How Do You Solve for Reactions in Shear and Moment Diagrams?

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Discussion Overview

The discussion revolves around solving for reactions in a beam and drawing the corresponding shear and moment diagrams. It includes aspects of problem-solving in structural analysis, specifically focusing on the application of Newton's laws and the interpretation of distributed loads.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant requests help with solving for reactions and drawing shear and moment diagrams.
  • Another participant emphasizes that the forum does not solve problems directly but encourages showing attempts and applying Newton's laws of equilibrium.
  • A participant reports calculated reactions of 52.44 kN at E and 39.56 kN at A, seeking guidance on the shear diagram for a 5 kN/m load.
  • A later reply suggests that the participant's earlier calculations may be affected by incorrect scaling in the drawing and provides a detailed explanation of how to approach the shear diagram, including the conversion of distributed loads into axial and shear components.
  • One participant claims to have resolved their confusion regarding the shear diagram and expresses gratitude for the assistance received.

Areas of Agreement / Disagreement

Participants generally agree on the approach to solving the problem using equilibrium principles, but there are variations in the calculations and interpretations of the shear diagram, indicating some level of disagreement or uncertainty in the details.

Contextual Notes

There are references to specific calculations and assumptions regarding the geometry of the beam and the distribution of loads, which may not be fully resolved or clarified in the discussion.

lunaticpimp12
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kindly help me with this problem.



problem-1.jpg






i can't solve for the reactions and draw the shear and moment diagrams. can you solve it for me?

pls help me. thanks in advance.
 
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Hi there, we don't solve problems here, we just try to help you solve them. Solve for the reactions using Newton's laws of equilbrium for forces in x direction, forces in y direction, and torques about any point, each of which must add to zero. The shear and moment diagrams , which then follow, are a bit time consuming. Please show some attempt, thanks.
 
i solve for the reactions and i got:

Reactions for E = 52.44 kN

Reactions for A = 39.56 kN

now my question is what is the shear diagram for the 5 kN/m. i got stuck when i draw it i got for the shear diagram of the reaction = 23.74 and i don't know how to draw for the 5kN/m. thanks.
 
lunaticpimp12 said:
i solve for the reactions and i got:

Reactions for E = 52.44 kN

Reactions for A = 39.56 kN

now my question is what is the shear diagram for the 5 kN/m. i got stuck when i draw it i got for the shear diagram of the reaction = 23.74 and i don't know how to draw for the 5kN/m. thanks.
Your reactions look good, but member AB is the hypotenuse of a 3:4:5 right triangle, which is not scaled on the drawing correctly, probably accounting for your error; the axial load is .6(R_A) = 23.74kN just up from A, and the shear load is .8(R_A) = 31.65kN at that point.

Drawing the shear diagram on AB, it starts at +31.65 at A, continues with that constant value up to the distributed 5kN/m load, at (2,1.5), that is, at 2.5m along the member. At this point, you need to change the 5kN/m load into its axial and shear components (4kN/m shear distributed load), and note that dV/dx = -q, that is, the slope of the shear diagram now becomes 4KN/m within that distributed 4 m long load segment, so at the end of it, the shear is now 15.65kN. Now painstakingly continue, cutting FBD's every now and then to check your numbers and your sanity.
 
i got it now it should be 3.2 kN/m and i got it all correctly. anyways thanks for the help.
 

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