Shear force and bending moment

In summary, the conversation discussed finding the shear force and bending moment diagrams for a load combination. There was confusion about the calculated values for the shear force and bending moment, and the participants discussed potential errors in the calculations. Ultimately, the correct values were determined and the process of finding the area under the shear force diagram to plot the bending moment diagram was explained.
  • #1
31
1

Homework Statement


I was asked to find the shear force diagram and bending moment diagram for this load combination...But , i have problem of getting the BMD now . i am not sure which part is wrong , can anyone point out ?

Homework Equations

The Attempt at a Solution


MA = 5(2)(1/2)(2/3 x 2) + [ (10x4x4) - (10x4x0.5x(2+(2/3)(4)) ] +2(10)
, thus MA = 73.33kNm.
For , VA , i gt (5 x 2/2) + (10x4/2) + 2 = 27kN
I have sketched the SFD as attached , but i have problem of finding the area below the SFD to get moment , how to do this ?
 

Attachments

  • 993.JPG
    993.JPG
    8.3 KB · Views: 567
  • 994.png
    994.png
    39.8 KB · Views: 566
Physics news on Phys.org
  • #2
I doubt that my MA = 5(2)(1/2)(2/3 x 2) + [ (10x4x4) - (10x4x0.5x(2+(2/3)(4)) ] +2(10)
, thus MA = 73.33kNm. is correct or not ?
Can someone help to check ?
 
  • #3
I get

$$ M_a = 5*2 + \frac{5 \cdot 2}{2} \cdot \frac{2}{3}\cdot 2 + \frac{10 \cdot 4}{2} \cdot (2+ \frac{1}{3} \cdot 4) = 83.33$$

and

$$V_a = 27$$
 
Last edited:
  • Like
Likes dss975599
  • #4
CivilSigma said:
I get

$$ M_a = 5*2 + \frac{5 \cdot 2}{2} \cdot \frac{2}{3}\cdot 2 + \frac{10 \cdot 4}{2} \cdot (2+ \frac{1}{3} \cdot 4) = 83.33$$

and

$$V_a = 27$$
why you ignore the moment caused by the 10kN force ?
 
  • #5
There seems to be something wrong with the original figure. If you have a region of 5 kN/m it should be a rectangular region, not a triangle. Same for the 10 kN/m. Or is the 5 kN/m at the peak of the triangle, and the load distribution is non- uniform?
 
  • Like
Likes dss975599
  • #6
dss975599 said:
why you ignore the moment caused by the 10kN force ?

If you mean the distributed 10 kN/m force, then I do not. It is the third value I am adding.

The total force of that triangular distribution is 10*4/2 , and the equivalent force will act at the centroid of the distribution, which is 1/3 * 4 from the left side of the triangle. Finally add 2m to find distance to the fixed support.

Chestermiller said:
There seems to be something wrong with the original figure. If you have a region of 5 kN/m it should be a rectangular region, not a triangle. Same for the 10 kN/m. Or is the 5 kN/m at the peak of the triangle, and the load distribution is non- uniform?
From my experience, the 5 kN/m force represents the peak of the triangle.
 
  • Like
Likes dss975599
  • #7
CivilSigma said:
If you mean the distributed 10 kN/m force, then I do not. It is the third value I am adding.

The total force of that triangular distribution is 10*4/2 , and the equivalent force will act at the centroid of the distribution, which is 1/3 * 4 from the left side of the triangle. Finally add 2m to find distance to the fixed support.From my experience, the 5 kN/m force represents the peak of the triangle.
OK. Then I'll give the problem a shot so we can compare.
 
  • Like
Likes dss975599
  • #8
CivilSigma said:
I get

$$ M_a = 5*2 + \frac{5 \cdot 2}{2} \cdot \frac{2}{3}\cdot 2 + \frac{10 \cdot 4}{2} \cdot (2+ \frac{1}{3} \cdot 4) = 83.33$$

and

$$V_a = 27$$
@dss975599 and @Chestermiller

I found a mistake in my calculation. The first term should be 10 * 2 , not 5*2 as in post #3. This will give a Ma= 93.3 kN*m.
I verified this using an online calculator.

https://skyciv.com/free-beam-calculator/
 
  • Like
Likes dss975599
  • #9
CivilSigma said:
@dss975599 and @Chestermiller

I found a mistake in my calculation. The first term should be 10 * 2 , not 5*2 as in post #3. This will give a Ma= 93.3 kN*m.
I verified this using an online calculator.

https://skyciv.com/free-beam-calculator/
Thanks for your answer , i have obtained the SFD as in the online calculator , but i have no idea to get the area under the SFD to get the BMd. Do you know how to get the area under the SFD so that i can plot BMD based on SFD ?
 
