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B |C E For this loading, calculate the reaction forces Ay and Ey acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-body diagram below. Note: Since Ax = 0, it has been omitted from the free-body diagram.)

B |C E For this loading, calculate the reaction forces Ay and Ey acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-body diagram below. Note: Since Ax = 0, it has been omitted from the free-body diagram.)

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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Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Let a=4.0 ft, b-8.0 ft, c-4.0 ft, d=3.0 ft, w = 6.5 kips/ft and P = 45 kips. Construct the shear-force and bending-moment diagrams on paper and use the results to answer the questions in the subsequent parts of this GO exercise. A a a B Part 2 b d For this loading, calculate the reaction forces Ay and Ey acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-body diagram below. (Note: Since Ax = 0, it has been omitted from the free-body diagram.) Save for Later B W Answers: Ay = i 37.575 eTextbook and Media W AV=P₂ b C V₁ - V₁ = Sw(x) dx C C с D + D d kips, Ey= M₂ E i E ZA x Ey X Incorrect If your answer for part (a) or (e) is incorrect, recall that concentrated loads create discontinuities in the shear-force diagram. Refer to the table below (Construction Rules for Shear-Force and Bending-Moment Diagrams), Rule 1. At each concentrated load, the shear diagram jumps by an amount equal to the concentrated load. The change in the diagram is in the same direction as the force arrow. From the FBD, observe that the positive direction for the reaction force is assumed to act upward at A. If your answer for part (b), (c), (d) or (f) is incorrect, recall that the change in the shear force between any two locations is equal to the area under the load diagram w between those same two points. Refer to the table, Rule 2. Also, note that since the arrows shown for the distributed load w point upward, the area under the load diagram is taken as a positive area. Thus, the change in the shear-force diagram is positive. Construction Rules for Shear-Force and Bending-Moment Diagrams Equation Load Diagram w Shear-Force Diagram V Rule 1: Concentrated loads create discontinuities in the shear-force diagram. Positive jump in shear force V P x 56.925 Attempts: 1 of 5 used Submit Answer V₂ % Ta Rule 2: The change in shear force is equal to the area under the distributed-load curve. Sw(x) dx- w(x) 4-S" AV-V₂-V₁-w) de V₂ Bending-Moment Diagram M Slope-V₁- MA kips. M₁ "A Slope V₂ 30 Ma Fa M₂
Transcribed Image Text:Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Let a=4.0 ft, b-8.0 ft, c-4.0 ft, d=3.0 ft, w = 6.5 kips/ft and P = 45 kips. Construct the shear-force and bending-moment diagrams on paper and use the results to answer the questions in the subsequent parts of this GO exercise. A a a B Part 2 b d For this loading, calculate the reaction forces Ay and Ey acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-body diagram below. (Note: Since Ax = 0, it has been omitted from the free-body diagram.) Save for Later B W Answers: Ay = i 37.575 eTextbook and Media W AV=P₂ b C V₁ - V₁ = Sw(x) dx C C с D + D d kips, Ey= M₂ E i E ZA x Ey X Incorrect If your answer for part (a) or (e) is incorrect, recall that concentrated loads create discontinuities in the shear-force diagram. Refer to the table below (Construction Rules for Shear-Force and Bending-Moment Diagrams), Rule 1. At each concentrated load, the shear diagram jumps by an amount equal to the concentrated load. The change in the diagram is in the same direction as the force arrow. From the FBD, observe that the positive direction for the reaction force is assumed to act upward at A. If your answer for part (b), (c), (d) or (f) is incorrect, recall that the change in the shear force between any two locations is equal to the area under the load diagram w between those same two points. Refer to the table, Rule 2. Also, note that since the arrows shown for the distributed load w point upward, the area under the load diagram is taken as a positive area. Thus, the change in the shear-force diagram is positive. Construction Rules for Shear-Force and Bending-Moment Diagrams Equation Load Diagram w Shear-Force Diagram V Rule 1: Concentrated loads create discontinuities in the shear-force diagram. Positive jump in shear force V P x 56.925 Attempts: 1 of 5 used Submit Answer V₂ % Ta Rule 2: The change in shear force is equal to the area under the distributed-load curve. Sw(x) dx- w(x) 4-S" AV-V₂-V₁-w) de V₂ Bending-Moment Diagram M Slope-V₁- MA kips. M₁ "A Slope V₂ 30 Ma Fa M₂
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