A shaft is subjected to fluctuating loads for which the normal torque (T) and bending moment (M) are 1000 N-m and 500 N-m respectively. If the combined shock and fatigue factor for bending is 1.5 and combined shock and fatigue factor for torsion is 2, then the equivalent twisting moment for the shaft is
A. 2000 N-m
B. 2050 N-m
C. 2100 N-m
D. 2136 N-m
Answer: Option D
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T₁= √(1.5x500) + (2x1000)² =2136 N-m
We know that equivalent twisting moment
T₁= √√(KXM)² + (K,XT)² Since T = 1000 N-m; M = 500 N-m; K=1.5 and K,= 2, therefore
T₁= √(1.5x500) + (2x1000)² =2136 N-m
THANK YOU
Teq = { (km * M)^2 + (kt * T)^2}^1/2
Sqrt [(2*1000)^2 + (1.5*500)^2].
Need full solutions
How?
Te=[(M*Kf)^2+(T*Kb)^2]^1/2
Explain me