Figure Ex. 14.4 gives a foundation beam with the vertical loads and moment acting thereon. The width of the beam is 0.70 m and depth 0.50 m. A uniform load of 16 kN/m (including the weight of the beam) is imposed on the beam. Draw (a) the base pressure distribution, (b) the shear force diagram, and

Question

Figure Ex. 14.4 gives a foundation beam with the vertical loads and moment acting thereon. The width of the beam is 0.70 m and depth 0.50 m. A uniform load of 16 kN/m (including the weight of the beam) is imposed on the beam. Draw (a) the base pressure distribution, (b) the shear force diagram, and (c) the bending moment diagram. The length of the beam is 8 m.

Accepted Answer

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,

[latex]R x=848 x=320 imes 7+400 imes 1+16 imes \frac{8^{2}}{2}-160=2992[/latex]

or [latex]x=\frac{2992}{848}=3.528 m[/latex]

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

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

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

[latex]q_{\min }^{\max }=\frac{\Sigma Q}{A} 1 \pm 6 \frac{e_{x}}{L}=\frac{848}{8 imes 0.7} 1 \pm \frac{6 imes 0.472}{8}=205.02 ext { or } 97.83 kN / m ^{2}[/latex]

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 [latex]R_{1}[/latex] (Fig. Ex. 14.4(a)) is explained below.

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

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

[latex]=\frac{68.48+77.9}{2}-16 imes 1=57.2 kN[/latex]

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

[latex]V=-320+57.2=-262.8 kN[/latex].

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 [latex]R_{1}=[/latex] moment due to force equal to the area abed + moment due to distributed load on beam ab

[latex]\begin{aligned}&=68.48 imes \frac{1}{2}+\frac{9.42}{2} imes \frac{1}{3}-16 imes \frac{1}{2} \&=27.8 kN - m\end{aligned}[/latex]

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

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

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