 500mm for x1, 200mm for x2,36.87° for α in equation (VI) to solve for M.

M(500mm)(384N)+(600Ncos36.87°+480N)(200mm)=0M(500mm)(0.001m1mm)(384N)+(480N+480N)(200mm)(0.001m1mm)=0M(0.500m)(384N)+(960N)(0.200m)=0

The above equation can be written as,

M192Nm+192Nm=0M=0

Therefore, the internal forces of shearing force is V=576N acting in the upward direction, axial force F=480N acting in the leftside of the frame, and the bending moment acting at that point is M=0.

(b)

To determine

The internal forces exerted at the point J to the left point C in a frame of 100mm.

(b)

Expert Solution
Check Mark

Answer to Problem 7.15P

The internal forces of shearing force is V=576N acting in the upward direction, axial force F=480N acting in the leftside of the frame, and the bending moment acting at that point is M=57.6Nm rotates in the counterclockwise direction.

Explanation of Solution

Sketch the free body diagram for the cable for the point AJ as shown in the Figure 4.

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 7.1, Problem 7.15P , additional homework tip  4

Write the equation of the axial force exerted at the axial point AJ of the frame from x direction.

Fx=0F+FB=0 (VII)

Here, the force exerted on the frame at the point E from x direction in equilibrium condition is Fx.

Write the equation of the axial force exerted at the axial point of the frame from y direction (Refer fig 4).

Fy=0V+ABxcosαF=0 (VIII)

Here, the axial force exerted on the pulley at point D is FD.

Write the equation of the moment of couple formed in the bending moment supported at the point J rotated in counterclockwise moment.

MJ=0Mx3A+(Bxcosα+F)x4=0 (IX)

Here, the moment of couple exerted at the point J in the frame is represented as MJ and moment of couple acting in the bending beam is M.

Conclusion:

Substitute 600N for FB and 36.87° for α in equation (VII) to solve for F.

F+(600N)cos36.87°=0F+480N=0F=480N

Substitute 384N for A, 600N for Bx, 480N for F, and 36.87° for α in equation (VIII) to solve for V.

V+384N(600N)cos36.87°480N=0V+384N480N480N=0V576N=0V=576N

Substitute 384N for A, 600N for Bx, 480N for F, 400mm for x3, 100mm for x4,36.87° for α in equation (IX) to solve for M.

M(400mm)(384N)+(600Ncos36.87°+480N)(100mm)=0M(400mm)(0.001m1mm)(384N)+(480N+480N)(100mm)(0.001m1mm)=0M(0.400m)(384N)+(960N)(0.100m)=0

The above equation can be written as,

M153.6Nm+96Nm=0M=57.6Nm

Therefore, the internal forces of shearing force is V=576N acting in the upward direction, axial force F=480N acting in the leftside of the frame, and the bending moment acting at that point is M=57.6Nm rotates in the counterclockwise direction.

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Chapter 7 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

Ch. 7.1 - A semicircular rod is loaded as shown. 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PROBLEM...Ch. 7.2 - Assuming the upward reaction of the ground on beam...Ch. 7.2 - Prob. 7.46PCh. 7.2 - Assuming the upward reaction of the ground on beam...Ch. 7.2 - Assuming the upward reaction of the ground on beam...Ch. 7.2 - Draw the shear and bending-moment diagrams for the...Ch. 7.2 - Draw the shear and bending-moment diagrams for the...Ch. 7.2 - Draw the shear and bending-moment diagrams for the...Ch. 7.2 - Draw the shear and bending-moment diagrams for the...Ch. 7.2 - Two small channel sections DF and EH have been...Ch. 7.2 - Solve Prob. 7.53 when = 60. 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