Question 6.4: A simply supported beam is subjected to a linearly varying l...
A simply supported beam is subjected to a linearly varying load having maximum intensity of w N/m at the extreme right end. If the beam has a wide-flange I-section¹ as shown in Figure 6.15, find out the loadintensity that may be applied provided the working stress is not to exceed 125 MPa in either tension or compression. Neglect self-weight of the beam.
Let us first draw the free-body diagram for the beam as shown in Figure 6.16.
Taking moment about B, we get R_{ A }=w N . Taking a section D–D at a distance x from left end [Figure 6.16(a)] and considering free-body diagram of the left side as in Figure 6.16(b), we can write for equilibrium:
V_x=w-\frac{1}{2}\left\lgroup\frac{x}{6} \right\rgroup w x=w-\frac{w x^2}{12} (1)
From Eq. (1), we can put V_x=0 to get x where shear force is 0 and consequently, moment is maximum.
V_x=0=w-\frac{w x^2}{12}
or x=\sqrt{12}=3.46 m
The general equation for moment is given by
M_x=w x-\frac{1}{2}\left\lgroup \frac{x}{6} \right\rgroup \frac{w}{3} \times x^2
or M_x=w x-\frac{w x^3}{36} Nm (2)
Putting x = 3.46 m in Eq. (2), we obtain M_x=M_{\max }=2.31 w Nm . Now, bending stress for the wide flange section is given by
\sigma=\frac{M y}{I}
where
I=\frac{150(250)^3}{12}-2\left[\frac{65(210)^3}{12}\right]=95 \times 10^6 mm ^4
Thus, the maximum stress occurs at y = ±125 mm as the neutral axis passes halfway through the section. Hence,
\sigma_{\max }=\frac{2.31 \times w \times 0.125}{95 \times 10^6 \times 10^{-12}} N / m ^2 \quad\left[\text { as } y=\frac{250}{2 \times 1000} m =0.125 m \right]
Equating this with working stress, we can write
125 \times 10^6=\frac{2.31 w \times 0.125}{95 \times 10^6 \times 10^{-12}}
or w = 41 kN/m
¹Quite frequently, these sections along with the other sections like ‘T’, ‘C’, Angle (‘L’) are used as beam sections. More popular use of such sections are done in engineering practice.
