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

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SUMMARY

The discussion focuses on solving for reactions and constructing shear and moment diagrams in structural analysis. The reactions calculated are 52.44 kN at point E and 39.56 kN at point A. The shear diagram begins at +31.65 kN at point A, transitioning through a distributed load of 5 kN/m, which is converted into its axial and shear components. The final shear value at the end of the distributed load segment is confirmed to be 15.65 kN.

PREREQUISITES
  • Understanding of Newton's laws of equilibrium for forces and torques
  • Knowledge of shear and moment diagram construction
  • Familiarity with distributed loads and their conversion to axial and shear components
  • Ability to perform Free Body Diagrams (FBDs) for verification
NEXT STEPS
  • Learn how to construct shear and moment diagrams for various loading conditions
  • Study the principles of converting distributed loads into point loads and their effects on shear
  • Explore the use of Free Body Diagrams (FBDs) in structural analysis
  • Investigate the application of the 3:4:5 triangle in structural member analysis
USEFUL FOR

Civil engineers, structural analysts, and students studying mechanics of materials will benefit from this discussion, particularly those focused on shear and moment diagram analysis.

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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