Question 14.4: Figure Ex. 14.4 gives a foundation beam with the vertical lo...

The steps to be followed are:

1 . Determine the resultant vertical force R of the applied loadings and its eccentricity with respect to the centers of the beam.

2. Determine the maximum and minimum base pressures.

3. Draw the shear and bending moment diagrams.

R = 320 + 400 + 16 x 8 = 848 kN.

Taking the moment about the right hand edge of the beam, we have,

or x=\frac{2992}{848}=3.528 m

e=4.0-3.528=0.472 rn to the right of center of the beam. Now from Eqs 12.39(a) and (b), using e_{y}=0,

q_{\max }=\frac{Q}{A} 1+6 \frac{e_{x}}{L}+6 \frac{e_{y}}{B} (12.39a)

q_{\min }=\frac{Q}{A} 1-6 \frac{e_{x}}{L}-6 \frac{e_{y}}{B} (12.39b)

Convert the base pressures per unit area to load per unit length of beam.

The maximum vertical load = 0.7 x 205.02 = 143.52 kN/m.

The minimum vertical load = 0.7 x 97.83 – 68.48 kN/m.

The reactive loading distribution is given in Fig. Ex. 14.4(b).

Shear force diagram

Calculation of shear for a typical point such as the reaction point R_{1} (Fig. Ex. 14.4(a)) is explained below.

Consider forces to the left of R_{1} (without 320 kN).

Shear force V = upward shear force equal to the area abed – downward force due to distributed load on beam ab

Consider to the right of reaction point R_{1} (with 320 kN).

V=-320+57.2=-262.8 kN.

In the same away the shear at other points can be calculated. Fig. Ex. 14.4(c) gives the complete shear force diagram.

Bending Moment diagram

Bending moment at the reaction point R_{1}= moment due to force equal to the area abed + moment due to distributed load on beam ab

The moments at other points can be calculated in the same way. The complete moment diagram is given in Fig. Ex. 14.4(d)