Homework Help is Here – Start Your Trial Now!
Engineering
Mechanical Engineering
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
See similar textbooks
icon
Related questions
Question
100%

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 ↑ ◄
Expert Solution
steps

Step by step

Solved in 3 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Strain Transformation
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Elements Of Electromagnetics
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY
SEE MORE TEXTBOOKS