Finding support reactions for shear and moment diagrams

Click For Summary

Discussion Overview

The discussion revolves around calculating support reactions for a beam in order to draw shear and moment diagrams. Participants are sharing their attempts at solving the problem, including their calculations and the assumptions made regarding the support types and loading conditions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant calculates the moment about point B and finds A = -50 kN, leading to a system of equations for support reactions C and D.
  • Another participant points out an error in the calculation of the moment due to the applied couple at A, suggesting it should not be multiplied by a moment arm.
  • A different calculation leads to A = 4 kN, with subsequent values for C and D being significantly different from earlier attempts.
  • There are corrections regarding the value of A and the interpretation of forces in the y-direction, with some participants suggesting to check the free body diagram (FBD) for accuracy.
  • One participant expresses confusion about the orientation of their image, which affects the clarity of their calculations.
  • Another participant acknowledges the negative value for D and suggests representing it as a downward force, while praising the overall quality of the shear and moment diagrams produced.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct values for the support reactions, as multiple conflicting calculations are presented. The discussion remains unresolved with various interpretations and corrections being proposed.

Contextual Notes

There are unresolved mathematical steps and differing assumptions regarding the moments and forces acting on the beam. The calculations depend heavily on the interpretation of the applied loads and support types.

Who May Find This Useful

Students preparing for exams in structural analysis or mechanics, as well as those interested in understanding the process of calculating support reactions and drawing shear and moment diagrams.

mwjrain
Messages
4
Reaction score
0
Draw the shear and moment diagrams for the beam in the figure. Clearly show all values, including the maximum shear and maximum moment and their positions on the diagrams. Assume the supports at A and C are rollers and D is a pin support. note there is a hinge at point B.

I am having trouble finding the support reactions.
My attempt is:
(Assuming positive Y direction is up)
Moment about b = 10(60)+10A-2(10)5=0
A=-50 kN

Forces in y direction = C+D-50-2(10)-5-18=0
C+D=93

Moment about D= C(12)+60(32)-(50)32-(20)(27)-5(16)-18(6)=0
C=34

D=93-34=59
 

Attachments

  • image.jpg
    image.jpg
    21.3 KB · Views: 517
Physics news on Phys.org
mwjrain said:
Draw the shear and moment diagrams for the beam in the figure. Clearly show all values, including the maximum shear and maximum moment and their positions on the diagrams. Assume the supports at A and C are rollers and D is a pin support. note there is a hinge at point B.

I am having trouble finding the support reactions.
My attempt is:
(Assuming positive Y direction is up)
Moment about b = 10(60)+10A-2(10)5=0
A=-50 kN

Forces in y direction = C+D-50-2(10)-5-18=0
C+D=93

Moment about D= C(12)+60(32)-(50)32-(20)(27)-5(16)-18(6)=0
C=34

D=93-34=59
Your work is generally very good, however, your error is in the calculation of the moment due to the applied couple at A. You are incorrectly multiplying that couple by a moment arm. The moment of a couple about any point is the couple itself.
 
So it would be
Moment about b = 60 +10A-2(10)5=0
A=4 kN

Moment about D = C(12)+60-50(32)-20(27)-5(16)-18(6)=0
C=189kN

Forces in y direction=C+D+4-(2)10-5-18=0
D=-150kN
 
mwjrain said:
So it would be
Moment about b = 60 +10A-2(10)5=0
A=4 kN
yes
Moment about D = C(12)+60-50(32) -20(27)-5(16)-18(6)=0
C=189kN
You still used A = -50 instead of A = +4
Forces in y direction=C+D+4-(2)10-5-18=0
D=-150kN
make necessary corrections. It is a good idea to look at the FBD of the beam between B and D to check results.

Welcome to PF!:smile:

Note: My lightweight monitor made it easy to turn it upside down.:wink:
 
Sorry about the image being upside down! I didn't realize that. And thank you for the warm welcome.

So
Moment about D= C12-18(6)-5(16)-2(10)27+4(32)+60=0
C=45

Forces in y =D+4+45-20-18-5=0
D=-6

Note (the new pictures going to be sideways. I can't figure out how to fix it on my iPad.) sorry!
 

Attachments

  • image.jpg
    image.jpg
    29.5 KB · Views: 535
mwjrain said:
Sorry about the image being upside down! I didn't realize that. And thank you for the warm welcome.

So
Moment about D= C12-18(6)-5(16)-2(10)27+4(32)+60=0
C=45

Forces in y =D+4+45-20-18-5=0
D=-6

Note (the new pictures going to be sideways. I can't figure out how to fix it on my iPad.) sorry!
You should show the negative 6 support reaction at D as a downward force of magnitude 6. Otherwise your reactions, shear, and moment diagrams are literally perfect, correctly curved, etc. , this is excellent work.
 
Thank you for all the help! Preparing for finals.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 6 ·
Replies
6
Views
9K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 12 ·
Replies
12
Views
10K