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clear
//
//Initilization of Variables
P=200*10**3 //N //Load
A_a=1000 //mm**2 //Area of Aluminium
A_s=800 //mm**2 //Area of steel
E_a=1*10**5 //N/mm**2 //Modulus of Elasticity of Aluminium
E_s=2*10**5 //N/mm**2 //Modulus of ELasticity of steel
sigma_a1=65 //N/mm**2 //stress in aluminium
sigma_s1=150 //N/mm**2 //Stress in steel
//Calculations
//Let P_a and P_s be the force in aluminium and steel pillar respectively
//Now,sum of forces in Vertical direction we get
//2*P_a+P_s=200 .........................................(1)
//By compatibility Equation dell_l_s=dell_l_a we get
//P_s=1.28*P_a ..........................................(2)
//Now substituting value of P_s in Equation 1 we get
P_a=200*3.28**-1 //KN
P_s=200-2*P_a //KN
//Stress developed in aluminium
sigma_a=P_a*10**3*A_a**-1 //N/mm**2
//Stress developed in steel
sigma_s=P_s*10**3*A_s**-1 //N/mm**2
//Part-2
//Let sigma_a1 and sigma_s1 be the stresses in Aluminium and steel due to Additional LOad
P_a1=sigma_a1*A_a //Load carrying capacity of aluminium
P_s1=1.28*P_a1
//Total Load carrying capacity
P1=2*P_a1+P_s1 //N
P_s2=sigma_s1*A_s //Load carrying capacity of steel
P_a2=P_s2*1.28**-1
//Total Load carrying capacity
P2=2*P_a2+P_s2
//Additional Load
P3=P1-P
//Result
printf("\n Stresses Developed in Each Pillar is:sigma_a %0.2f N/mm**2",sigma_a)
printf("\n :sigma_s %0.2f N/mm**2",sigma_s)
printf("\n Additional Load taken by pillars is %0.2f N",P3)
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