Question

PROBLEM ( 2.98 ) For ( P=100 mathrm{kN} ), determine the minimum plate thickness ( t ) required if the allowable stress is ( 125 mathrm{MPa} ). SOLUTION At the hole: [ egin{aligned} r_{A} & =20 mathrm{~mm} quad d_{A}=88-40=48 mathrm{~mm} \ frac{2 r_{A}}{D_{A}} & =frac{2(20)}{88}=0.455 end{aligned} ] From Fig. 2.60a, [ egin{aligned} K & =2.20 \ sigma_{max } & =frac{K P}{A_{ ext {net }}}=frac{K P}{d_{A} t} quad herefore quad t=frac{K P}{d_{A} sigma_{max }} \ t & =frac{(2.20)left(100 imes 10^{3} mathrm{~N} ight)}{(0.048 mathrm{~m})left(125 imes 10^{6} mathrm{~Pa} ight)}=36.7 imes 10^{-3} mathrm{~m}=36.7 mathrm{~mm} end{aligned} ] At the fillet: [ egin{array}{ll} D=88 mathrm{~mm}, & d_{B}=64 mathrm{~mm} quad frac{D}{d_{B}}=frac{88}{64}=1.375 \ r_{B}=15 mathrm{~mm} & frac{r_{B}}{d_{B}}=frac{15}{64}=0.2344 end{array} ] From Fig. 2.60b, [ egin{aligned} K & =1.70 \ sigma_{max } & =frac{K P}{A_{min }}=frac{K P}{d_{B} t} \ t & =frac{K P}{d_{B} sigma_{max }}=frac{(1.70)left(100 imes 10^{3} mathrm{~N} ight)}{(0.064 mathrm{~m})left(125 imes 10^{6} mathrm{~Pa} ight)}=21.25 imes 10^{-3} mathrm{~m}=21.25 mathrm{~mm} end{aligned} ] The larger value is the required minimum plate thickness. [ t=36.7 mathrm{~mm} ]

          PROBLEM ( 2.98 )
For ( P=100 mathrm{kN} ), determine the minimum plate thickness ( t ) required if the allowable stress is ( 125 mathrm{MPa} ).
SOLUTION
At the hole:
[
egin{aligned}
r_{A} & =20 mathrm{~mm} quad d_{A}=88-40=48 mathrm{~mm} \
frac{2 r_{A}}{D_{A}} & =frac{2(20)}{88}=0.455
end{aligned}
]
From Fig. 2.60a,
[
egin{aligned}
K & =2.20 \
sigma_{max } & =frac{K P}{A_{	ext {net }}}=frac{K P}{d_{A} t} quad 	herefore quad t=frac{K P}{d_{A} sigma_{max }} \
t & =frac{(2.20)left(100 	imes 10^{3} mathrm{~N}
ight)}{(0.048 mathrm{~m})left(125 	imes 10^{6} mathrm{~Pa}
ight)}=36.7 	imes 10^{-3} mathrm{~m}=36.7 mathrm{~mm}
end{aligned}
]
At the fillet:
[
egin{array}{ll}
D=88 mathrm{~mm}, & d_{B}=64 mathrm{~mm} quad frac{D}{d_{B}}=frac{88}{64}=1.375 \
r_{B}=15 mathrm{~mm} & frac{r_{B}}{d_{B}}=frac{15}{64}=0.2344
end{array}
]
From Fig. 2.60b,
[
egin{aligned}
K & =1.70 \
sigma_{max } & =frac{K P}{A_{min }}=frac{K P}{d_{B} t} \
t & =frac{K P}{d_{B} sigma_{max }}=frac{(1.70)left(100 	imes 10^{3} mathrm{~N}
ight)}{(0.064 mathrm{~m})left(125 	imes 10^{6} mathrm{~Pa}
ight)}=21.25 	imes 10^{-3} mathrm{~m}=21.25 mathrm{~mm}
end{aligned}
]
The larger value is the required minimum plate thickness.
[
t=36.7 mathrm{~mm}
]

$PROBLEM ( 2.98 ) For ( P=100 mathrmkN ), determine the minimum plate thickness ( t ) required if the allowable stress is ( 125 mathrmMPa ). SOLUTION At the hole: [ eginaligned rA =20 mathrm mm quad dA=88-40=48 mathrm mm frac2 rADA =frac2(20)88=0.455 endaligned ] From Fig. 2.60a, [ eginaligned K =2.20 sigmamax =fracK PA ext net =fracK PdA t quad herefore quad t=fracK PdA sigmamax t =frac(2.20)left(100 imes 10^3 mathrm N ight)(0.048 mathrm m)left(125 imes 10^6 mathrm Pa ight)=36.7 imes 10^-3 mathrm m=36.7 mathrm mm endaligned ] At the fillet: [ eginarrayll D=88 mathrm mm, dB=64 mathrm mm quad fracDdB=frac8864=1.375 rB=15 mathrm mm fracrBdB=frac1564=0.2344 endarray ] From Fig. 2.60b, [ eginaligned K =1.70 sigmamax =fracK PAmin =fracK PdB t t =fracK PdB sigmamax =frac(1.70)left(100 imes 10^3 mathrm N ight)(0.064 mathrm m)left(125 imes 10^6 mathrm Pa ight)=21.25 imes 10^-3 mathrm m=21.25 mathrm mm endaligned ] The larger value is the required minimum plate thickness. [ t=36.7 mathrm mm ]$

Added by Ibrahim A.

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