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In the image below they used P=F but in the second picture they solved by using F=P and i dont know why. Can you please explain whats the difference between these two?

In the image below they used P=F but in the second picture they solved by using F=P and i dont know why. Can you please explain whats the difference between these two?

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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In the image below they used P=F but in the second picture they solved by using F=P and i dont know why. Can you please explain whats the difference between these two?

thank you

2.5 m 3.5 m P B A SOLUTION 4.0 m Use bar AB as a free body. 8= PROBLEM 2.14 The 4-mm-diameter cable BC is made of a steel with E = 200 GPa. Knowing that the maximum stress in the cable must not exceed 190 MPa and that the elongation of the cable must not exceed 6 mm, find the maximum load P that can be applied as shown. FBCLBC AE Considering allowable stress, o=190×10° Pa A== d² = 4 +)ΣΜΑ =0: LBC =√√6² +4² = 7.2111 m V π 4 FBC : FBC=A=(190×106)(12.566×106) = 2.388×10³ N A Considering allowable elongation, 8=6×10-³ m :. FBC P=0.9509FBC = 4 (7.2111 FBC)=0 3.5P-( -(6) (0.004)² = 12.566×10 m² Smaller value governs. FBC = 2.091×10³ N AE8 _ (12.566×10¯)(200×10³)(6×10¯³) LBC 7.2111 P = 2.091×10³N P=0.9509FBC = (0.9509) (2.091×10³) =1.988×10³ N FBC - Ах P= 1.988 KN
Transcribed Image Text:2.5 m 3.5 m P B A SOLUTION 4.0 m Use bar AB as a free body. 8= PROBLEM 2.14 The 4-mm-diameter cable BC is made of a steel with E = 200 GPa. Knowing that the maximum stress in the cable must not exceed 190 MPa and that the elongation of the cable must not exceed 6 mm, find the maximum load P that can be applied as shown. FBCLBC AE Considering allowable stress, o=190×10° Pa A== d² = 4 +)ΣΜΑ =0: LBC =√√6² +4² = 7.2111 m V π 4 FBC : FBC=A=(190×106)(12.566×106) = 2.388×10³ N A Considering allowable elongation, 8=6×10-³ m :. FBC P=0.9509FBC = 4 (7.2111 FBC)=0 3.5P-( -(6) (0.004)² = 12.566×10 m² Smaller value governs. FBC = 2.091×10³ N AE8 _ (12.566×10¯)(200×10³)(6×10¯³) LBC 7.2111 P = 2.091×10³N P=0.9509FBC = (0.9509) (2.091×10³) =1.988×10³ N FBC - Ах P= 1.988 KN
A 125 mm B A SOLUTION Free body member AC: SB = SBD &c = 8CE 225 mm = B FBD FBD LBD EBD ABD 225 mm C SB 150 mm E FCEY PROBLEM 2.25 Link BD is made of brass (E = 105 GPa) and has a cross-sectional area of 240 mm². Link CE is made of aluminum (E = 72 GPa) and has a cross- sectional area of 300 mm². Knowing that they support rigid member ABC, determine the maximum force P that can be applied vertically at point A if the deflection of A is not to exceed 0.35 mm. (1.55556P)(0.225) (105 x 10°)(240×106) FCELCE = (0.55556P)(0.150) ECE ACE (72 x 10°) (300 × 106) Deformation Diagram: From the deformation diagram, Sc +ΣΜ c +EMB Apply displacement limit. S = 0.35 x 10-³₁ P = 14.7381 x 10³ N = 0: 0.350P-0.225FBD = 0 = 13.8889 × 10 P = 3.8581 x 10 P Slope: = 0: 0.125P-0.225FCE =0 m = 23.748 x 10-⁹P FRD = 1.55556P 0 = &B +c_17.7470×10 P LBC SA=SB + LABO = 13.8889×10 ⁹P+(0.125)(78.876×10-⁹P) = 23.748×10 P = FCE=0.55556P 0.225 -= 78.876×10 F P=14.74 kNA
Transcribed Image Text:A 125 mm B A SOLUTION Free body member AC: SB = SBD &c = 8CE 225 mm = B FBD FBD LBD EBD ABD 225 mm C SB 150 mm E FCEY PROBLEM 2.25 Link BD is made of brass (E = 105 GPa) and has a cross-sectional area of 240 mm². Link CE is made of aluminum (E = 72 GPa) and has a cross- sectional area of 300 mm². Knowing that they support rigid member ABC, determine the maximum force P that can be applied vertically at point A if the deflection of A is not to exceed 0.35 mm. (1.55556P)(0.225) (105 x 10°)(240×106) FCELCE = (0.55556P)(0.150) ECE ACE (72 x 10°) (300 × 106) Deformation Diagram: From the deformation diagram, Sc +ΣΜ c +EMB Apply displacement limit. S = 0.35 x 10-³₁ P = 14.7381 x 10³ N = 0: 0.350P-0.225FBD = 0 = 13.8889 × 10 P = 3.8581 x 10 P Slope: = 0: 0.125P-0.225FCE =0 m = 23.748 x 10-⁹P FRD = 1.55556P 0 = &B +c_17.7470×10 P LBC SA=SB + LABO = 13.8889×10 ⁹P+(0.125)(78.876×10-⁹P) = 23.748×10 P = FCE=0.55556P 0.225 -= 78.876×10 F P=14.74 kNA
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