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Knowledge Expert
· commented
· 1 Months ago
∝ = 0°
Shear strain s=cotϕ+tan(ϕ−α)s=cotϕ+tan(ϕ−α)
⇒s=cotϕ+tanϕ⇒s=cotϕ+tanϕ
for minimum value, dsdϕ=0dsdϕ=0
⇒−cosec2ϕ+sec2ϕ=0⇒−1sin2ϕ+1cos2ϕ=0⇒cos2ϕ−sin2ϕ=0⇒cos2ϕ=0⇒ϕ=π4⇒−cosec2ϕ+sec2ϕ=0⇒−1sin2ϕ+1cos2ϕ=0⇒cos2ϕ−sin2ϕ=0⇒cos2ϕ=0⇒ϕ=π4
again d2sdϕ2=−2cosecϕ×(−cosecϕ.cotϕ)+2secϕ×(secϕ.tanϕ)d2sdϕ2=−2cosecϕ×(−cosecϕ.cotϕ)+2secϕ×(secϕ.tanϕ)
at =π4,d2sdϕ2=2cosec2(π4)cot(π4)+2 sec2π4tanπ4=π4,d2sdϕ2=2cosec2(π4)cot(π4)+2 sec2π4tanπ4
= 2 × 2 × 1 + 2 × 2 × 1 = 8 > 0
So, minimum value at ϕ=π4ϕ=π4
then S(min)=cotπ4+tanπ4=1+1=2
Thanks and Regards
Team TopRankers