Answer to Question #227858 in Chemical Engineering for Gideon Etison

Question

a short reinforced concrete column of section 200mm ×220mm is to be reinforce with 4 number steel bars and is required to carry n axial load of 850kn. the stress in the concrete must not exceed 7n/mm^2 and the stress in the steel must not exceed 150n/mm^2. determine the diameter required for the bars and the subsequent stresses occuring in the concrete and the steel under the specified load.[ young's modulus; concrete =14kn/mm^2, steel =210kn/mm^2]

Answer 1

$\delta_{st}=\delta_{co}=\delta$

$(\frac{F\cdot l}{S\cdot E})_{co}=(\frac{F\cdot l}{S\cdot E})_{st}$

$\frac{\sigma_{co}\cdot l}{14\cdot10^9}=\frac{\sigma_{st}\cdot l}{210\cdot10^9}$

$14\cdot \sigma_{st}=210\cdot \sigma_{co}$

When $\sigma_{st}=150\cdot 10^6 Pa$

$14\cdot 150\cdot 10^6=210\cdot \sigma_{co} \to \sigma_{co}=10\cdot 10^6 Pa>6\cdot 10^6 Pa$ (not okay!)

When $\sigma_{co}=6\cdot 10^6 Pa$

$14\cdot \sigma_{st}=210\cdot 6\cdot 10^6 \to \sigma_{st}=90\cdot 10^6 Pa<150\cdot 10^6 Pa$ (okay!)

Use $\sigma_{st}=90\cdot 10^6 Pa; \sigma_{co}=6\cdot 10^6 Pa$

$F_{st}+F_{co}=850000$

$\sigma_{st}\cdot S_{st}+\sigma_{co}\cdot S_{co}=850000$

$90\cdot 10^6\cdot S_{st}+6\cdot 10^6\cdot(0.2\cdot 0.22-S_{st})=850000$

$84\cdot S_{st}=0.85-0.264 \to S_{st}=0.006976 m^2=6976m^2$

For one steel bar

$S_{0st}=6976/4=1744 mm^2$

$\frac{\pi\cdot D^2}{4}=1744 \to D\approx 47.1 mm$