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-rwxr-xr-x764/CH4/EX4.10.b/result4_10.txt44
-rwxr-xr-x764/CH4/EX4.10.b/solution4_10.sce44
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diff --git a/764/CH4/EX4.10.b/result4_10.txt b/764/CH4/EX4.10.b/result4_10.txt
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+-->//(Design against Static Load) Example 4.10
+
+-->//Refer Fig.4.29
+
+-->//Distance between the axis of the column and the load e (mm)
+
+-->e = 500
+ e =
+
+ 500.
+
+-->//Tensile yield strength of FeE200 Syt (N/mm2)
+
+-->Syt = 200
+ Syt =
+
+ 200.
+
+-->//Factor of safety fs
+
+-->fs = 4
+ fs =
+
+ 4.
+
+-->//Load supported by the column P (kN)
+
+-->P = 25
+ P =
+
+ 25.
+
+-->//Ratio of outer diameter to inner diameter ratio
+
+-->ratio = 0.8
+ ratio =
+
+ 0.8
+
+
+Outer diameter(d0) = 160.000000 mm
+
+Inner diameter(di) = 128.000000 mm
+ \ No newline at end of file
diff --git a/764/CH4/EX4.10.b/solution4_10.sce b/764/CH4/EX4.10.b/solution4_10.sce
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+
+//Function to round-up a value such that it is divisible by 5
+function[v] = round_five(w)
+ v = ceil(w)
+ rem = pmodulo(v,5)
+ if (rem ~= 0)
+ v = v + (5 - rem)
+ end
+endfunction
+
+//Obtain path of solution file
+path = get_absolute_file_path('solution4_10.sce')
+//Obtain path of data file
+datapath = path + filesep() + 'data4_10.sci'
+//Clear all
+clc
+//Execute the data file
+exec(datapath)
+//Calculate the permissible tensile stress sigmat (N/mm2)
+sigmat = Syt/fs
+//Let the outer diameter of the cross-section be 1mm d0
+d0 = 1
+//Calculate the direct compressive stress D (N/mm2)
+D = (P * 1000)/((%pi/4)*((1 - (ratio^2))*(d0^2)))
+//Calculate the value of y (mm)
+y = d0/2
+//Calculate the second moment of area I (mm4)
+I = (%pi/64)*((1 - (ratio^4))*(d0^4))
+//Calculate the tensile stress due to bending moment B (N/mm2)
+B = (P * 1000 * e * y)/I
+//Coefficients of the resultant cubic equation
+p = [sigmat 0 D (-1 * B)]
+r = roots(p)
+real_part = real(r)
+for i = 1:1:3
+ if(real_part(i)>0)
+ d0 = real_part(i)
+ break
+ end
+end
+d0 = round_five(d0)
+//Print results
+printf('\nOuter diameter(d0) = %f mm\n',d0)
+printf('\nInner diameter(di) = %f mm\n',(0.8 * d0))