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+// Problem 5.15,Page no.137
+
+clc;clear;
+close;
+
+B=20 //cm //width of timber
+D=30 //cm //depth of timber
+d=25 //cm //depth of steel plate
+b=1.2 //cm //width of steel plate
+sigma_s=90 //N/mm**2 //Bending stress in steel
+sigma_t=6 //N/mm**2 //Bending stress in timber
+m=20 //Ratio of modulus of elasticity of of steel to timber
+
+//Calculation
+
+//Equivalent width of wood section,when 1.2 cm wide steel plate is replaced by steel plate is
+b_1=1.2*20 //cm
+d_1=25 //cm //depth of wood section
+y_1=d*2**-1 //cm //C.G of timber section
+y_2=D*2**-1 //cm //C.G of steel section
+
+Y_bar=(2*d*b_1*y_1+D*B*y_2)*(2*d*b_1+D*B)**-1 //cm //Distance of C.G from Bottom edge
+I=B*D**3*12**-1+B*D*(y_2-Y_bar)**2+2*(b_1*d_1**3*12**-1+b_1*d_1*(Y_bar-y_1)**2) //M.I of equivalent timber section about N.A
+Y=30-Y_bar //distance of C.G from top of equivalent wood section
+
+//Thus max stress will occur at top and that in steel will occur at bottom
+//sigma_s=m*Y_bar*Y**-1*sigma_t 
+
+//After simplifying we get
+//sigma_s=15.99*sigma_t
+
+sigma_t=sigma_s*15.99**-1 //N/mm**2 //Max stress in Equivalent timber section
+
+Z_t=I*Y**-1 //Section modulus of equivalent section
+M=sigma_t*Z_t*10**-5*100 //Moment of resistance of beam
+
+//Result
+printf("Position of N.A is %.2f cm",Y_bar)
+printf("\n Moment of Resistance of beam is %.2f kN-m",M)