The steel bar has a 20 x 10 mm rectangular cross section and is welded along section a-a. The weld material has a tensile yield
strength of 325 MPa and a shear yield strength of 200 MPa, and the bar material has a tensile yield strength of 350 MPa. An overall factor of safety of at least 2.0 is required. Find the largest load P that can be applied, to satisfy all criteria.
1 answer:
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Answer:
a). Work transfer = 527.2 kJ
b). Heat Transfer = 197.7 kJ
Explanation:
Given:
= 5 Mpa
= 1623°C
= 1896 K
= 0.05
Also given
Therefore, = 1
R = 0.27 kJ / kg-K
= 0.8 kJ / kg-K
Also given :
Therefore, =
= 0.1182 MPa
a). Work transfer, δW =
= 527200 J
= 527.200 kJ
b). From 1st law of thermodynamics,
Heat transfer, δQ = ΔU+δW
=
=
=
= 197.7 kJ