A Steel wire hangs vertically under its weight. What is the greatest length it can have if the allowable tensile stress o, =200 kN/m3.(ans:-2500 m) Solution: MPa? Density of steel y=80 O; =200 MPa= 200*103 kN/m² ; L y=80 kN/m³. Wt. of wire P=(Tt/4)(D²)(L)* y c/s area of wire A=(t/4)*D² Of = PIA solving above eq. L=2500m
A Steel wire hangs vertically under its weight. What is the greatest length it can have if the allowable tensile stress o, =200 kN/m3.(ans:-2500 m) Solution: MPa? Density of steel y=80 O; =200 MPa= 200*103 kN/m² ; L y=80 kN/m³. Wt. of wire P=(Tt/4)(D²)(L)* y c/s area of wire A=(t/4)*D² Of = PIA solving above eq. L=2500m
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![Example: 2
A Steel wire hangs vertically under its weight. What is
the greatest length it can have if the allowable tensile
Density of steel y=80
stress
o, =200 MPa?
kN/m3.(ans:-2500 m)
Solution:
0, =200 MPa= 200*103 kN/m² :
L
y=80 kN/m³.
Wt. of wire P=(t/4)(D²)(L)* y
c/s area of wire A=(t/4)*D²
Of = PIA
solving above eq. L=2500m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F25353e16-904b-46a0-b1c3-5114be9b80d3%2F77b02eb6-bf71-43c8-838f-9e728e08757f%2Fp78zb71_processed.png&w=3840&q=75)
Transcribed Image Text:Example: 2
A Steel wire hangs vertically under its weight. What is
the greatest length it can have if the allowable tensile
Density of steel y=80
stress
o, =200 MPa?
kN/m3.(ans:-2500 m)
Solution:
0, =200 MPa= 200*103 kN/m² :
L
y=80 kN/m³.
Wt. of wire P=(t/4)(D²)(L)* y
c/s area of wire A=(t/4)*D²
Of = PIA
solving above eq. L=2500m
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