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 0.24 mm the values of elastic modulus for steel is 221 MPa. Take H1 = 4 cm, H2 = 9 cm, top bar (R) is 2 x 2 cm and bottom bar (Q) is 6 x 6 cm. Also, determine the stress-induced in each section of the bar. %3D Solution: R Cm X R Cm Steel bar i) Magnitude Force, P = H1 cm ii) Stress-induced in the top bar = Q cm x Qcm Steel bar H2 cm iii) Stress-induced in the bottom bar =

Elements Of Electromagnetics
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ISBN:9780190698614
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 0.24 mm the values of elastic modulus for steel is 221 MPa. Take H1 = 4 cm, H2
is 2 x 2 cm and bottom bar (Q) is 6 x 6 cm. Also, determine the stress-induced in each section of the
bar.
= 9 cm, top bar (R)
IP
Solution:
R cm x RCm
Steel bar
i) Magnitude Force, P =
H1 cm
ii) Stress-induced in the top bar =
Q cm x Q cm
Steel bar
H2 cm
iii) Stress-induced in the bottom bar =
Transcribed Image Text: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 0.24 mm the values of elastic modulus for steel is 221 MPa. Take H1 = 4 cm, H2 is 2 x 2 cm and bottom bar (Q) is 6 x 6 cm. Also, determine the stress-induced in each section of the bar. = 9 cm, top bar (R) IP Solution: R cm x RCm Steel bar i) Magnitude Force, P = H1 cm ii) Stress-induced in the top bar = Q cm x Q cm Steel bar H2 cm iii) Stress-induced in the bottom bar =
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