Answer
$$\text{stress of steel}\quad \sigma_{s} =-8.34 \mathrm{ksi}$$
$$\text{stress of cement}\quad \sigma_{c} =-1.208 \mathrm{ksi} $$
Work Step by Step
$\text{we start solution by Let}\ \\\quad P_{c}\text { = portion of axial force stand by concrete. }$
$\quad \quad P_{s}=\text { portion stand by the six steel rods. }$
$$\delta=\frac{P_{c} L}{E_{c} A_{c}} \quad P_{c}=\frac{E_{c} A_{c} \delta}{L}$$
$$\delta=\frac{P_{s} L}{E_{s} A_{s}} \quad P_{s}=\frac{E_{s} A_{s} \delta}{L}$$
$$P=P_{c}+P_{s}=\left(E_{c} A_{c}+E_{s} A_{s}\right) \frac{\delta}{L}$$
$$ \varepsilon =\frac{\delta}{L}=\frac{-P}{E_{c} A_{c}+E_{s} A_{s}} $$
$$ A_{s} =6 \frac{\pi}{4} d_{s}^{2}=\frac{6 \pi}{4}(1.125 \mathrm{in} .)^{2}=5.9641 \mathrm{in}^{2} $$
$$ A_{c} =\frac{\pi}{4} d_{c}^{2}-A_{s}=\frac{\pi}{4}(18 \mathrm{in} .)^{2}-5.9641 \mathrm{in}^{2} $$
$$ =248.51 \mathrm{in}^{2} $$
$$ L =4.5 \mathrm{ft}=54 \mathrm{in} $$
$$ \varepsilon =\frac{-350 \times 10^{3} \mathrm{lb}}{\left(4.2 \times 10^{6} \mathrm{psi}\right)\left(248.51 \mathrm{in}^{2}\right)+\left(29 \times 10^{6} \mathrm{psi}\right)\left(5.9641 \mathrm{in}^{2}\right)}=-2.8767 \times 10^{-4} $$
$$\sigma_{s} =E_{s} \varepsilon=\left(29 \times 10^{6} \mathrm{psi}\right)\left(-2.8767 \times 10^{-4}\right)=-8.3424 \times 10^{3} \mathrm{psi} $$
$$\therefore \sigma_{s} =-8.34 \mathrm{ksi}$$
$$ \sigma_{c} =E_{c} \varepsilon=\left(4.2 \times 10^{6} \mathrm{psi}\right)\left(-2.8767 \times 10^{-4}\right)=1.20821 \times 10^{3} \mathrm{psi}$$
$$\therefore \sigma_{c} =-1.208 \mathrm{ksi} $$
