Question 6.8: A simply supported beam made of rolled steel joist (I-sectio...

(a) Let us show the beam section in Figure 6.23.

Clearly, \bar{I} is the second moment of area about neutral axis and is found as

\bar{I}=35060(10)^4+2\left[\frac{1}{12}(300)(20)^3+(300)(20)(225+10)^2\right]=1.0137 \times 10^9  mm ^4

For the beam with point load at the midspan, we know that M_{\max }=w L / 4 Therefore, maximum bending stress, \sigma_{\max } is

\sigma_{\max }=\frac{M_{\max }}{Z}

so,    125=\frac{(w)(5000)}{4} \times \frac{25}{1.0137\left(10^9\right)}

or          W = 413.76 kN

(b) Let us show the beam with plates riveted on the flanges in Figure 6.24. Let a be the minimum length of 300 mm plates riveted to the beam.

In Figure 6.24(a), the beam with two numbers of 300 mm × 20 mm plates riveted is shown; while in Figure 6.24(b), the bending moment diagram of the beam is depicted. Clearly,

M_{\max }=\sigma_{\max } Z=125 \times \frac{1.0137 \times 10^9}{245}=517.19 \times 10^6  N  mm

If the maximum bending stress in the flanges of the joist is to be limited to 125 MPa, then it corresponds to the bending moment M, which is given by

M=\sigma Z=125 \times \frac{35060 \times 10^4}{225}=194.78 \times 10^6  N  mm

Thus, in the base joist (i.e., without reinforcement by the plates), the bending moment is to be limited to 194.78 \left(10^6\right) N mm. Therefore, if a is the minimum length of two numbers of 300 mm × 20 mm plates, then from the bending moment diagram, we have

\frac{a / 2}{2.5}=\frac{517.19-194.78}{517.19}

or            a = 1.56 m

Thus, the minimum length of the plates is 1.56 m.