Question 12.3: Draw the shear and bending-moment diagrams for the simply su...

From the free-body diagram of the entire beam, we determine the magnitude of the reactions at the supports.

R_{A} = R_{B} = \frac{1}{2} wL

Next, we draw the shear diagram. Close to the end A of the beam, the shear is equal to R_{A}, that is, to \frac{1}{2} wL, as we can check by considering as a free body a very small portion of the beam. Using Eq. (12.6), we then determine the shear V at any distance x from A; we write

V_{D} – V_{C} = -\int_{x_{C}}^{x_{D}} {w  dx}                               (12.6)

V- V_{A} = -\int_{0}^{x} {w  dx} = -wx

V = V_{A} – wx = \frac{1}{2} wL – wx = w(\frac{1}{2} L – x)

The shear curve is thus an oblique straight line which crosses the x axis at x = L/2 (Fig. 12.13a). Considering, now, the bending moment, we first observe that M_{A} = 0. The value M of the bending moment at any distance x from A may then be obtained from Eq. (12.8); we have

M_{D} – M_{C} =\int_{x_{C}}^{x_{D}} {V  dx}                               (12.8)

M – M_{A} =\int_{0}^{x} {V  dx}

M = \int_{0}^{x} { w(\frac{1}{2} L – x)  dx}= \frac{1}{2} w(L x – x^{2})

The bending-moment curve is a parabola. The maximum value of the bending moment occurs when x = L/2, since V (and thus dM/dx) is zero for that value of x. Substituting x = L/2 in the last equation, we obtain M_{max} = wL^{2}/8 (Fig. 12.13b).