The modulus of elasticity is 44.4GPa
E<u>xplanation:</u>
Given
original length L1=10cm
New length L2=10.036 cm
Force F=16000N
Stress of the bar σ=Force/Area
Change in length ΔL=L2-L1=10.036-10=0.036 cm
Strain is obtained by dividing the change in length by original length
Strain ε=ΔL/L1
=0.036/10=0.0036
Modulus of elasticity=stress/strain
=σ/ε
Pa
Answer:
a) 0.978
b) 0.9191
c) 1.056
d) 0.849
Explanation:
Given data :
Stiffness of each bolt = 1.0 MN/mm
Stiffness of the members = 2.6 MN/mm per bolt
Bolts are preloaded to 75% of proof strength
The bolts are M6 × 1 class 5.8 with rolled threads
Pmax =60 kN, Pmin = 20kN
<u>a) Determine the yielding factor of safety</u>
------ ( 1 )
Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N
Input the given values into the equation above
equation 1 becomes ( np ) = = 0.978
note : values above are derived values whose solution are not basically part of the required solution hence they are not included
<u>b) Determine the overload factor of safety</u>
------- ( 2 )
Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N
input values into equation 2 above
hence : = 0.9191
<u>C) Determine the factor of safety based on joint separation</u>
Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,
input values into equation above
Hence = 1.056
<u>D) Determine the fatigue factor of safety using the Goodman criterion.</u>
nf = 0.849
attached below is the detailed solution .
