Finding Reactions in a Pinned and Roller Supported Beam

  • Thread starter Thread starter andycampbell1
  • Start date Start date
  • Tags Tags
    Frames
Click For Summary

Discussion Overview

The discussion revolves around determining the reactions at various supports of a beam (pinned at A and roller supported at B and C) under a specific loading condition. Participants explore methods to calculate these reactions using principles of static equilibrium, including moment calculations and force summations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The initial poster presents their calculations for reactions at the pin P and support C, arriving at +4.00 kN and +6.00 kN respectively, using moment and force balance equations.
  • andycampbell1 suggests taking moments about points A or B to find reactions on the left-hand side of the beam.
  • The initial poster attempts to calculate the vertical reactions at A and B, referencing a textbook for expected values of +3.87 kN and +13.12 kN.
  • Another participant points out that the initial moment calculation omitted the force at the pin P, which is necessary for accurate results.
  • The initial poster later successfully calculates the reactions at A and B as +3.87 kN and +13.12 kN, respectively, after revising their moment equation to include all forces.

Areas of Agreement / Disagreement

Participants generally agree on the methods to approach the problem, but there is no consensus on the initial calculations due to differing interpretations of the moment equations and the inclusion of forces. The discussion reflects a progression from uncertainty to successful calculation, but no definitive agreement on the initial steps is reached.

Contextual Notes

Some calculations depend on the correct application of moment equations and the inclusion of all forces acting on the structure. The initial poster's approach to splitting the structure for analysis is noted as a valid method, though it introduces complexity in ensuring all forces are accounted for.

andycampbell1
Messages
34
Reaction score
0

Homework Statement


The beam ABC in attachment has a pinned support at A and Roller supports at B and C. P is an internal pin. I have to determine the reactions at A, B and C and at the pin P when the beam is loaded as shown in the attachment
structuralmech.jpg


Homework Equations


\SigmaV=O \SigmaH=0



The Attempt at a Solution


I have split the problem in half and started with the right hand side. I have worked out the reaction at the pin as +4.00 kN and the reaction at C as +6.00 kN. I have done this by taking the moments of P as (10x3)-VC*5=0. I then worked out the sum of the vertical which = 0 to get the reaction at the pin that was VP-VC-10= 0 meaning that VP =+4.00 kN. I have worked out the horizontals on the left by doing the sum of the horizontals = 0, \SigmaH=0
=HA-15cos60=0 HA= +7.5kN. What I would like to know is how to find out the vertical reaction at A and B Thanks
 
Physics news on Phys.org
andycampbell1: Nice work, so far. Hint 1: For the left-hand side, take moments about point A or B.
 
I have tried taking the moments of A. The book I am working from gives me the answers for VA as +3.87kN and VB as +13.12kN. For taking moments about A I tried this

\SigmaMA=0
=2.5x15sin(60)-VBx4=0.
I have also tried adding VBx4 and adding the value of the pin and dividing by the length of 5 but I cannot come up with what the book states as the answer. Would I be as well to keep this as 2 separate structures for taking the moments of A or using the whole structure?
 
Keeping this as two separate structures for taking moments is good, if you wish. Either way you prefer is fine. Your MA summation, above, omitted VP. You cannot omit forces on your structure (or on your sectioned-off portion of a structure). Also, see the last paragraph of post https://www.physicsforums.com/showpost.php?p=2946515".
 
Last edited by a moderator:
Hi finally managed to get it, this is what I done

\SigmaMA=0
=(2.5x15sin60)+(VPx5)-VBx4)=0
=(32.48)+(20)-(VBx4)=0
VB=+13.12 kN

\SigmaV=0
=VA+VB-15sin60-VP=0
= 0 + 13.12 - 12.99 - 4 = 0
VA = +3.87 kN

Thanks for the links they have helped me.
 

Similar threads

Calculating Support Reactions for a Beam with Inclined Resting Position
  • · Replies 4 ·
Replies
4
Views
4K
Identifying Pin Joints & Rollers - Structural Analysis Homework Help
  • · Replies 20 ·
Replies
20
Views
5K
Find reactions at supports and bending/shear/axial diagrams
  • · Replies 3 ·
Replies
3
Views
4K
The number of reactions in pinned support
  • · Replies 13 ·
Replies
13
Views
2K
Why Replace Pinned Support with Roller Guide for Qualitative Influence Lines?
  • · Replies 1 ·
Replies
1
Views
2K
Solving AB Beam Reactions with Hyperstatic System
  • · Replies 1 ·
Replies
1
Views
2K
Why is Only One Reaction Shown at the Pinned Support?
  • · Replies 12 ·
Replies
12
Views
2K
Does Angling a pin support on a truss change anything?
  • · Replies 7 ·
Replies
7
Views
5K
Engineering Find the support reactions on a beam
  • · Replies 3 ·
Replies
3
Views
2K
Vertical reactions of pin supports of a weightless Beam with a UDL applied
  • · Replies 2 ·
Replies
2
Views
5K