Draw diagrams of normal force, shear force and bending moment for the cranked cantilever beam shown in Fig. P.3.5. Insert all principal values.

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Accepted Answer

Again, as in P.3.4, it is unnecessary to calculate the support reactions if forces and moments to the right of any section are considered.

Normal force
There are no horizontal forces applied to the beam so that the normal force in both AB and CD is zero. The normal force in BC is due to the component, resolved parallel to BC, of the force at C produced by the distributed load on CD, that is

[latex] N_{ BC }=+4 imes 1 \cos 60^{\circ}=+2 kN ( ext { tension }) [/latex]

The normal force diagram is therefore as shown in Fig. S.3.5(a).

Shear force
Since there is a uniformly distributed load on CD the shear force distribution will be linear and vary from zero at D to + 4×1=+ 4 kN at C. In BC the shear force is equal to the component, perpendicular to BC, of the load at C from CD, that is

[latex] S_{ BC }=+4 imes 1 \sin 60^{\circ}=+3.46 kN [/latex]

In AB the shear force at any section is given by

[latex] S_{ AB }=+4 imes 1+10=+14 kN [/latex]

The distribution of shear force is then as shown in Fig. S.3.5(b).

Bending moment
The bending moment at any section a distance x, say, from D is given by

[latex] M_{ CD }=-4 \frac{x^{2}}{2} [/latex]

When [latex] x=0, M_{ CD }=0 ext { and when } x=1 m , M_{ CD }=-2 kNm [/latex] and the distribution is parabolic. In BC and AB there are no applied distributed loads so that the bending moment distributions will be linear. Then

[latex]\begin{aligned}&M_{ C }=-2 kNm ( ext { already calculated }) \&M_{ B }=-4 imes 1 imes 1.5=-6 kNm \&M_{ A }=-4 imes 1 imes 2.5-10 imes 1=-20 kNm\end{aligned}[/latex]

The complete distribution is then as shown in Fig. S.3.5(c).

Question 3.5: Draw diagrams of normal force, shear force and bending momen...

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