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Problem 3. For problem 2, determine the absolute maximum shear stress developed in the bolt shank at point A. Represent the results for absolute maximum shear stress on a properly oriented element located at this point. Draw the three Mohr's circles for the state of stress at this point. Locate the points of maximum, intermediate, and minimum normal stresses on the circles; also locate the point of absolute maximum shear stress.

Problem 3. For problem 2, determine the absolute maximum shear stress developed in the bolt shank at point A. Represent the results for absolute maximum shear stress on a properly oriented element located at this point. Draw the three Mohr's circles for the state of stress at this point. Locate the points of maximum, intermediate, and minimum normal stresses on the circles; also locate the point of absolute maximum shear stress.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Problem 2 is answered and the solution is provided. Problem 3 need to be answer

Solution: Let's see the free body diagram Torque at 0= Forcex distance To=18x 6 inch =108 16-inch So B Torsional Equation → - 1816-Force 2 inch 108 tb-irch (torque) Torque at section of AB = 108 16-inch Moment at section of AB = 2x18 - 36 16-inch An Figure 2, At A point Bending stress will be tensile & compressive at B. F-1 #7 (10-25 )4 *(0,25) 23.468 ksi CA= 100 ok x 2x (0.2514 (25) Fig. 2 181b 6 inch B6 0.25 inch TA = 35.2 ksi (i). Stress condition at A Now, As we know the standard relation 05/0₂= * ± √(√2-9) ²³+5 3+9 ± √(1-9) I/₂ = 23.468 +0 + 23-468 2 [(23.167 J²+(35.2) ² T= 117 + 37.16 = 48.8 ksi 2= 11-7-37-10 2=-25.4 ksi So tan (26) = 2xy x- tan (20p) = 2X 35-2 23.468 35.8° Op 0₂= Op ±90 P₂ = 35-8-90 0₂=-54-20% Tmay = - = 488 -(-25.4) 2 Tray=3711 ksi Problem 3. For problem 2, determine the absolute maximum shear stress developed in the bolt shank at point A. Represent the results for absolute maximum shear stress on a properly oriented element located at this point. Draw the three Mohr's circles for the state of stress at this point. Locate the points of maximum, intermediate, and minimum normal stresses on the circles; also locate the point of absolute maximum shear stress. A 23-4684 35.2 ksi - Cry B 2 in. A Txy = Sin 20= 1 or -1 9-02 Sin 20 Ⓒ=45° or 135° -Inge is at 45° from principle plane. 0-54.2 +45 9--9-2⁰ Similarly for point B Only shear stress will act Bending stress will be o. Shear stress due to torsion =√xx 9. Tv x(0.25/4 Shear stress due to vertical lood at B. Tv = £ x ²3 108 X0-128 #x(0.252 Tv=-0.5ksi T. -35-2-0.5 = 34.7 ksi q/02 = °J34.7 १२ = 35.2 ksi नूं 34.7ksi 2-34.7 kst tan (20p) = -2034-7 O 6 in. B [Tay - Tag (-ve sign because of -downward load) 8 Top=-45° B itself is at a plane where may shear is. 18 lb 34:7481
Transcribed Image Text:Solution: Let's see the free body diagram Torque at 0= Forcex distance To=18x 6 inch =108 16-inch So B Torsional Equation → - 1816-Force 2 inch 108 tb-irch (torque) Torque at section of AB = 108 16-inch Moment at section of AB = 2x18 - 36 16-inch An Figure 2, At A point Bending stress will be tensile & compressive at B. F-1 #7 (10-25 )4 *(0,25) 23.468 ksi CA= 100 ok x 2x (0.2514 (25) Fig. 2 181b 6 inch B6 0.25 inch TA = 35.2 ksi (i). Stress condition at A Now, As we know the standard relation 05/0₂= * ± √(√2-9) ²³+5 3+9 ± √(1-9) I/₂ = 23.468 +0 + 23-468 2 [(23.167 J²+(35.2) ² T= 117 + 37.16 = 48.8 ksi 2= 11-7-37-10 2=-25.4 ksi So tan (26) = 2xy x- tan (20p) = 2X 35-2 23.468 35.8° Op 0₂= Op ±90 P₂ = 35-8-90 0₂=-54-20% Tmay = - = 488 -(-25.4) 2 Tray=3711 ksi Problem 3. For problem 2, determine the absolute maximum shear stress developed in the bolt shank at point A. Represent the results for absolute maximum shear stress on a properly oriented element located at this point. Draw the three Mohr's circles for the state of stress at this point. Locate the points of maximum, intermediate, and minimum normal stresses on the circles; also locate the point of absolute maximum shear stress. A 23-4684 35.2 ksi - Cry B 2 in. A Txy = Sin 20= 1 or -1 9-02 Sin 20 Ⓒ=45° or 135° -Inge is at 45° from principle plane. 0-54.2 +45 9--9-2⁰ Similarly for point B Only shear stress will act Bending stress will be o. Shear stress due to torsion =√xx 9. Tv x(0.25/4 Shear stress due to vertical lood at B. Tv = £ x ²3 108 X0-128 #x(0.252 Tv=-0.5ksi T. -35-2-0.5 = 34.7 ksi q/02 = °J34.7 १२ = 35.2 ksi नूं 34.7ksi 2-34.7 kst tan (20p) = -2034-7 O 6 in. B [Tay - Tag (-ve sign because of -downward load) 8 Top=-45° B itself is at a plane where may shear is. 18 lb 34:7481
Problem 2. The bolt is fixed to its support at C. If a force of 18 lb is applied to the wrench to tighten it, determine the principal stresses developed in the bolt C shank at point A. Represent the results on a properly oriented element located at this point. Also draw a Mohr's circle and locate the points of maximum and minimum principal stresses, and maximum in-plane shear stress. The shank has a diameter of 0.25 in. B 2 in. A 6 in. 18 lb
Transcribed Image Text:Problem 2. The bolt is fixed to its support at C. If a force of 18 lb is applied to the wrench to tighten it, determine the principal stresses developed in the bolt C shank at point A. Represent the results on a properly oriented element located at this point. Also draw a Mohr's circle and locate the points of maximum and minimum principal stresses, and maximum in-plane shear stress. The shank has a diameter of 0.25 in. B 2 in. A 6 in. 18 lb
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