Area Of the Inclined Plane  
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The figure shows a steel bar of height 100 millimeters. A section inclined at an angle of 55 degrees with respect to the horizontal is shown. The load on the left end of the bar is 40 kilo newton directed leftwards. The load on the inclined surface has magnitude 40 kilo newton and is directed rightwards. The load on the inclined surface is resolved into a vector of magnitude 22.943 kilo newton parallel to the inclined surface and a vector of magnitude 32.766 kilo newton perpendicular to the inclined surface. The normal to the surface is at an angle of theta which equals 35 degrees with respect to the horizontal.

To compute the stresses acting on section a-a, the area of the inclined plane must also be determined.

The bar height is 100 mm. Use the angle θ to compute the length of the inclined surface:

length of plane a-a=100 mmcos 35∘          =122.077 mm

The bar is 25 mm thick; therefore, the inclined plane area is:

Aa-a=(122.077 mm)(25 mm)       =3,051.936 mm2