Question

PROBLEM ( 2.81 ) A vibration isolation unit consists of two blocks of hard rubber bonded to a plate ( A B ) and to rigid supports as shown. Knowing that a force of magnitude ( P=25 mathrm{kN} ) causes a deflection ( delta=1.5 mathrm{~mm} ) of plate ( A B ), determine the modulus of rigidity of the rubber used. SOLUTION [ egin{aligned} F & =frac{1}{2} P=frac{1}{2}left(25 imes 10^{3} mathrm{~N} ight)=12.5 imes 10^{3} mathrm{~N} \ au & =frac{F}{A}=frac{left(12.5 imes 10^{3} mathrm{~N} ight)}{(0.15 mathrm{~m})(0.1 mathrm{~m})}=833.33 imes 10^{3} mathrm{~Pa} \ delta & =1.5 imes 10^{-3} mathrm{~m} quad h=0.03 mathrm{~m} \ gamma & =frac{delta}{h}=frac{1.5 imes 10^{-3}}{0.03}=0.05 \ G & =frac{ au}{gamma}=frac{833.33 imes 10^{3}}{0.05}=16.67 imes 10^{6} mathrm{~Pa} end{aligned} ] [ G=16.67 mathrm{MPa} ]

          PROBLEM ( 2.81 )
A vibration isolation unit consists of two blocks of hard rubber bonded to a plate ( A B ) and to rigid supports as shown. Knowing that a force of magnitude ( P=25 mathrm{kN} ) causes a deflection ( delta=1.5 mathrm{~mm} ) of plate ( A B ), determine the modulus of rigidity of the rubber used.
SOLUTION
[
egin{aligned}
F & =frac{1}{2} P=frac{1}{2}left(25 	imes 10^{3} mathrm{~N}
ight)=12.5 	imes 10^{3} mathrm{~N} \
	au & =frac{F}{A}=frac{left(12.5 	imes 10^{3} mathrm{~N}
ight)}{(0.15 mathrm{~m})(0.1 mathrm{~m})}=833.33 	imes 10^{3} mathrm{~Pa} \
delta & =1.5 	imes 10^{-3} mathrm{~m} quad h=0.03 mathrm{~m} \
gamma & =frac{delta}{h}=frac{1.5 	imes 10^{-3}}{0.03}=0.05 \
G & =frac{	au}{gamma}=frac{833.33 	imes 10^{3}}{0.05}=16.67 	imes 10^{6} mathrm{~Pa}
end{aligned}
]
[
G=16.67 mathrm{MPa}
]

$PROBLEM ( 2.81 ) A vibration isolation unit consists of two blocks of hard rubber bonded to a plate ( A B ) and to rigid supports as shown. Knowing that a force of magnitude ( P=25 mathrmkN ) causes a deflection ( delta=1.5 mathrm mm ) of plate ( A B ), determine the modulus of rigidity of the rubber used. SOLUTION [ eginaligned F =frac12 P=frac12left(25 imes 10^3 mathrm N ight)=12.5 imes 10^3 mathrm N au =fracFA=fracleft(12.5 imes 10^3 mathrm N ight)(0.15 mathrm m)(0.1 mathrm m)=833.33 imes 10^3 mathrm Pa delta =1.5 imes 10^-3 mathrm m quad h=0.03 mathrm m gamma =fracdeltah=frac1.5 imes 10^-30.03=0.05 G =frac augamma=frac833.33 imes 10^30.05=16.67 imes 10^6 mathrm Pa endaligned ] [ G=16.67 mathrmMPa ]$

Added by Ibrahim A.

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