Question
The steel pipe is supported on the circular base plate and concrete pedestal. If the thickness of the pipe is $t=5 \mathrm{mm}$ and the base plate has a radius of $150 \mathrm{mm}$ determine the factors of safety against failure of the steel and concrete. The applied force is $500 \mathrm{kN}$, and the normal failure stresses for steel and concrete are $\left(\sigma_{ \text {fail } )_{\text {st }}}=350 \mathrm{MPa}\right.$ and $\left(\sigma_{ \text {fail } )_{\text {con }}}=25$ MPa, respectively. \right.
Step 1
The formula for the area of a circle is $A = \pi r^2$. Given that the radius of the base plate is $150 \, \text{mm}$, we can substitute this into the formula to get the area of the concrete slab. \[A = \pi (150)^2 = 70650 \, \text{mm}^2\] Show more…
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The steel pipe is supported on the circular base plate and concrete pedestal. If the normal failure stress for the steel is $\left(\sigma_{ \text {fail } )_{\text {st }}}=350 \text { MPa, determine the minimum thickness } t\right.$ of the pipe if it supports the force of $500 \mathrm{kN}$. Use a factor of safety against failure of $1.5 .$ Also, find the minimum radius $r$ of the base plate so that the minimum factor of safety against failure of the concrete due to bearing is $2.5 .$ The failure bearing stress for concrete is $\left(\sigma_{\text {fail }}\right)_{\text {con }}=25$ MPa.
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