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A steel rod of cross-sectional areaIm^(2) is acted upon by forces shown in the fig. Determine the total elongation of the bar. (Take Y=2.0xxK=8.33xx10^(-11)N//n^(2)) |
Answer» Solution :The ACTION of forces on each part of rod is SHOWN in fig. We know that the EXTENSION due to external force F is given by `Deltal_(AB)=(FL)/(AY)` `Deltal_(AB)=((60xx10^(3))xx1.5)/(1xx2xx10^(11))=4.5xx10^(-7)m` `Deltal_(BC)=((70xx10^(3))xx1)/(1xx2xx10^(11))=3.5xx10^(-7)m and Deltal_(CD)=((50xx10^(3))xx2)/(1xx2xx10^(11))=5.0xx10^(-7)m` The total extension `Delta=Delta_(AB)+Delta_(BC)+Delta_(CD)` `=4.5xx10^(-7)+3.5xx10^(-7)+5.0xx10^(-7)=13xx10^(-7)m` |
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