From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:27:19 +0530
Subject: initial commit / add all books

---
 3772/CH5/EX5.8/Ex5_8.sce | 32 ++++++++++++++++++++++++++++++++
 1 file changed, 32 insertions(+)
 create mode 100644 3772/CH5/EX5.8/Ex5_8.sce

(limited to '3772/CH5/EX5.8/Ex5_8.sce')

diff --git a/3772/CH5/EX5.8/Ex5_8.sce b/3772/CH5/EX5.8/Ex5_8.sce
new file mode 100644
index 000000000..189a866ea
--- /dev/null
+++ b/3772/CH5/EX5.8/Ex5_8.sce
@@ -0,0 +1,32 @@
+// Problem 5.8,Page no.127
+
+clc;clear;
+close;
+
+L=3 //m //span of beam
+t=20 //mm //Thickness of steel
+D=200 //mm //overall depth 
+B=140 //mm //overall width
+b=100 //mm //width of inner rectangle
+d=160 //mm //depth of inner rectangle
+w=77 //KN/mm**2
+sigma=100 //N/mm**2 //Bending stress
+//Calculations
+V=((D*10**-3*B*10**-3)-(d*10**-3*b*10**-3)) //m**3 //Volume of rectangular box
+W=V*3*w //KN //Weight of Beam
+I=(B*D**3-b*d**3)*12**-1 //mm**4 //M.I of beam section
+
+//Now using the relation,M*I**-1=sigma*y**-1
+
+y=200 //mm //distance from farthest fibre
+M=I*sigma*2*y**-1 //N*mm
+
+//M=3000*W+2772*3000*2**-1
+//After sub values in above equation we get
+
+W=((59.2*10**6-2772*3000*2**-1)*(3000)**-1)*10**-3 //KN //Max concentrated Load at free end
+
+F=W+2.772*2**-1 //KN //shear force at half length
+
+//Result
+printf("The shear force at half length is %.2f kN",F)
-- 
cgit