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// Problem 16.4,Page no.368
clc;clear;
close;
t=18 //mm //thickness of plates
sigma_t=100 //MPa //Tensile stress //Notification has been changed
sigma_s=70 //MPa //Shearing stress //Notification has been changed
//Calculations
d=6*t**0.5 //mm //Diameter of rivet //Answer is in correct in textbook
s=%pi*4**-1*d**2*10**-6*sigma_s*10**6 //N //Strength of one rivet in single shear //Answer is in correct in textbook
//Consider strip of joint equal to %pitch p
//S_1=(p-d)*t*10**-3*sigma_t*10**6 //Strength of plate against tearing along 1-1
//After substituting values and further simplifying we get
//S_1=1800*p-45900 (Equation 1)
//S_2=(p-d)*t*10**-3*sigma_t*10**6+s //Strength of plate against tearing along 1-1
//After substituting values and further simplifying we get
//S_1=1800*p-56050.64 (Equation 2)
//But the value of Equation 2 is smaller than Equation 1
//Strength of rivets in single shear is
S=4*s
//Equating Equation 2 to shearing value
//1800*p-56050.64=S
p=(S+56050.64)*18000**-1 //cm //%pitch of rivet
//Result
printf("Diameter of rivets is %.2f",d);printf(" mm")
printf("\n pitch of rivet is %.2f",p);printf(" cm")
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