Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
10th Edition
ISBN: 9780077591670
Author: BUDYNAS
Publisher: MCG
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Chapter 3, Problem 1P

3–1* to 3–4 Sketch a free-body diagram of each element in the figure. Compute the magnitude and direction of each force using an algebraic or vector method, as specified.

Chapter 3, Problem 1P, 31 to 34 Sketch a free-body diagram of each element in the figure. Compute the magnitude and

Expert Solution & Answer
Check Mark
To determine

The free body diagram of each element of the given figure.

The magnitude and direction of each force.

Answer to Problem 1P

The free body diagram of element OB is shown in Figure (1).

The free body diagram of element BC is shown in Figure (2).

The free body diagram of pin joint C is shown in Figure (3).

The free body diagram of pin joint O is shown in Figure (4).

The magnitude of force at O is 66.66lbf_ in the positive y direction.

The magnitude of force at B is 33.33lbf_ in the positive y direction.

The magnitude of force at C is 33.33lbf_ in the negative y direction.

Explanation of Solution

Figure (1) shows the free body diagram of the element OB.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 1P , additional homework tip  1

Figure (1)

Write the net moment at point O on the link OB.

    FBV×l1W×l2=0                                                        (I)

Here, perpendicular distance between point O and FBV is l1, perpendicular distance between point O and the applied force is l2, force acting at point A is W.

Write the net reaction force in the vertical direction on the link OB.

    FOV+FBVW=0                                                        (II)

Here, force acting at point O is FOV, force acting at point B is FBV.

Write the net reaction force in the horizontal direction on the link OB.

    FOHFBH=0                                                             (III)

Here, force acting at point O is FOH, force acting at point B is FBH.

Figure (2) shows the free body diagram of element BC.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 1P , additional homework tip  2

Figure (2)

Write the net reaction force in the horizontal direction on the link BC.

    FBHFCH=0                                               (IV)

Here, force acting at point B is FBH, force acting at point C is FCH.

Write the net reaction force in the vertical direction on the link OB.

    FCVFBV=0                                                (V)

Here, force acting at point C is FCV, force acting at point B is FBV.

Figure (3) shows the free body diagram of pin joint C.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 1P , additional homework tip  3

Figure (3)

Write the expression of net reaction force at pin joint C in horizontal direction.

    FCH=0                                               (VI)

Here, force acting at point C in the horizontal direction is FCH.

The force acting in the vertical direction at point C is balanced by the reaction force by link BC at point C.

Figure (4) shows the free body diagram of the pivot point O.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 1P , additional homework tip  4

Figure (4)

Write the expression of net reaction force at pin joint O in horizontal direction.

    FOH=0                                               (VII)

Here, force acting at point O in the horizontal direction is FOH.

The force acting in the vertical direction at point O is balanced by the reaction force by link OB at point O.

Conclusion

Substitute 100lbf for W in Equation (I).

    FBV×(18in×1ft12in)(100lbf×6in×1ft12in)=0FBV×18ft12=600lbfft12FBV=60018lbfFBV=33.33lbf

Thus, the vertical component of the force at point B is 33.33lbf.

Substitute 33.33lbf for FBV and 100lbf for W in Equation (II).

    FOV+33.33lbf100lbf=0FOV=(10033.33)lbfFOV=66.66lbf

Thus, the vertical component of the force at point O is 66.66lbf.

Substitute 33.33lbf for FBV in Equation (V).

    FCV33.33lbf=0FCV=33.33lbf

Thus, the vertical component of the force at point C is 33.33lbf.

Substitute 0 for FCH in Equation (IV).

    FBH0=0FBH=0

Thus, the horizontal component of the force at point B is 0.

Substitute 0 for FBH in Equation (III).

    FOH0=0FOH=0

Thus, the horizontal component of the force at point O is 0.

Thus, the magnitude of force at O is 66.66lbf_ in the positive y direction.

Thus, the magnitude of force at B is 33.33lbf_ in the positive y direction.

Thus, the magnitude of force at C is 33.33lbf_ in the negative y direction.

