written 7.1 years ago by | • modified 3.2 years ago |
Question (contd.) find out the stress produced at extreme fibre of the column section. Take E=200GPa
written 7.1 years ago by | • modified 3.2 years ago |
Question (contd.) find out the stress produced at extreme fibre of the column section. Take E=200GPa
written 7.1 years ago by | • modified 7.0 years ago |
Given data:
D=300mmd=220mml=4mt=40mmLoad(P)=100 kNeccentricity e=40 mmE=200GPa=2×105MPa
To find: Extreme stress
For a hollow circular column
Area of the column,
A=π4(D2−d2)=π4(3002−2202)=32672.56 mm2
Moment of inertai of the column section
I=π64(D4=d4)=π64(3004−2204)=282.62×106mm
Section Modulus (z) =Iy=282.62×106150=1.884×106mm3
Since the column is pinned at both ends
Left = L = 4m =4000mm
Extereme Stress:
Maximum bending moment = P.e sec(12√pEI)
Let us determine the angle
L2√PEI=40002√100×1032×105×282.62×106=0.08 radian=4∘58′sec4∘58′=1.003
maximum BM=100×103×40×1.003
BM=4.01×106N−mm_
maximum compressive Stress Tmax
σmax=Tmax=PA+MZ=1000×10332672.56+4.01×1061.884×106[σmax=5.189 N/mm2]_