A member is formed by connecting two steel bars shown below. Assuming that the bars are prevented from buckling sideways, calculate the magnitude force "P" that will cause the total length of the member to decrease 0.79 mm the values of elastic modulus for steel is 212 MPa. Take H1 = 7 cm, H2 = 9 cm, top bar (R) is 2 x 2 cm and bottom bar (Q) is 5 x 5 cm. Also, determine the stress-induced in each section of the bar. %3D %3D Solution: R cm x R Cm Steel bar i) Magnitude Force, P = Hi cm ii) Stress-induced in the top bar = Q cm x Qcm Šteel bar H2 cm iii) Stress-induced in the bottom bar =

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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A member is formed by connecting two steel bars shown below. Assuming that the bars
are prevented from buckling sideways, calculate the magnitude force "P" that will cause
the total length of the member to decrease 0.79 mm the values of elastic modulus for
steel is 212 MPa. Take H1 = 7 cm, H2 = 9 cm, top bar (R) is 2 x 2 cm and bottom bar
(Q) is 5 x 5 cm. Also, determine the stress-induced in each section of the bar.
on
Solution:
R cm x RCm
Steel bar
i) Magnitude Force, P =
Hi cm
ii) Stress-induced in the top bar =
Q cm x Qcm
Steel bar
H2 cm
iii) Stress-induced in the bottom bar =
Transcribed Image Text:e-Learning Portal Courses - Reports e-Services Academic Departments ETC - CIMS - A member is formed by connecting two steel bars shown below. Assuming that the bars are prevented from buckling sideways, calculate the magnitude force "P" that will cause the total length of the member to decrease 0.79 mm the values of elastic modulus for steel is 212 MPa. Take H1 = 7 cm, H2 = 9 cm, top bar (R) is 2 x 2 cm and bottom bar (Q) is 5 x 5 cm. Also, determine the stress-induced in each section of the bar. on Solution: R cm x RCm Steel bar i) Magnitude Force, P = Hi cm ii) Stress-induced in the top bar = Q cm x Qcm Steel bar H2 cm iii) Stress-induced in the bottom bar =
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