Shear force and bending moment

AI Thread Summary
The discussion revolves around calculating the shear force diagram (SFD) and bending moment diagram (BMD) for a given load combination. The initial calculations for the moment and shear force were questioned, particularly regarding the interpretation of triangular and rectangular load distributions. Participants identified errors in the original calculations and clarified the need to account for the moments caused by distributed loads correctly. The conversation also included methods for integrating the shear force to derive the bending moment, with some users sharing their equations and results for verification. The thread emphasizes the importance of accurately understanding load distributions and their effects on structural analysis.
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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 ?
 

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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 ?
 
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

$$V_a = 27$$
 
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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 ?
 
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?
 
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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.
 
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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.
 
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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/
 
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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
 

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  • #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
 

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  • #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
 
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  • #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
 
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  • #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)?
 

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