Question

PROBLEM 2.15 A single axial load of magnitude ( P=58 mathrm{kN} ) is applied at end ( C ) of the brass rod ( A B C ). Knowing that ( E=105 mathrm{GPa} ), determine the diameter ( d ) of portion ( B C ) for which the deflection of point ( C ) will be ( 3 mathrm{~mm} ). SOLUTION [ egin{array}{l} delta_{C}=sum frac{P_{i} L_{i}}{A_{i} E}=frac{P}{E}left{frac{L_{A B}}{A_{A B}}+frac{L_{B C}}{A_{B C}} ight} \ frac{L_{B C}}{A_{B C}}=frac{E delta_{C}}{P}-frac{L_{A B}}{A_{A B}}=frac{left(105 cdot 10^{9} ight)left(3 cdot 10^{-3} ight)}{58 cdot 10^{3}}-frac{1.2}{frac{pi}{4}(0.030)^{2}}=3.7334 cdot 10^{3} mathrm{~m}^{-1} \ A_{B C}=frac{L_{B C}}{3.7334 cdot 10^{3}}=frac{0.8}{3.7334 cdot 10^{3}}=214.28 cdot 10^{-6} mathrm{~m}^{2} \ A_{B C}=frac{pi}{4} d_{B C}^{2} herefore d_{B C}=sqrt{frac{4 A_{B C}}{pi}}=sqrt{frac{(4)left(214.28 cdot 10^{-6} ight)}{pi}}=16.52 cdot 10^{-3} mathrm{~m} \ =16.52 mathrm{~mm} \ end{array} ]

          PROBLEM 2.15

A single axial load of magnitude ( P=58 mathrm{kN} ) is applied at end ( C ) of the brass rod ( A B C ). Knowing that ( E=105 mathrm{GPa} ), determine the diameter ( d ) of portion ( B C ) for which the deflection of point ( C ) will be ( 3 mathrm{~mm} ).

SOLUTION

[
egin{array}{l}
delta_{C}=sum frac{P_{i} L_{i}}{A_{i} E}=frac{P}{E}left{frac{L_{A B}}{A_{A B}}+frac{L_{B C}}{A_{B C}}
ight} \
frac{L_{B C}}{A_{B C}}=frac{E delta_{C}}{P}-frac{L_{A B}}{A_{A B}}=frac{left(105 cdot 10^{9}
ight)left(3 cdot 10^{-3}
ight)}{58 cdot 10^{3}}-frac{1.2}{frac{pi}{4}(0.030)^{2}}=3.7334 cdot 10^{3} mathrm{~m}^{-1} \
A_{B C}=frac{L_{B C}}{3.7334 cdot 10^{3}}=frac{0.8}{3.7334 cdot 10^{3}}=214.28 cdot 10^{-6} mathrm{~m}^{2} \
A_{B C}=frac{pi}{4} d_{B C}^{2} 	herefore d_{B C}=sqrt{frac{4 A_{B C}}{pi}}=sqrt{frac{(4)left(214.28 cdot 10^{-6}
ight)}{pi}}=16.52 cdot 10^{-3} mathrm{~m} \
=16.52 mathrm{~mm} \
end{array}
]

$PROBLEM 2.15 A single axial load of magnitude ( P=58 mathrmkN ) is applied at end ( C ) of the brass rod ( A B C ). Knowing that ( E=105 mathrmGPa ), determine the diameter ( d ) of portion ( B C ) for which the deflection of point ( C ) will be ( 3 mathrm mm ). SOLUTION [ eginarrayl deltaC=sum fracPi LiAi E=fracPEleftfracLA BAA B+fracLB CAB C ight fracLB CAB C=fracE deltaCP-fracLA BAA B=fracleft(105 cdot 10^9 ight)left(3 cdot 10^-3 ight)58 cdot 10^3-frac1.2fracpi4(0.030)^2=3.7334 cdot 10^3 mathrm m^-1 AB C=fracLB C3.7334 cdot 10^3=frac0.83.7334 cdot 10^3=214.28 cdot 10^-6 mathrm m^2 AB C=fracpi4 dB C^2 herefore dB C=sqrtfrac4 AB Cpi=sqrtfrac(4)left(214.28 cdot 10^-6 ight)pi=16.52 cdot 10^-3 mathrm m =16.52 mathrm mm endarray ]$

Added by Ibrahim A.

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