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diff --git a/3862/CH11/EX11.2/Ex11_2.sce b/3862/CH11/EX11.2/Ex11_2.sce
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+clear
+//
+
+//A material has strength in tension, compression and shear as 30N/mm2, 90 N/mm2 and 25 N/mm2, respectively. If a specimen of diameter 25 mm is tested in tension and compression identity the failure surfaces and loads. 
+
+//variable declaration
+
+//In tension: Let axial direction be x direction. Since it is uniaxial loading, py  = 0, q = 0 and only px exists.when the material is subjected to full tensile stress, px = 30 N/mm^2.
+
+pt=(30)
+pc=(90)
+ps=(25)
+
+d=(25)
+px=(30)         //N/mm^2
+py=0
+q=0
+p1=(px+py)/2+sqrt((((px-py)/2)**2)+(q**2))
+
+p2=(px+py)/2-sqrt((((px-py)/2)**2)+(q**2))
+
+qmax=(px-py)/2
+
+//Hence failure criteria is normal stress p1
+
+A=%pi*(d**2)/4
+
+//Corresponding load P is obtained by
+p=p1
+P=p1*A
+
+printf("\n (a) P= %0.2f  N",P)
+
+//In case of compression test,
+
+px=-pc
+
+P=-px*A
+
+printf("\n (b) P= %0.2f  N compressive",(-P))
+
+//at this stage
+
+qmax=sqrt((((px-py)/2**2))+(q**2))
+
+printf("\n Material fails because of maximum shear and not by axial compression.")
+qmax=25
+px=2*qmax
+
+P=px*A
+printf("\n P= %0.0f  N",P)
+printf("\n The plane of qmax is at 45° to the plane of px. ")