Explanation The below figure shows the free body diagram of wrench. Figure-(1) Write the expression for moment about A. ∑ M A = 0 F × l = F B × l A B . (I) Here, the force on the wrench is F , the length of point A from applied force is l , the force at point B is F B and the length of point A from point B is l AB . Write the expression for component of force F B in x- direction. F B x = F B cos α . (II) Here, the angle made by force from horizontal is α . Write the expression for component of force F B in y- direction. F B y = F B sin α (III) Write the expression for equilibrium in horizontal direction. R A x − F B x = 0 . (IV) Here, the horizontal component of reaction at A is R A x . Write the expression for equilibrium of forces in vertical direction. R A y − F − F B y = 0 . (V) Here, the vertical component of reaction at A is R Ay Write the expression for net reaction at A. R = ( R A x ) 2 + ( R A y ) 2 (VI) Write the expression for moment. M = F × l (VII) Here, the length of arm of moment is l . Write the expression for throat area of the weld. A = 2 × 1.414 π h r . (VIII) Write the expression for unit second moment area. J u = 2 π r 3 . (IX) Write the expression for polar moment area. J = 2 × 0.707 h J u (X) Write the expression for primary shear stress. τ ′ = R A A (XI) Write the expression for secondary shear stress. τ ″ = M C J (XII) Write the expression for maximum shear stress. τ max = ( τ ′ ) 2 + ( τ ″ ) 2 (XIII) Write the expression for leg size using maximum stress relation. τ max = τ allow (XIV) Conclusion: Substitute 16 in for l , 3 in for l AB and 100 lbf for F in Equation (I). ( 100 lbf ) ( 16 in ) = F B × ( 3 in ) F B = ( 100 lbf ) ( 16 in 3 in ) F B = 533.3 lbf Substitute 533.33 lbf for F B and 60 ° for α in Equation (II). F B x = ( 533.33 lbf ) cos ( 60 ° ) = ( 533.33 lbf ) × 0.5 = 266.665 lbf Substitute 533.33 lbf for F B and 60 ° for α in Equation (III). F B y = ( 533.33 lbf ) sin ( 60 ° ) = ( 533.33 lbf ) × 0.866 = 461.86 lbf Substitute 266.665 lbf for F B x in Equation (IV). R A x − 266.665 lbf = 0 R A x = 266.665 lbf Substitute 461.86 lbf for F B y and 100 lbf for F in Equation (V). R Ay − 100 lbf − 461.86 lbf = 0 R Ay = 461.86 lbf + 100 lbf R Ay = 561.86 lbf Substitute 266.665 lbf for R A x and 561.86 lbf for R A y in Equation (VI). R A = ( 266.665 lbf ) 2 + ( 561

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

10th Edition

ISBN: 9780073398204

Author: Richard G Budynas, Keith J Nisbett

Publisher: McGraw-Hill Education

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Chapter 9, Problem 50P

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The thickness of weld.

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9-25 The weldment shown in the figure is subjected to an alternating force F. The hot-rolled steel to bar has a thickness h and is of AISI 1010 steel. The vertical support is likewise AISI 1010 HR 9-28 steel. The electrode is given in the table below. Estimate the fatigue load F the bar will carry if three fillet welds are used. Problem Number b 9-25 9-27 50 mm 50 mm d 50 mm 5 mm 30 mm 5 mm h Electrode E6010 E7010 |-b Problems 9-25 to 9-28

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9-34 The attachment shown in the figure is made of 1018 HR steel 12 mm thick. The static force is 100 kN. The member is 75 mm wide. Specify the weldment (give the patter, electrode number, type of weld, length of weld, and leg size). 100- 1018 HRV 1018 HR Dimensions in millimeters. 37.5 dia. 75 dia -12 TO 225 F = 100 KN Problem 9-34

Chapter 9 Solutions

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

Ch. 9 - 91 to 94 The figure shows a horizontal steel bar...Ch. 9 - 91 to 94 The figure shows a horizontal steel bar...Ch. 9 - 91 to 94 The figure shows a horizontal steel bar...Ch. 9 - 91 to 94 The figure shows a horizontal steel bar...Ch. 9 - 95 to 98 For the weldments of Probs. 91 to 94, the...Ch. 9 - 95 to 98 For the weldments of Probs. 91 to 94, the...Ch. 9 - 95 to 98 For the weldments of Probs. 91 to 94, the...Ch. 9 - 95 to 98 For the weldments of Probs. 91 to 94, the...Ch. 9 - 99 to 912 The materials for the members being...Ch. 9 - Prob. 10P

Ch. 9 - Prob. 11P Ch. 9 - 99 to 912 The materials for the members being...Ch. 9 - 913 to 916 A steel bar of thickness h is welded to...Ch. 9 - 913 to 916 A steel bar of thickness h is welded to...Ch. 9 - 913 to 916 A steel bar of thickness h is welded to...Ch. 9 - Prob. 16P Ch. 9 - Prob. 17P Ch. 9 - 917 to 920 A steel bar of thickness h, to be used...Ch. 9 - 917 to 920 A steel bar of thickness h, to be used...Ch. 9 - 917 to 920 A steel bar of thickness h, to be used...Ch. 9 - Prob. 21P Ch. 9 - 921 to 924 The figure shows a weldment just like...Ch. 9 - Prob. 23P Ch. 9 - Prob. 24P Ch. 9 - 9-25 to 9-28 The weldment shown in the figure is...Ch. 9 - 9-25 to 9-28 The weldment shown in the figure is...Ch. 9 - Prob. 27P Ch. 9 - 925 to 928 The weldment shown in the figure is...Ch. 9 - The permissible shear stress for the weldment...Ch. 9 - Prob. 30P Ch. 9 - 9-30 to 9-31 A steel bar of thickness h is...Ch. 9 - In the design of weldments in torsion it is...Ch. 9 - Prob. 33P Ch. 9 - Prob. 34P Ch. 9 - The attachment shown carries a static bending load...Ch. 9 - The attachment in Prob. 935 has not had its length...Ch. 9 - Prob. 37P Ch. 9 - Prob. 39P Ch. 9 - Prob. 40P Ch. 9 - Prob. 42P Ch. 9 - 9-43 to 9-45 A 2-in dia. steel bar is subjected to...Ch. 9 - 9-43 to 9-45 A 2-in dia. steel bar is subjected to...Ch. 9 - Prob. 45P Ch. 9 - Prob. 46P Ch. 9 - Find the maximum shear stress in the throat of the...Ch. 9 - The figure shows a welded steel bracket loaded by...Ch. 9 - Prob. 49P Ch. 9 - Prob. 50P Ch. 9 - Prob. 51P Ch. 9 - Brackets, such as the one shown, are used in...Ch. 9 - For the sake of perspective it is always useful to...Ch. 9 - Hardware stores often sell plastic hooks that can...Ch. 9 - For a balanced double-lap joint cured at room...

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The thickness of weld.

Solution Summary: The following figure shows the free body diagram of wrench. Write the expression for moment about A.

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