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Hi there, can you please explain the + and - in the bending flexure equation in the red box. I tried solving the question myself, i got every steps correct but all my answer are wrong because of the "+" and "-" sign. How do i know or what do i look at to know if it -Mc/I or +Mc/I

Hi there, can you please explain the + and - in the bending flexure equation in the red box. I tried solving the question myself, i got every steps correct but all my answer are wrong because of the "+" and "-" sign. How do i know or what do i look at to know if it -Mc/I or +Mc/I

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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Hi there, can you please explain the + and - in the bending flexure equation in the red box. I tried solving the question myself, i got every steps correct but all my answer are wrong because of the "+" and "-" sign. How do i know or what do i look at to know if it -Mc/I or +Mc/I.

SOLUTION 1.5 in. TDB Disin Lin 1.5%. A CROSS CECTION 3 in MIN 10.75 Downward force. C1 Www T0.5. 2.Sin DOWNWARD FORCE: 2,1 -1.70in, 2.30th 1.5 in. C We choose the smaller value. Upward force. 1.501 e Section a-a X = 12.75 in ³ 7.5 in² UPWARD FORLE 1.20 At D: op= X = ΣAỹ _ (1 × 3)(0.5) + 2(3 × 0.25)(2.5) ΣΑ (1 x 3) + 2(3x 0.75) A = 7.5 in² Fall = +5 ksi 4,- (-126² +4²) = Σ Ad 12 I = 10.825 inª 2.30in P + 5 ksi= At E: σE = - 12 ksi = 0.75 in. t ↓ 0.75 in. M = P(1.5 in.+1.70 in.) = (3.20 in.)P P A = 1.700 in. P 7.5 + 5 = P(+0.6359) P Mc A I + We choose the smaller value. Fall = -12 ksi (3)(1)² + (3 × 1)(1.70 – 0.5)² + Mc I P 7.5 - 12 = P(-0.5466) (3.20)P(1.70) 10.825 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. PROBLEM 4.113 A vertical rod is attached at point A to the cast iron hanger shown. Knowing that the allowable stresses in the hanger are all = +5 ksi and all -12 ksi, determine the largest downward force and the largest upward force that can be exerted by the rod. (3.20)P(2.30) 10.825 564 PROBLEM 4.113 (Continued) At D: OD = M = P(1.5 in. + 1.70 in.) = (3.20 in.)P P -12 ksi = P 7.5 -12 = P(-0.6359) At E: σE = + + 12 (1.5)(3) ³. (1.5)(3)³ + (1.5x3)(2.5-1.70)² P A +5 ksi= Mc I (3.20)P(1.70) 10.825 Mc P (3.20)P(2.30) 7.5 10.825 +5= P(+0.5466) P = 7.86 kips ↓ P = 21.95 kips ↓ P = 7.86 kips ◄ P = 18.87 kips ↑ P = 9.15 kips ↑ P = 9.15 kips ↑ ◄
Transcribed Image Text:SOLUTION 1.5 in. TDB Disin Lin 1.5%. A CROSS CECTION 3 in MIN 10.75 Downward force. C1 Www T0.5. 2.Sin DOWNWARD FORCE: 2,1 -1.70in, 2.30th 1.5 in. C We choose the smaller value. Upward force. 1.501 e Section a-a X = 12.75 in ³ 7.5 in² UPWARD FORLE 1.20 At D: op= X = ΣAỹ _ (1 × 3)(0.5) + 2(3 × 0.25)(2.5) ΣΑ (1 x 3) + 2(3x 0.75) A = 7.5 in² Fall = +5 ksi 4,- (-126² +4²) = Σ Ad 12 I = 10.825 inª 2.30in P + 5 ksi= At E: σE = - 12 ksi = 0.75 in. t ↓ 0.75 in. M = P(1.5 in.+1.70 in.) = (3.20 in.)P P A = 1.700 in. P 7.5 + 5 = P(+0.6359) P Mc A I + We choose the smaller value. Fall = -12 ksi (3)(1)² + (3 × 1)(1.70 – 0.5)² + Mc I P 7.5 - 12 = P(-0.5466) (3.20)P(1.70) 10.825 PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. PROBLEM 4.113 A vertical rod is attached at point A to the cast iron hanger shown. Knowing that the allowable stresses in the hanger are all = +5 ksi and all -12 ksi, determine the largest downward force and the largest upward force that can be exerted by the rod. (3.20)P(2.30) 10.825 564 PROBLEM 4.113 (Continued) At D: OD = M = P(1.5 in. + 1.70 in.) = (3.20 in.)P P -12 ksi = P 7.5 -12 = P(-0.6359) At E: σE = + + 12 (1.5)(3) ³. (1.5)(3)³ + (1.5x3)(2.5-1.70)² P A +5 ksi= Mc I (3.20)P(1.70) 10.825 Mc P (3.20)P(2.30) 7.5 10.825 +5= P(+0.5466) P = 7.86 kips ↓ P = 21.95 kips ↓ P = 7.86 kips ◄ P = 18.87 kips ↑ P = 9.15 kips ↑ P = 9.15 kips ↑ ◄
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