1) (35 pts) For the curved lifting bar, calculate the stresses at A, B, and the centroid of the section.
1 answer:
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Answer:
critical stress required for the propagation is 27.396615 × N/m²
Explanation:
given data
specific surface energy = 0.90 J/m²
modulus of elasticity E = 393 GPa = 393 × N/m²
internal crack length = 0.6 mm
to find out
critical stress required for the propagation
solution
we will apply here critical stress formula for propagation of internal crack
( σc ) = .....................1
here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 × m
so now put value in equation 1 we get
( σc ) =
( σc ) =
( σc ) = 27.396615 × N/m²
so critical stress required for the propagation is 27.396615 × N/m²
Answer:
Final length= 746.175 mm
Explanation:
Given that Length of aluminium at 223 C is 750 mm.As we know that when temperature of material increases or decreases then dimensions of material also increases or decreases respectively with temperature.
Here temperature of aluminium decreases so the final length of aluminium decreases .
As we know that
Now by putting the values
ΔL=3.82 mm
So final length =750-3.82 mm
Final length= 746.175 mm