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Chapter 3 Solutions

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

Ch. 3 - Repeat Prob. 37 using singularity functions...Ch. 3 - Repeat Prob. 38 using singularity functions...Ch. 3 - For a beam from Table A9, as specified by your...Ch. 3 - A beam carrying a uniform load is simply supported...Ch. 3 - For each of the plane stress states listed below,...Ch. 3 - Repeat Prob. 315 for: (a)x = 28 MPa, y = 7 MPa, xy...Ch. 3 - Repeat Prob. 315 for: a) x = 12 kpsi, y = 6 kpsi,...Ch. 3 - For each of the stress states listed below, find...Ch. 3 - Repeat Prob. 318 for: (a)x = 10 kpsi, y = 4 kpsi...Ch. 3 - The state of stress at a point is x = 6, y = 18, z...Ch. 3 - The state of stress at a point is x = 6, y = 18, z...Ch. 3 - Repeat Prob. 320 with x = 10, y = 40, z = 40, xy =...Ch. 3 - A 34-in-diameter steel tension rod is 5 ft long...Ch. 3 - Repeat Prob. 323 except change the rod to aluminum...Ch. 3 - A 30-mm-diameter copper rod is 1 m long with a...Ch. 3 - A diagonal aluminum alloy tension rod of diameter...Ch. 3 - Repeat Prob. 326 with d = 16 mm, l = 3 m, and...Ch. 3 - Repeat Prob. 326 with d = 58 in, l = 10 ft, and...Ch. 3 - Electrical strain gauges were applied to a notched...Ch. 3 - Repeat Prob. 329 for a material of aluminum. 3-29...Ch. 3 - The Roman method for addressing uncertainty in...Ch. 3 - Using our experience with concentrated loading on...Ch. 3 - The Chicago North Shore Milwaukee Railroad was an...Ch. 3 - For each section illustrated, find the second...Ch. 3 - 3-35 to 3-38 For the beam illustrated in the...Ch. 3 - 3-35 to 3-38 For the beam illustrated in the...Ch. 3 - 3-35 to 3-38 For the beam illustrated in the...Ch. 3 - 3-35 to 3-38 For the beam illustrated in the...Ch. 3 - The figure illustrates a number of beam sections....Ch. 3 - A pin in a knuckle joint canning a tensile load F...Ch. 3 - Repeat Prob. 3-40 for a = 6 mm, b = 18 mm. d = 12...Ch. 3 - For the knuckle joint described in Prob. 3-40,...Ch. 3 - The figure illustrates a pin tightly fitted into a...Ch. 3 - For the beam shown, determine (a) the maximum...Ch. 3 - A cantilever beam with a 1-in-diameter round cross...Ch. 3 - Consider a simply supported beam of rectangular...Ch. 3 - In Prob. 346, h 0 as x 0, which cannot occur. If...Ch. 3 - 348 and 349 The beam shown is loaded in the xy and...Ch. 3 - The beam shown is loaded in the xy and xz planes....Ch. 3 - Two steel thin-wall tubes in torsion of equal...Ch. 3 - Consider a 1-in-square steel thin-walled tube...Ch. 3 - The thin-walled open cross-section shown is...Ch. 3 - 3-53 to 3-55 Using the results from Prob. 3-52,...Ch. 3 - 3-53 to 3-55 Using the results from Prob. 3-52,...Ch. 3 - 3-53 to 3-55 Using the results from Prob. 3-52,...Ch. 3 - Two 300-mm-long rectangular steel strips are...Ch. 3 - Using a maximum allowable shear stress of 70 Mpa,...Ch. 3 - Repeat Prob. 357 with an allowable shear stress of...Ch. 3 - Using an allowable shear stress of 50 MPa,...Ch. 3 - A 20-mm-diameter steel bar is to be used as a...Ch. 3 - A 2-ft-long steel bar with a 34-in diameter is to...Ch. 3 - A 40-mm-diameter solid steel shaft, used as a...Ch. 3 - Generalize Prob. 3-62 for a solid shaft of...Ch. 3 - A hollow steel shaft is to transmit 4200 N m of...Ch. 3 - The figure shows an endless-bell conveyor drive...Ch. 3 - The conveyer drive roll in the figure for Prob....Ch. 3 - Consider two shafts in torsion, each of the same...Ch. 3 - 3-68 to 3-71 A countershaft two V-belt pulleys is...Ch. 3 - 3-68 to 3-71 A countershaft two V-belt pulleys is...Ch. 