Wednesday, 5 April 2017

SHEAR FORCE AND BENDING MOMENT DIAGRAM FOR SIMPLY SUPPORTED BEAM WITH AN ECCENTRIC POINT LOAD

SHEAR FORCE AND BENDING MOMENT DIAGRAM FOR SIMPLY SUPPORTED BEAM WITH AN ECCENTRIC POINT LOAD


Today we will see here the concept to draw shear force and bending moment diagrams for a simply supported beam with an eccentric point load with the help of this post.

 
Let us see the following figure, we have one beam AB of length L and beam is resting or supported freely on the supports at its both ends. And therefore, displayed beam indicates the simply supported beam AB and length L.

Let us consider that one point load W is acting over the beam at distance of a from the end A or at a distance of b from the end B as displayed in following figure. Load W is acting at point C which is not the midpoint of the loaded beam AB, but also point C is at distance of a from the end A or at a distance of b from the end B.

First of all we will remind the important points for drawing shear force and bending moment diagram. Now always remember that we will have to first determine the reaction forces at each support.

Let us consider that RA and RB are the reaction forces at end support A and end support B respectively and we will use concept of equilibrium in order to determine the value of these reaction forces.

ƩFX =0, ƩFY =0, ƩM =0,
RA + RB –W =0
RA + RB = W
ƩMB =0
RA * L – W*b = 0
RA =Wb/L

ƩMA =0
W*a- RB*L = 0
RB = Wa/L

Now we have values of reaction forces at end A and end B and it is as mentioned above. Let us determine now the value of shear force and bending moment at all critical points.

 
Let us consider one section XX at a distance x from the end A between A and C as displayed in following figure. Let us assume that FX is shear force at section XX and bending moment is MX at section XX.

Shear force diagram

As we have assumed here section XX at a distance x from end A between A and C and therefore loaded beam AB will be divided in two portion and let us consider the left portion of the beam. Shear force at section XX will be equivalent to the resultant of forces acting on beam to the left side of the section.
FX = RA
FX = Wb/L

Shear force will be positive here and we can refer the post sign conventions for shear force and bending moment in order to understand the sign of shear force which is determined here.

As we have already discussed that shear force will be constant between two vertical loads or we can also say that if there is no load between two points then shear force will be constant and will be represented by a horizontal line.

Therefore we can say here that shear force between A and C will be constant and it will have positive value of Wb/L.

Now suppose if we have considered section XX between C and B and at a distance x from the end A. Again we will have to remind the basics of shear force and bending moment diagrams and we will be able to calculate shear force at section XX.

FX= Wb/L - W
FX= W (b/L-1) = W (b-L/L)
FX= W (-a)/L
FX= -Wa/L

Therefore, we can say here that shear force between C and B will be constant and it will have negative value i.e. -Wa/L. We will have to note it here that shear force at point C is changing from + Wb/L to – Wa/L.

Now we have information here in respect of shear force at all critical points i.e.
Shear force at point A, FA= + Wb/L
Shear force at point B, FB = -Wa/L
Shear force at point C, FC = Shear force changes from + Wb/L to – Wa/L.

 
We can now draw here shear force diagram and it is displayed here in following figure.

Bending moment diagram

As we have considered above one section XX at a distance x from the end A between A and C and hence bending moment at section XX i.e. MX will be determined as mentioned here.
MX = RA *x = Wb*x/L
MX = Wbx/L

Bending moment at point A, x=0
MA = 0

Bending moment at point C, x=a
MC = Wab/L

Now suppose if we have considered section XX between C and B and at a distance x from the end A. Again we will have to remind the basics of shear force and bending moment diagrams and we will be able to calculate bending moment equation for the section XX and it could be written as mentioned here.

MX = RA*x –W (x-a)
MX = Wbx/L –W (x-a)

Bending moment at point C, x=a
MC = Wab/L

Bending moment at point B, x=L
MB = Wb –W (L-a)
MB = Wb –W (a+b-a)
MB = Wb –Wb = 0
MB = 0

Therefore we have bending moment for all critical points and we have also information about the equation followed by bending moment i.e. bending moment equation stated above is following the linear equation and therefore we can draw here the BMD i.e. bending moment diagram and we have drawn it as displayed in above figure.

Do you have any suggestions? Please write in comment box

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google

We will see another important topic i.e. shear force and bending moment diagrams for a simply supported beam with uniform distributed load in the category of strength of material. 

Also read

Archivo del blog

Copyright © ENGINEERING MADE EASY | Powered by Blogger | Designed by Dapinder