Question

PROBLEM ( 2.14 ) The 4-mm-diameter cable ( B C ) is made of a steel with ( E=200 mathrm{GPa} ). Knowing that the maximum stress in the cable must not exceed ( 190 mathrm{MPa} ) and that the elongation of the cable must not exceed ( 6 mathrm{~mm} ), find the maximum load ( mathbf{P} ) that can be applied as shown. SOLUTION [ L_{B C}=sqrt{6^{2}+4^{2}}=7.2111 mathrm{~m} ] Use bar ( A B ) as a free body. [ egin{aligned} +) Sigma M_{A} & =0: quad 3.5 P-(6)left(frac{4}{7.2111} F_{B C} ight)=0 \ P & =0.9509 F_{B C} end{aligned} ] Considering allowable stress, ( sigma=190 imes 10^{6} mathrm{~Pa} ) [ egin{array}{l} A=frac{pi}{4} d^{2}=frac{pi}{4}(0.004)^{2}=12.566 imes 10^{-6} mathrm{~m}^{2} \ sigma=frac{F_{B C}}{A} quad herefore quad F_{B C}=sigma A=left(190 imes 10^{6} ight)left(12.566 imes 10^{-6} ight)=2.388 imes 10^{3} mathrm{~N} end{array} ] Considering allowable elongation, ( delta=6 imes 10^{-3} mathrm{~m} ) [ delta=frac{F_{B C} L_{B C}}{A E} quad herefore quad F_{B C}=frac{A E delta}{L_{B C}}=frac{left(12.566 imes 10^{-6} ight)left(200 imes 10^{9} ight)left(6 imes 10^{-3} ight)}{7.2111}=2.091 imes 10^{3} mathrm{~N} ] Smaller value governs. ( F_{B C}=2.091 imes 10^{3} mathrm{~N} ) [ P=0.9509 F_{B C}=(0.9509)left(2.091 imes 10^{3} ight)=1.988 imes 10^{3} mathrm{~N} quad P=1.988 mathrm{kN} ]

          PROBLEM ( 2.14 )
The 4-mm-diameter cable ( B C ) is made of a steel with ( E=200 mathrm{GPa} ). Knowing that the maximum stress in the cable must not exceed ( 190 mathrm{MPa} ) and that the elongation of the cable must not exceed ( 6 mathrm{~mm} ), find the maximum load ( mathbf{P} ) that can be applied as shown.
SOLUTION
[
L_{B C}=sqrt{6^{2}+4^{2}}=7.2111 mathrm{~m}
]
Use bar ( A B ) as a free body.
[
egin{aligned}
+) Sigma M_{A} & =0: quad 3.5 P-(6)left(frac{4}{7.2111} F_{B C}
ight)=0 \
P & =0.9509 F_{B C}
end{aligned}
]
Considering allowable stress, ( sigma=190 	imes 10^{6} mathrm{~Pa} )
[
egin{array}{l}
A=frac{pi}{4} d^{2}=frac{pi}{4}(0.004)^{2}=12.566 	imes 10^{-6} mathrm{~m}^{2} \
sigma=frac{F_{B C}}{A} quad 	herefore quad F_{B C}=sigma A=left(190 	imes 10^{6}
ight)left(12.566 	imes 10^{-6}
ight)=2.388 	imes 10^{3} mathrm{~N}
end{array}
]
Considering allowable elongation, ( delta=6 	imes 10^{-3} mathrm{~m} )
[
delta=frac{F_{B C} L_{B C}}{A E} quad 	herefore quad F_{B C}=frac{A E delta}{L_{B C}}=frac{left(12.566 	imes 10^{-6}
ight)left(200 	imes 10^{9}
ight)left(6 	imes 10^{-3}
ight)}{7.2111}=2.091 	imes 10^{3} mathrm{~N}
]
Smaller value governs. ( F_{B C}=2.091 	imes 10^{3} mathrm{~N} )
[
P=0.9509 F_{B C}=(0.9509)left(2.091 	imes 10^{3}
ight)=1.988 	imes 10^{3} mathrm{~N} quad P=1.988 mathrm{kN}
]

$PROBLEM ( 2.14 ) The 4-mm-diameter cable ( B C ) is made of a steel with ( E=200 mathrmGPa ). Knowing that the maximum stress in the cable must not exceed ( 190 mathrmMPa ) and that the elongation of the cable must not exceed ( 6 mathrm mm ), find the maximum load ( mathbfP ) that can be applied as shown. SOLUTION [ LB C=sqrt6^2+4^2=7.2111 mathrm m ] Use bar ( A B ) as a free body. [ eginaligned +) Sigma MA =0: quad 3.5 P-(6)left(frac47.2111 FB C ight)=0 P =0.9509 FB C endaligned ] Considering allowable stress, ( sigma=190 imes 10^6 mathrm Pa ) [ eginarrayl A=fracpi4 d^2=fracpi4(0.004)^2=12.566 imes 10^-6 mathrm m^2 sigma=fracFB CA quad herefore quad FB C=sigma A=left(190 imes 10^6 ight)left(12.566 imes 10^-6 ight)=2.388 imes 10^3 mathrm N endarray ] Considering allowable elongation, ( delta=6 imes 10^-3 mathrm m ) [ delta=fracFB C LB CA E quad herefore quad FB C=fracA E deltaLB C=fracleft(12.566 imes 10^-6 ight)left(200 imes 10^9 ight)left(6 imes 10^-3 ight)7.2111=2.091 imes 10^3 mathrm N ] Smaller value governs. ( FB C=2.091 imes 10^3 mathrm N ) [ P=0.9509 FB C=(0.9509)left(2.091 imes 10^3 ight)=1.988 imes 10^3 mathrm N quad P=1.988 mathrmkN ]$

Added by Ibrahim A.

Question

Please give Ace some feedback