3 - 3-68 to 3-71 A countershaft two V-belt pulleys is...Ch. 3 - A countershaft carrying two V-belt pulleys is...Ch. 3 - A gear reduction unit uses the countershaft shown...Ch. 3 - Prob. 73PCh. 3 - Prob. 74PCh. 3 - Prob. 75PCh. 3 - Prob. 76PCh. 3 - Prob. 77PCh. 3 - Prob. 78PCh. 3 - Prob. 79PCh. 3 - The cantilevered bar in the figure is made from a...Ch. 3 - Repeat Prob. 3-80 with Fx = 0, Fy = 175 lbf, and...Ch. 3 - Repeat Prob. 3-80 with Fx = 75 lbf, Fy= 200 lbf,...Ch. 3 - For the handle in Prob. 3-80, one potential...Ch. 3 - The cantilevered bar in the figure is made from a...Ch. 3 - Repeat Prob. 3-84 with Fx = 300 lbf, Fy = 250 lbf,...Ch. 3 - Repeat Prob. 3-84 with Fx = 300 lbf, Fy = 250 lbf,...Ch. 3 - Repeat Prob. 3-84 for a brittle material,...Ch. 3 - Repeat Prob. 3-84 with Fx = 300 lbf, Fy = 250 lbf,...Ch. 3 - Repeat Prob. 3-84 with Fx = 300 lbf, Fy = 250 lbf,...Ch. 3 - The figure shows a simple model of the loading of...Ch. 3 - Develop the formulas for the maximum radial and...Ch. 3 - Repeat Prob. 391 where the cylinder is subject to...Ch. 3 - Develop the equations for the principal stresses...Ch. 3 - 3-94 to 3-96 A pressure cylinder has an outer...Ch. 3 - 3-94 to 3-96 A pressure cylinder has an outer...Ch. 3 - 3-94 to 3-96A pressure cylinder has an outer...Ch. 3 - 3-97 to 3-99 A pressure cylinder has an outer...Ch. 3 - 3-97 to 3-99 A pressure cylinder has an outer...Ch. 3 - 3-97 to 3-99 A pressure cylinder has an outer...Ch. 3 - An AISI 1040 cold-drawn steel tube has an OD = 50...Ch. 3 - Repeat Prob. 3-100 with an OD of 2 in and wall...Ch. 3 - Prob. 102PCh. 3 - Prob. 103PCh. 3 - A thin-walled cylindrical Steel water storage tank...Ch. 3 - Repeat Prob. 3-104 with the tank being pressurized...Ch. 3 - Find the maximum shear stress in a 512-in-diameter...Ch. 3 - The maximum recommended speed for a...Ch. 3 - An abrasive cutoff wheel has a diameter of 5 in,...Ch. 3 - A rotary lawnmower blade rotates at 3500 rev/min....Ch. 3 - 3110 to 3115 The table lists the maximum and...Ch. 3 - Prob. 111PCh. 3 - Prob. 112PCh. 3 - 3110 to 3115 The table lists the maximum and...Ch. 3 - Prob. 114PCh. 3 - Prob. 115PCh. 3 - 3116 to 3119 The table gives data concerning the...Ch. 3 - Prob. 117PCh. 3 - Prob. 118PCh. 3 - 3116 to 3119 The table gives data concerning the...Ch. 3 - A utility hook was formed from a round rod of...Ch. 3 - A utility hook was formed from a round rod of...Ch. 3 - The steel eyebolt shown in the figure is loaded...Ch. 3 - For Prob. 3122 estimate the stresses at the inner...Ch. 3 - Repeat Prob. 3122 with d = 14 in, Ri = 12 in, and...Ch. 3 - Repeat Prob. 3123 with d = 14 in, Ri = 12 in, and...Ch. 3 - Shown in the figure is a 12-gauge (0.1094-in) by...Ch. 3 - Repeat Prob. 3126 with a 10-gauge (0.1406-in)...Ch. 3 - Prob. 128PCh. 3 - The cast-iron bell-crank lever depicted in the...Ch. 3 - Prob. 130PCh. 3 - Prob. 131PCh. 3 - A cast-steel C frame as shown in the figure has a...Ch. 3 - Two carbon steel balls, each 30 mm in diameter,...Ch. 3 - A carbon steel ball with 25-mm diameter is pressed...Ch. 3 - Repeat Prob. 3134 but determine the maximum shear...Ch. 3 - A carbon steel ball with a 30-mm diameter is...Ch. 3 - An AISI 1018 steel ball with 1-in diameter is used...Ch. 3 - An aluminum alloy cylindrical roller with diameter...Ch. 3 - A pair of mating steel spur gears with a 0.75-in...Ch. 3 - 3140 to 3142 A wheel of diameter d and width w...Ch. 3 - 3140 to 3142 A wheel of diameter d and width w...Ch. 3 - 3140 to 3142 A wheel of diameter d and width w...
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