  • #10
@Chestermiller @CivilSigma here's my working ... At x = 6m , i get M = -61kNm , but not -8kNm as provided by the online calculator , which part of my working is wrong ? for x = 0 and x = 2 , i gt the M value same as the online calculator
 

Attachments

  • 995.JPG
    995.JPG
    15.4 KB · Views: 511
  • 996.JPG
    996.JPG
    18.3 KB · Views: 521
  • #11
here's my trying ... Can anyone point out which part of my working is wrong ? I have been looking at this for the whole day
 

Attachments

  • DSC_1433.JPG
    DSC_1433.JPG
    34.3 KB · Views: 508
  • #12
I can't make out what you have written in the photo. But, first of all, for the direction of the moment you have drawn in your original figure, I get +93.33 kNm (in agreement with @CivilSigma) rather than your -93.33. For the shear force, I get the following:

##V=-1.25x^2+27## for (0<x<2)
##V=1.25x^2-15x+47## for (2<x<6)
I integrate to get the moment variation $$M=93.33-\int_0^x{V(x')dx'}$$
This gives me:
##M=93.33-27x+\frac{1.25x^3}{3}## for (0<x<2)
##M=110-47x+7.5x^2-\frac{1.25x^3}{3}## for (2<x<6)

This gives values for M of 42.67 kNm at x = 2 and 8 kNm at x = 6
 
  • Like
Likes dss975599
  • #13
Chestermiller said:
I can't make out what you have written in the photo. But, first of all, for the direction of the moment you have drawn in your original figure, I get +93.33 kNm (in agreement with @CivilSigma) rather than your -93.33. For the shear force, I get the following:

##V=-1.25x^2+27## for (0<x<2)
##V=1.25x^2-15x+47## for (2<x<6)
I integrate to get the moment variation $$M=93.33-\int_0^x{V(x')dx'}$$
This gives me:
##M=93.33-27x+\frac{1.25x^3}{3}## for (0<x<2)
##M=110-47x+7.5x^2-\frac{1.25x^3}{3}## for (2<x<6)

This gives values for M of 42.67 kNm at x = 2 and 8 kNm at x = 6
##V=1.25x^2-15x+47## for (2<x<6)## ,may i know how do you get this ?
 
  • #14
Chestermiller said:
I can't make out what you have written in the photo. But, first of all, for the direction of the moment you have drawn in your original figure, I get +93.33 kNm (in agreement with @CivilSigma) rather than your -93.33. For the shear force, I get the following:

##V=-1.25x^2+27## for (0<x<2)
##V=1.25x^2-15x+47## for (2<x<6)
I integrate to get the moment variation $$M=93.33-\int_0^x{V(x')dx'}$$
This gives me:
##M=93.33-27x+\frac{1.25x^3}{3}## for (0<x<2)
##M=110-47x+7.5x^2-\frac{1.25x^3}{3}## for (2<x<6)

This gives values for M of 42.67 kNm at x = 2 and 8 kNm at x = 6
and how do you get ##M=110-47x+7.5x^2-\frac{1.25x^3}{3}## for (2<x<6) ? By integrating ##V=1.25x^2-15x+47## for (2<x<6) , i gt ##M=0.42x^3-7.5x^2+47x## for (2<x<6)...
 
  • #15
For the distributed loading, I got

##w=2.5x## for (0<x<2)
##w=2.5(6-x)=15-2.5x## for (2<x<6)

For the shear force, I integrated $$\frac{dV}{dx}=-w(x)$$subject to the initial condition V(0)=27

For the moment, I integrated $$\frac{dM}{dx}=-V (x)$$ subject to the initial condition M(0)=93.33
 
  • Like
Likes dss975599 and CivilSigma
  • #16
Chestermiller said:
For the distributed loading, I got

##w=2.5x## for (0<x<2)
##w=2.5(6-x)=15-2.5x## for (2<x<6)

For the shear force, I integrated $$\frac{dV}{dx}=-w(x)$$subject to the initial condition V(0)=27

For the moment, I integrated $$\frac{dM}{dx}=-V (x)$$ subject to the initial condition M(0)=93.33
how do you get 2.5(6-x) ? I don't understand
 
  • #17
dss975599 said:
how do you get 2.5(6-x) ? I don't understand
What is the equation for the straight line passing through the points (2,10) and (6,0)?
 

1. What is shear force and bending moment?

Shear force is the internal force that is perpendicular to the length of an object, while bending moment is the internal force that causes an object to bend or twist.

2. How are shear force and bending moment related?

Shear force and bending moment are related by the formula M = F x d, where M is the bending moment, F is the shear force, and d is the distance from the point of interest to the applied force.

3. What is the significance of shear force and bending moment in structural analysis?

Shear force and bending moment are important factors in understanding how a structure will behave under different loads. They help determine the strength and stability of a structure and are used to design safe and efficient structures.

4. How are shear force and bending moment calculated?

Shear force and bending moment can be calculated using equations of equilibrium and free body diagrams. They can also be determined experimentally through testing.

5. What are some common applications of shear force and bending moment analysis?

Shear force and bending moment analysis are used in a variety of fields, such as civil engineering, mechanical engineering, and materials science. They are commonly used in the design of buildings, bridges, and other structures to ensure their safety and structural integrity.

Suggested for: Shear force and bending moment

Replies
2
Views
703
Replies
2
Views
843
Replies
9
Views
1K
Replies
2
Views
2K
Replies
4
Views
899
Replies
5
Views
894
Replies
2
Views
2K
Replies
5
Views
2K
Replies
4
Views
1K
Replies
3
Views
990
Back
Top