Shear force and bending moment

[dss975599]

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 ?

 

[dss975599]

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 ?

 

[CivilSigma]

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]

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 ?

 

[Chestermiller]

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?

 

[CivilSigma]

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.

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.

 

[Chestermiller]

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.

 

[CivilSigma]

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/

 

[dss975599]

@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 ?

 

[dss975599]

@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

 

[dss975599]

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

 

[Chestermiller]

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

 

[dss975599]

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 ?

 

[dss975599]

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)...

 

[Chestermiller]

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

 

[dss975599]

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

 

[Chestermiller]

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)?