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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /764/CH4/EX4.19.b | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '764/CH4/EX4.19.b')
| -rwxr-xr-x | 764/CH4/EX4.19.b/result4_19.txt | 49 | ||||
| -rwxr-xr-x | 764/CH4/EX4.19.b/solution4_19.sce | 29 |
2 files changed, 78 insertions, 0 deletions
diff --git a/764/CH4/EX4.19.b/result4_19.txt b/764/CH4/EX4.19.b/result4_19.txt new file mode 100755 index 000000000..7c458f537 --- /dev/null +++ b/764/CH4/EX4.19.b/result4_19.txt @@ -0,0 +1,49 @@ +-->//(Design against Static Load) Example 4.19
+
+-->//Refer Fig.4.65 and Fig.4.66
+
+-->//Tensile yield strength of 45C8 Syt (N/mm2)
+
+-->Syt = 380
+ Syt =
+
+ 380.
+
+-->//Factor of safety fs
+
+-->fs = 3.5
+ fs =
+
+ 3.5
+
+-->//All dimensions in mm
+
+-->bi = 90
+ bi =
+
+ 90.
+
+-->bo = 30
+ bo =
+
+ 30.
+
+-->h = 120
+ h =
+
+ 120.
+
+-->Ro = 170
+ Ro =
+
+ 170.
+
+-->Ri = 50
+ Ri =
+
+ 50.
+
+
+
+The load carrying capacity(P) = 94828.181768 N
+
\ No newline at end of file diff --git a/764/CH4/EX4.19.b/solution4_19.sce b/764/CH4/EX4.19.b/solution4_19.sce new file mode 100755 index 000000000..d7fa73adc --- /dev/null +++ b/764/CH4/EX4.19.b/solution4_19.sce @@ -0,0 +1,29 @@ +
+//Obtain path of solution file
+path = get_absolute_file_path('solution4_19.sce')
+//Obtain path of data file
+datapath = path + filesep() + 'data4_19.sci'
+//Clear all
+clc
+//Execute the data file
+exec(datapath)
+//Calculate the permissible stresses for steel parts sigmat (N/mm2)
+sigmat = Syt/fs
+//Calculate the eccentricity e (mm)
+Rn = (((bi + bo)/2)*h)/((((bi * Ro) - (bo * Ri))/h)*log(Ro/Ri)-(bi - bo))
+R = Ri + ((h*(bi + (2 * bo)))/(3 * (bi + bo)))
+e = R - Rn
+hi = Rn - Ri
+A = (h * (bi + bo))/2
+//Assume the value load P to be 1N
+P = 1
+//Calculate the bending moment Mb (N-mm)
+Mb = R * P
+//Calculate bending stress at the inner fibre B (N/mm2)
+B = (Mb * hi)/(A * e * Ri)
+//Calculate the direct tensile stress T (N/mm2)
+T = P/A
+//Calculate the actual load carrying capacity P (N)
+P = sigmat/(B + T)
+//Print results
+printf('\nThe load carrying capacity(P) = %f N\n',P)
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