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

Question

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.

Accepted Answer

Let us first draw the free-body diagram for the beam as shown in Figure 6.16.

Taking moment about B, we get [latex] R_{ A }=w N [/latex]. 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:

[latex] V_x=w-\frac{1}{2}\left\lgroup\frac{x}{6} ight group w x=w-\frac{w x^2}{12} [/latex] (1)

From Eq. (1), we can put [latex] V_x=0 [/latex] to get x where shear force is 0 and consequently, moment is maximum.

[latex] V_x=0=w-\frac{w x^2}{12} [/latex]

or [latex] x=\sqrt{12}=3.46 m [/latex]

The general equation for moment is given by

[latex] M_x=w x-\frac{1}{2}\left\lgroup \frac{x}{6} ight group \frac{w}{3} imes x^2 [/latex]

or [latex] M_x=w x-\frac{w x^3}{36} Nm [/latex] (2)

Putting x = 3.46 m in Eq. (2), we obtain [latex] M_x=M_{\max }=2.31 w Nm [/latex]. Now, bending stress for the wide flange section is given by

[latex] \sigma=\frac{M y}{I} [/latex]

where

[latex] I=\frac{150(250)^3}{12}-2\left[\frac{65(210)^3}{12} ight]=95 imes 10^6 mm ^4 [/latex]

Thus, the maximum stress occurs at y = ±125 mm as the neutral axis passes halfway through the section. Hence,

[latex] \sigma_{\max }=\frac{2.31 imes w imes 0.125}{95 imes 10^6 imes 10^{-12}} N / m ^2 \quad\left[ ext { as } y=\frac{250}{2 imes 1000} m =0.125 m ight] [/latex]

Equating this with working stress, we can write

[latex] 125 imes 10^6=\frac{2.31 w imes 0.125}{95 imes 10^6 imes 10^{-12}} [/latex]

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.

Question 6.4: A simply supported beam is subjected to a linearly varying l...

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