Question 18.2: Determine the shape factor for the I-section beam shown in F...

Again, as in Ex. 18.1, the elastic and plastic neutral axes coincide with the centroid, \mathrm{G}, of the section.

In the fully plastic condition the stress distribution in the beam is that shown in Fig. 18.5(b). The total direct force in the upper flange is

\sigma_{\mathrm{Y}} b t_{\mathrm{f}} \quad(\text { compression) }

and its moment about \mathrm{G} z is

\sigma_{\mathrm{Y}} b t_{\mathrm{f}}\left(\frac{d}{2}-\frac{t_{\mathrm{f}}}{2}\right) \equiv \frac{\sigma_{\mathrm{Y}} b t_{\mathrm{f}}}{2}\left(d-t_{\mathrm{f}}\right)        (i)

Similarly the total direct force in the web above \mathrm{G} z is

\sigma_{\mathrm{Y}} t_{\mathrm{W}}\left(\frac{d}{2}-t_{\mathrm{f}}\right) \quad \text { (compression) }

and its moment about \mathrm{G} z is

\sigma_{\mathrm{Y}} t_{\mathrm{w}}\left(\frac{d}{2}-t_{\mathrm{f}}\right) \frac{1}{2}\left(\frac{d}{2}-t_{\mathrm{f}}\right) \equiv \frac{\sigma_{\mathrm{Y}} t_{\mathrm{w}}}{8}\left(d-2 t_{\mathrm{f}}\right)^{2}      (ii)

The lower half of the section is in tension and contributes the same moment about \mathrm{G} z so that the total plastic moment, M_{\mathrm{P}}, of the complete section is given by

M_{\mathrm{P}}=\sigma_{\mathrm{Y}}\left[b t_{\mathrm{f}}\left(d-t_{\mathrm{f}}\right)+\frac{1}{4} t_{\mathrm{w}}\left(d-2 t_{\mathrm{f}}\right)^{2}\right]     (iii)

Comparing Eqs (18.5)

M_{\mathrm{P}}=\sigma_{\mathrm{Y}}Z_{\mathrm{P}}        (18.5)

and (iii) we see that Z_{\mathrm{P}} is given by

Z_{\mathrm{P}}=b t_{\mathrm{f}}\left(d-t_{\mathrm{f}}\right)+\frac{1}{4} t_{\mathrm{W}}\left(d-2 t_{\mathrm{f}}\right)^{2}      (iv)

Alternatively we could have obtained Z_{\mathrm{P}} from Eq. (18.6).

Z_{\mathrm{P}}={\frac{A({{{\bar{y}_{1}}}}+{{{\bar{y}_{2}}}})}{2}}      (18.6)

The second moment of area, I, of the section about the common neutral axis is

I=\frac{b d^{3}}{12}-\frac{\left(b-t_{\mathrm{w}}\right)\left(d-2 t_{\mathrm{f}}\right)^{3}}{12}

so that the elastic modulus Z_{\mathrm{e}} is given by

Z_{\mathrm{e}}=\frac{I}{d / 2}=\frac{2}{d}\left[\frac{b d^{3}}{12}-\frac{\left(b-t_{\mathrm{w}}\right)\left(d-2 t_{\mathrm{f}}\right)^{3}}{12}\right]       (v)

Substituting the actual values of the dimensions of the section in Eqs (iv) and (v) we obtain

Z_{\mathrm{P}}=150 \times 12(300-12)+\frac{1}{4} \times 8(300-2 \times 12)^{2}=6.7 \times 10^{5} \mathrm{~mm}^{3}

and

Z_{\mathrm{e}}=\frac{2}{300}\left[\frac{150 \times 300^{3}}{12}-\frac{(150-8)(300-24)^{3}}{12}\right]=5.9 \times 10^{5} \mathrm{~mm}^{3}

Therefore from Eq. (18.7)

f={\frac{M_{\mathrm{P}}}{M_{\mathrm{Y}}}}={\frac{\sigma _{\mathrm{Y}}Z_{\mathrm{P}}}{\sigma_{\mathrm{Y}}Z_{\mathrm{e}}}}={\frac{Z_{\mathrm{P}}}{Z_{\mathrm{e}}}}      (18.7)

f=\frac{M_{\mathrm{P}}}{M_{\mathrm{Y}}}=\frac{Z_{\mathrm{P}}}{Z_{\mathrm{e}}}=\frac{6.7 \times 10^{5}}{5.9 \times 10^{5}}=1.14

and we see that the fully plastic moment is only 14 \% greater than the moment at initial yielding.