A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of Syt = 60 kpsi and Syc = 75 kpsi. Using
the ductile Coulomb-Mohr theory, determine the factor of safety for the states of plane stress.
1 answer:
Answer:
2.135
Explanation:
Lets make use of these variables
Ox 16.5 kpsi, and Oy --14,5 kpsi
To determine the factor of safety for the states of plane stress. We have to first understand the concept of Coulomb-Mohr theory.
Mohr–Coulomb theory is a mathematical model describing the response of brittle materials such as concrete, or rubble piles, to shear stress as well as normal stress.
Please refer to attachment for the step by step solution.
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The asymptotes of the open loop transfer are:
- Horizontal: y = 0
- Vertical: x = -10 and x = -100
The open loop transfer function is given as:
f(s) = 100(s + 1)/((s + 10)(s + 100))
Set the numerator of the function to 0.
So, we have:
f(s) = 0/((s + 10)(s + 100))
Evaluate
f(s) = 0
This means that, the vertical asymptote is y = 0
Set the denominator of the function to 0.
(s + 10)(s + 100) 0
Split
s + 10 = 0 and s + 100 = 0
Solve for s
s = -10 and s = -100
This means that, the horizontal asymptotes are s = -10 and s = -100
See attachment for the graph of the asymptotes
Read more about asymptotes at:
#SPJ1
