- #1
freespirit
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Two solid cylinderical rods AB & BC are welded together at B & loaded as shown. Knowing that d(1)=30mm & d(2)=50mm, find the average normal stress in the midsection of (a)rod AB, (b) rod BC.
B____________C
125kN--->| |
A______________| |
60kN<---| | |
| | |
|______________| |
|<- 0.9m------>| |
125kN--->|____________|
|<---1.2m--->|
stress = P/A
(a) A= pie r ^2=pie (30mm/2)^2=.000707 m^2
stress=P/A= (60x10^3 N)/(.000707)=84.9MPa
(b) A= pie r^2)= pie(50mm/2)^2=.001963
stress= P/A= (250x10^3)/.001963=106.1 MPa
the stress for (b) is wrong; it's suppose to be 96.8MPa Can someone show me what I have done wrong and explain why it should be done differently? Thanks
B____________C
125kN--->| |
A______________| |
60kN<---| | |
| | |
|______________| |
|<- 0.9m------>| |
125kN--->|____________|
|<---1.2m--->|
stress = P/A
(a) A= pie r ^2=pie (30mm/2)^2=.000707 m^2
stress=P/A= (60x10^3 N)/(.000707)=84.9MPa
(b) A= pie r^2)= pie(50mm/2)^2=.001963
stress= P/A= (250x10^3)/.001963=106.1 MPa
the stress for (b) is wrong; it's suppose to be 96.8MPa Can someone show me what I have done wrong and explain why it should be done differently? Thanks