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Page 429. Sample Problem 6.2.Solve the problem completely;but this time,placethe three applied forces from2.5 kips, 1 kip and 2.5 kips to the new values 3 kips, 2 kipsand1kip.Keep in mind that this change will result inmany other changes across the board.Discuss all practical issues in your findings

Page 429. Sample Problem 6.2.Solve the problem completely;but this time,placethe three applied forces from2.5 kips, 1 kip and 2.5 kips to the new values 3 kips, 2 kipsand1kip.Keep in mind that this change will result inmany other changes across the board.Discuss all practical issues in your findings

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
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Page 429. Sample Problem 6.2.Solve the problem completely;but this time,placethe three applied forces from2.5 kips, 1 kip and 2.5 kips to the new values 3 kips, 2 kipsand1kip.Keep in mind that this change will result inmany other changes across the board.Discuss all practical issues in your findings

2.5 kips с 3 kips (6) -2--3--3--2- 3 kips 6 kip-ft b-3.5 in. 3.5 in. 1 kip -0.5 kip I (1.5) D I--4 11.25 in 2.5 kips Fig. 2 Section of beam having depth d 05 kip (-1.5) 7.5 kip-ft E Fig. 1 Free-body diagram of beam with shear and bending moment diagrams. (-6) 3 kips 6 kip-ft 4 in. x 12 in. nominal size Fig. 3 Design cross section. MODELING: Maximum Shear and Bending Moment. The free-body diagram is used to determine the reactions and draw the shear and bending-moment dia- grams in Fig. 1. We note that ANALYSIS: **Design Based on Allowable Normal Stress. We first express the elas- tic section modulus S in terms of the depth d (Fig. 2). We have -3 kips 1=hd¹ S====bd² = (3.5)d² = 0.5833d² M S= T= V For Mx = 90 kip-in. and = 1800 psi, we write 3 V 2 A M d² = 85.7 = 7.5 kip-ft = 90 kip-in. = 3 kips We have satisfied the requirement that ≤ 1800 psi. Check Shearing Stress. For Vx = 3 kips and d = 9.26 in., we find 0.5833d²= == 90 x 10' lb-in. 1800 psi d = 9.26 in. 3 3000 lb 2 (3.5 in.) (9.26 in.) 3 V 2 A Since = 120 psi, the depth d= 9.26 in. is not acceptable and we must redesign the beam on the basis of the requirement that T ≤ 120 psi. Design Based on Allowable Shearing Stress. Since we now know that the allowable shearing stress controls the design, we write T = 138.8 psi 120 psi 3 3000 lb 2 (3.5 in.)d d = 10.71 in. 4 The normal stress is, of course, less than = 1800 psi, and the depth of 10.71 in. is fully acceptable. REFLECT and THINK: Since timber is normally available in nominal depth increments of 2 in., a 4 x 12-in. standard size timber should be used. The actual cross section would then be 3.5 x 11.25 in. (Fig. 3).
Transcribed Image Text:2.5 kips с 3 kips (6) -2--3--3--2- 3 kips 6 kip-ft b-3.5 in. 3.5 in. 1 kip -0.5 kip I (1.5) D I--4 11.25 in 2.5 kips Fig. 2 Section of beam having depth d 05 kip (-1.5) 7.5 kip-ft E Fig. 1 Free-body diagram of beam with shear and bending moment diagrams. (-6) 3 kips 6 kip-ft 4 in. x 12 in. nominal size Fig. 3 Design cross section. MODELING: Maximum Shear and Bending Moment. The free-body diagram is used to determine the reactions and draw the shear and bending-moment dia- grams in Fig. 1. We note that ANALYSIS: **Design Based on Allowable Normal Stress. We first express the elas- tic section modulus S in terms of the depth d (Fig. 2). We have -3 kips 1=hd¹ S====bd² = (3.5)d² = 0.5833d² M S= T= V For Mx = 90 kip-in. and = 1800 psi, we write 3 V 2 A M d² = 85.7 = 7.5 kip-ft = 90 kip-in. = 3 kips We have satisfied the requirement that ≤ 1800 psi. Check Shearing Stress. For Vx = 3 kips and d = 9.26 in., we find 0.5833d²= == 90 x 10' lb-in. 1800 psi d = 9.26 in. 3 3000 lb 2 (3.5 in.) (9.26 in.) 3 V 2 A Since = 120 psi, the depth d= 9.26 in. is not acceptable and we must redesign the beam on the basis of the requirement that T ≤ 120 psi. Design Based on Allowable Shearing Stress. Since we now know that the allowable shearing stress controls the design, we write T = 138.8 psi 120 psi 3 3000 lb 2 (3.5 in.)d d = 10.71 in. 4 The normal stress is, of course, less than = 1800 psi, and the depth of 10.71 in. is fully acceptable. REFLECT and THINK: Since timber is normally available in nominal depth increments of 2 in., a 4 x 12-in. standard size timber should be used. The actual cross section would then be 3.5 x 11.25 in. (Fig. 3).
2.5 kips 1 kip 2.5 kips 2 ft 3 ft 3 ft2 -10 ft- Sample Problem 6.2 3.5 in. BA timber beam AB of span 10 ft and nominal width 4 in. (actual width = 3.5 in.) is to support the three concentrated loads shown. Knowing that for the grade of timber used all = 1800 psi and Tall = 120 psi, determine the minimum required depth d of the beam. STRATEGY: A free-body diagram with the shear and bending-moment diagrams is used to determine the maximum shear and bending moment. The resulting design must satisfy both allowable stresses. Start by assuming that one allowable stress criterion governs, and solve for the required depth d. Then use this depth with the other criterion to deter- mine if it is also satisfied. If this stress is greater than the allowable, revise the design using the second criterion. (continued)
Transcribed Image Text:2.5 kips 1 kip 2.5 kips 2 ft 3 ft 3 ft2 -10 ft- Sample Problem 6.2 3.5 in. BA timber beam AB of span 10 ft and nominal width 4 in. (actual width = 3.5 in.) is to support the three concentrated loads shown. Knowing that for the grade of timber used all = 1800 psi and Tall = 120 psi, determine the minimum required depth d of the beam. STRATEGY: A free-body diagram with the shear and bending-moment diagrams is used to determine the maximum shear and bending moment. The resulting design must satisfy both allowable stresses. Start by assuming that one allowable stress criterion governs, and solve for the required depth d. Then use this depth with the other criterion to deter- mine if it is also satisfied. If this stress is greater than the allowable, revise the design using the second criterion. (continued)
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