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diff --git a/3862/CH11/EX11.7/Ex11_7.sce b/3862/CH11/EX11.7/Ex11_7.sce
new file mode 100644
index 000000000..7faeff669
--- /dev/null
+++ b/3862/CH11/EX11.7/Ex11_7.sce
@@ -0,0 +1,53 @@
+clear
+//
+//variable declaration
+//Let the principal plane make anticlockwise angle theta with the plane of px with y-axis. Then
+
+px=(-50)                    //N/mm^2
+py=(100)                    //N/mm^2
+q=(75)                     //N/mm^2
+
+
+p1=(px+py)/2+sqrt((((px-py)/2)**2)+(q**2))
+
+printf("\n (i) p1= %0.2f  N/mm^2",p1)
+
+p2=(px+py)/2-sqrt((((px-py)/2)**2)+(q**2))
+
+printf("\n p2= %0.2f  N/mm^2",p2)
+
+qmax=sqrt((((px-py)/2**2))+(q**2))
+
+printf("\n (ii) qmax= %0.2f  N/mm^2",qmax)
+
+//let theta be the inclination of principal stress to the plane of px.
+
+
+theta1=(atan((2*q)/(px-py))*180)/(%pi*2)                        
+
+printf("\n theta= %0.2f  °  clockwise",theta1)
+
+//Plane of maximum shear makes 45° to it 
+
+theta2=theta1+45
+printf("\n theta2= %0.2f  °",theta2)
+
+//Normal stress on this plane is given by
+
+pn=((px+py)/2)+((px-py)/2)*cos(2*theta2*%pi/180)+q*sin(2*theta2*%pi/180)
+
+pt=qmax
+
+//Resultant stress
+p=sqrt((pn**2)+(pt**2))
+
+printf("\n p= %0.2f  N/mm^2",p)
+
+//Let ‘p’ make angle phi to tangential stress (maximum shear stress plane). 
+
+phi=atan(pn/pt)*180/%pi 
+
+printf("\n phi= %0.1f  °",phi)
+
+//there is mistake in book
+printf("\n mitake in book answer is wrong")