SHEAR FORCE AND BENDING MOMENT DIAGRAM FOR OVERHANGING BEAM

We were discussing shear force and bending moment diagrams for a simply supported beam with a point load acting at midpoint of the loaded beam and shear force and bending moment diagrams for a simply supported beam with uniform distributed load during our previous posts.

We have also seen shear force and bending moment diagrams for a simply supported beam with an eccentric point load and shear force and bending moment diagrams for a simply supported beam with uniform varying load in our recent post.

Today we will see here the concept to draw shear force and bending moment diagrams for an overhanging beam with the help of this post.

Let us see the following figure, we have one beam AC of length L and beam is resting or supported freely on the supports A and B as displayed in following figure. As we can see from below figure that beam is extended beyond the support B and such type of beam will be termed as overhanging beam.

We have already discussed about the various types of beam in strength of materials and we have studied there that overhanging beam is basically defined as a beam where end portion of beam is extended beyond the supports. In such types of beams, one end or also both ends of the beam might be extended beyond the supports.

Let us consider that distance between two supports is a and length of overhanging portion is b and therefore L= a + b

Let us consider that overhanging beam AC is loaded with uniformly distributed load with rate of loading w per unit length over the entire length as displayed in following figure.

First of all we will remind here 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 R_A and R_B are the reaction forces at end support A and support B respectively and we will use the concept of equilibrium in order to determine the value of these reaction forces.

Before going ahead, we must have to understand here the uniformly distributed loading and we can refer the post different types of load acting on the beam in order to understand the uniformly distributed load.

During determination of the total load, total uniformly distributed load will be converted in to point load by multiplying the rate of loading i.e. w (N/m) with the span of load distribution i.e. L and will be acting over the midpoint of the length of the uniformly load distribution.

ƩF_X =0, ƩF_Y =0, ƩM =0,

R_A + R_B – w. L=0

R_A + R_B = w. L

ƩM_A =0

R_B * a – w. L*(L/2) = 0

R_B * a = w. L²/2

R_B = w L²/2a

R_A = w. L - w L²/2a

R_A = w. L (1-L/2a)

Now we have values of reaction forces at support A and support 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 between A and B at a distance x from end A. Now we will have two portion of the beam AB i.e. left portion and right portion. Let us deal with the left portion here. Let us assume that F_X is shear force at section XX and bending moment is M_X at section XX.

Shear force diagram

As we have assumed here section XX between A and B at a distance x from end A 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. 

F_X = R_A – w. x

Force acting to the left of the section and in upward direction will be considered as positive and force acting to the left of the section but in downward direction will considered as negative 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.

Now we have information here in respect of shear force equation which is following linear equation and therefore we will determine shear force at all critical points i.e. 

Shear force at point A, x= 0

F_A= R_A

F_A= w. L (1-L/2a) = w. L (a-b)/2a

Shear force at point B, x= a

F_B = R_A – w. a

F_B = w. L (a-b)/2a - w. a

F_B = w. L/2- w. L. b/2a – w. a

After calculation, we will have negative value for shear force at support B and therefore we can say that there will be a point between A and B where shear force will be zero and we can secure the value of that point by following equation. 

Let us think that the point is D, where shear force will be zero or F_D=0

R_A – w. x = 0

x = R_A / w

Shear force will be zero at a distance, x = R_A / w from the end A.

Let us consider section XX between B and C at a distance x from the end A. let us determine the value of shear force at support B and C.

Let us also determine the value of x for zero value of shear force, we can determine as mentioned here

F_X = R_A – w. a + R_B – w(x-a)

Shear force at point B, x= a

F_B = R_A – w. a + R_B = w. L – w. a

F_B = w. (L – a)

This value of F_B will be positive as L is greater than a

Shear force at point C, x= L

F_C = R_A – w. a + R_B – w (L-a)

F_C = w. L – w. a – w (L-a)

F_C = 0

Now we have information about shear force at all critical points as determined above and we can draw shear force diagram as displayed here in following figure.

Bending moment diagram

As we have assumed here section XX between A and B at a distance x from end A, therefore let us determine here the equation for bending moment at any section between A and B will be as mentioned here.

M_X will be determined as mentioned here.

M_X = R_A. x – w.x²/2

Bending moment at end A, x = 0

M_A = 0

Bending moment at point D will be maximum as shear force is zero at this point and we can determine the value of maximum bending moment by putting the value of x = R_A/w in above equation of bending moment.

Bending moment at point B, x= a

M_B = R_A. a – w.a²/2

M_B = w. L (a-b)/2– w.a²/2

M_B = w. (a + b) (a-b)/2– w.a²/2

M_B = w. (a² - b²) /2– w.a²/2

M_B = - w. b²/2

Bending moment at point B will be negative.

As we can see here from above equation that bending moment at any section between A and B is following parabolic law and hence we can draw the bending moment diagram as displayed in above figure.

Let us consider section XX between B and C at a distance x from the end A. let us determine the equation for bending moment at any section between B and C.

M_X = R_A. x – w.x²/2 + R_B (x-a)

Bending moment at point C, x= L

M_C = w. L² (a-b)/2a – w.L²/2 + wL² (L-a)/2a

M_C = (w. L² /2a) [a-b + L-a] – w.L²/2

M_C = (w. L² /2a) a – w.L²/2

M_C = w. L² /2 – w.L²/2

M_C = 0

Now we have values for bending moment at each critical point and we can see here from above equation that bending moment at any section between B and C is following parabolic law and hence we can draw the bending moment diagram as displayed in above figure.

Point of contraflexure

Point of contraflexure is basically defined as the point at which bending moment will be zero after changing it sign from negative to positive or from positive to negative.

Let us try to secure the point of contraflexure and we can easily say that it will be between A and B as bending moment is changing its sign from positive to negative between A and B and hence point of contraflexure will be located somewhere between A and B.

0 ₌ R_A. x – w.x²/2

R_A = w. x/2

x/2 = R_A / w

x = 2 (R_A / w)

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. in the category of strength of material.

Also read

Shear force and bending moment diagrams for a simply supported beam with uniform varying load from zero at one end to w per unit length at the other end

Cements and its types

Clinker grinding process in cement manufacturing unit

Clinker storage system