Calculating a force on a member (Statics)

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Discussion Overview

The discussion revolves around calculating the forces acting on members in a static structure, specifically focusing on members AB, BC, and BD. Participants explore methods for determining these forces using free body diagrams and equilibrium principles.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the angle α and the force Ax, questioning how this relates to finding the force in member BC.
  • Another participant suggests cutting member AC to introduce an internal force in member BC and recommends taking moments about point A to solve for this force.
  • A subsequent post reiterates the cutting method but seeks clarification on whether calculating the force in member AB is necessary.
  • A participant asserts that finding the force in member AB is irrelevant for determining the force in member BC and explains the equilibrium condition after cutting the member.
  • Another participant proposes that there may be a typo in the problem statement, suggesting it should ask for the force in BD instead of BC, and notes that BC experiences both axial and shear forces, along with a bending moment.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of calculating the force in member AB for determining the force in member BC. There is also a suggestion of a potential typo in the problem statement regarding which force is being asked for, indicating a lack of consensus on the problem's intent.

Contextual Notes

Some assumptions about the structure's equilibrium and the definitions of forces acting on the members are not explicitly stated, which may affect the clarity of the discussion.

bardia sepehrnia
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upload_2019-2-16_1-45-29.png


I isolated the member ABC and drew the free body diagram:
upload_2019-2-16_1-46-37.png


α is then calculated using inverse tan: Tan-1=(6.25+15)/50=23.03
Then force of member BD on the joint can be found by sum of all moments around point A.
upload_2019-2-16_1-52-35.png

Then Ax is calculated which is equal to BD×Cos(α)=235.2×Cos(23.03) Ax=216.48
Then I draw free body diagram of the block with joint A:
upload_2019-2-16_1-58-49.png

Then P=Ax=216.48.
I cannot to the second part of this question and calculate the force on the member BC!?
I can calculate Ay in first free body diagram however would that at all help me find the force on Member BC? Is the resultant force of vectors Ay and Ax the force acting on member AB?
 

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To find the force in member BC, try cutting the member AC just after point B by a bit.

Then for equilibrium to take place, there will be a new internal force introduced when you "cut". This force is the internal force of member BC. Take moments about point A and then solve for this force.
 
CivilSigma said:
To find the force in member BC, try cutting the member AC just after point B by a bit.

Then for equilibrium to take place, there will be a new internal force introduced when you "cut". This force is the internal force of member BC. Take moments about point A and then solve for this force.

I don't understand. Can you elaborate? Am I not supposed to firs calculate the force on AB member or is that irrelevant?
 
It's irrelevant to find force in member AB.

You need to "cut" the rod somewhere between point B and C if you want to find force BC.

We know there is a force in member BC. Imagine you cut the member somewhere from in between point B and C - call it point W.

Now you have two rods: A-B-W and W-C. Since the original rod was under equilibrium, the internal force resulting from "this cut" in member A-B-W and W-C must be equal in magnitude and opposite in direction. This internal force also represents the axial force in member BC.


Here are a few links that may be helpful :

http://www2.hawaii.edu/~takebaya/lessons/method-of-sections.pdf
 
I really think that there is a typo in the problem, that it means to ask for the force in BD rather than the force in BC, which is only a part of the member ABC. BC has both axial and shear forces, as well as a varying bending moment. But anyway, as suggested, cut ABC just to the right of B, then look at the FBD of BC to find the force, or axial and shear components of that force.
 

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