Problem 5-7

Determine the reactions at the support and draw the diagrams for shear force and bending moment due to the imposed load on the compound beam shown in figure 5-7(a). Support at A is roller, B is internal pin and D is fixed support.

Figure 5-7(a)

Solution:

The above compound beam has one internal hinge at point B which connects two parts AB and BD. First we draw the free body diagram for each part of the compound beam as given below. Roller at A will have only one vertical reaction V_A, the hinge at B will have two reactions (horizontal H_B, Vertical V_B) and the fixed support will have the reactions H_D, V_D and M_D.

Figure 5-7(b)

Now we calculate the support reactions of beam AB. It is very clear that beam AB can be treated as simply supported beam having three unknown reactions which can be easily calculated by applying the equations of static equilibrium. The horizontal reaction at B i.e. H_B will be zero because there is no horizontal force acting on the beam.

Consider the vertical equilibrium of AB. The two vertical reactions will be equal as the uniform load is symmetrically applied on AB. Therefore V_A = V_B = 2 x 3/2 = 3 kN acting vertically upward. The internal hinge at B will transfer the reaction V_B at B in the opposite direction on beam BD.

Now we consider the equilibrium of beam BD and calculate the unknown reactions at the support D. there is no horizontal force on beam BD also.

Σ V = 0; (equilibrium in vertical direction)

V_D - 3 - 4 = 0; Therefore V_D = 7 kN

Σ M_D = 0; (considering rotational equilibrium)

- M_D + 3 x 3 + 4 x 1 = 0. Therefore M_D = 13 kNm.

Now we can easily draw the SFD and BMD of each part as shown in the figure 5-7(c) and figure 5-7(d) given below.

Figure 5-7(c)

We follow the standard sign convention of shear force and bending moment. We advise you to see our solved examples Problem 5-1 and Problem 5-2 for more explanation about different steps of calculation of shear force and bending moment and plotting the diagrams. Beam AB is simply supported with uniform load on full span. The reactions will be 3 kN each and the S.F. will be +3 kN at A and -3 kN at B with zero S.F. at the centre. It will be a straight line from A to B. The B.M. will have zero values at support A and B and maximum value at the centre which will be equal to M_max = w L² / 8 = 2 x 3² / 8 = 2.25 kNm.

The B.M. will be sagging in nature and it is plotted positive. The BMD will be parabolic from A to B as shown in the figure shown above.

Now consider beam BD. Beam BD is a cantilever with B as the free end and D as the fixed end. The S.F. value will be = - 3 kN from B to C and it will be = - 3 - 4 = - 7 kN from C to D. The SFD is drawn in the figure 5-7(d). B.M. at B will be zero; B.M. at C will be = - 3 x 1 = - 3 kNm;

B.M.at D will be = - 3 x 2 - 4 x 1 = - 10 kNm. The B.M. will be hogging in nature. Hogging moments are plotted as negative. The BMD is plotted in figure 5-7(d).

Figure 5-7(d)

Please refer to our page for more explanation about the nature of curve for bending moment diagram

You can visit the following links of solved examples on Bending moment and shear force calculations and plotting of diagrams

Example 5-2

Example 5-3

Exmple 5-1

Example 3-1 Calculation of Truss member forces

You can also use following calculators for Bending moment and shear force

Bending moment calculator for Cantilever

Bending Moment calculator for Simply supported beam

Bending Moment calculator for Overhanging beam

Bending Moment calculator for Fixed beam

More Solved Examples
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