Strength of Materials Lecture-1 Example 1.2 Rigid bar ABC is supported by a pin at A and axial member (1), which has a cross-sectional area of 540 mm2. The weight of rigid bar ABC can be neglected. (Note: 1 kN = 1,000 N.) > Determine the normal stress in member (1) if a load of P = 8 kN is applied at C. > If the maximum normal stress in member (1) must be limited to 50 MPa, what is the maximum load magnitude P that may be applied to the rigid bar at C? (1) 1.6 m (1) 1.6 m 22m 22m Free-body diagram of rigid bar ABC.

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Strength of Materials
Lecture-1
Example 1.2
Rigid bar ABC is supported by a pin at A and axial member (1), which has a cross-sectional area
of 540 mm2. The weight of rigid bar ABC can be neglected. (Note: 1 kN = 1,000 N.)
> Determine the normal stress in member (1) if a load of P = 8 kN is applied at C.
> If the maximum normal stress in member (1) must be limited to 50 MPa, what is the
maximum load magnitude P that may be applied to the rigid bar at C?
(1)
1.6 m
(1)
1.6 m
F.
22m
22 m
Free-body diagram of rigid bar ABC
Solution:
Transcribed Image Text:Ce A classroom.google.com/c/NDA4NJK5ODA4MZA5/m/NDA4ODK1MDMXNZA4/details Strength of Materials Lecture-1 Example 1.2 Rigid bar ABC is supported by a pin at A and axial member (1), which has a cross-sectional area of 540 mm2. The weight of rigid bar ABC can be neglected. (Note: 1 kN = 1,000 N.) > Determine the normal stress in member (1) if a load of P = 8 kN is applied at C. > If the maximum normal stress in member (1) must be limited to 50 MPa, what is the maximum load magnitude P that may be applied to the rigid bar at C? (1) 1.6 m (1) 1.6 m F. 22m 22 m Free-body diagram of rigid bar ABC Solution:
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