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authorprashantsinalkar2018-02-03 11:01:52 +0530
committerprashantsinalkar2018-02-03 11:01:52 +0530
commit7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch)
tree449d555969bfd7befe906877abab098c6e63a0e8 /3864/CH4/EX4.24/Ex4_24.sce
parentd1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff)
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+clear
+//
+
+//Initilization of Variables
+
+L=4000 //mm //span
+
+//Rectangular Cross-section
+b=100 //mm //Width
+d=200 //mm //Thickness
+
+F_per=10 //N/mm**2 //Max Bending stress
+q_max=0.6 //N/mm**2 //Shear stress
+
+//Calculations
+
+//If the Load W is in KN/m
+
+//Max shear Force
+//F=w*l*2**-1 //KN
+//After substituting values and further simplifying we get
+//M=2*w //KN-m
+
+//Max Load from Consideration of moment
+//M=1*6**-1*b*d**2*F_per
+//After substituting values and further simplifying we get
+w=(1*6**-1*b*d**2*F_per)*(2*10**6)**-1 //KN/m
+
+//Max Load from Consideration of shear stress
+//q_max=1.5*F*(b*d)**-1 //N
+//After substituting values and further simplifying we get
+F=q_max*(1.5)*b*d //N
+
+//If w is Max Load in KN/m,then
+//2*w*1000=8000
+//After Rearranging and Further simplifying we get
+w2=8000*(2*1000)**-1 //KN/m
+
+//Result
+printf("\n Uniformly Distributed Load Beam can carry is %0.2f KN/m",w)