Shear force and Bending Moment diagram for cantilever

Problem 5-1

Determine the values and draw the diagrams for shear force and bending moment due to the imposed load on cantilever shown in figure 5-1(a)

Figure 5-1(a)

Solution:

The cantilever is a beam which has one end free and the other is fixed. All the reaction components will be experienced only on the fixed end. In the case of cantilever shown in figure 5-1(a), there is no horizontal force and hence the fixed support will have only two reaction components (one vertical Dy and the other moment MD) as shown in the free body diagram in figure 5-1(b).

The support reactions of cantilever can be easily determined by applying the equations of static equilibrium as follows;

ΣFy = 0;   (equilibrium in y-direction)

Dy - 2 - 4 x 2 = 0;

Therefore Dy = 10 kN;

Considering z-axis passing through D and taking moment of all the forces about point D (clock-wise -ve, anticlock-wise +ve).

ΣMz = 0;

2 x 4 + 4 x 2 x (1+1) - MD = 0;

Which yields, MD = 24 kNm

Figure 5-1(b)

Shearing force calculations

In case of point load acting at a point, we should calculate shear force on both sides (left and right) of the point.

FA -left = 0;

FA -right = -2kN;

FB = -2 kN;

FC = -10 kN;

FD = -10 kN;

The shear force diagram (SFD) is drawn in figure 5-1(b) which shows constant value of shear force from A to B and thereafter it increases with a constant rate upto C and then from C to D it is again constant.

Bending moment calculations

MA = 0;

MB =  -2 x 1 =  -2 kNm;

MC = -2 x 3 - 4 x 2 x 1 =  -14 kNm;

MD = -2 x 4 - 4 x 2 x (1+1) = -24 kNm;

The bending moment diagram (BMD) is drawn in figure 5-1(b) which shows that it is straight line from A to B, parabolic from B to C and again straight line from C to D. Throughout the span the bending moment is hogging in nature. The maximum value of bending moment is at the fixed support and equal to 24 kNm.

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