
The following are some typical section and equations for some of the most common members:Assume that the Square Shape is a 6" x 6", therefore, to find the Section Modulus (Sx): Sx = d ^{ 3 } / 6 = 6 ^{ 3 } / 6 = 36 inch ^{ 3 } , for xx through the Center. Once we know the Section Modulus, the Moment of Inertia can be computer using the Equation: Ix = Sx x c = 36 x 3 = 108 inch ^{ 4 } or the Equation Ix = d ^{ 4 } / 12 may be used. or Sx = d ^{ 3 } / 3 = 6 ^{ 3 } / 3 = 72 inch ^{ 3 } , for xx through the Bottom Edge. Assume that the Square Shape is a 4" wide x 6" high, therefore, to find the Section Modulus (Sx): Sx = bd ^{ 2 } / 6 = 4 x 6 ^{ 2 } / 6 = 24 inch ^{ 3 } , for xx through the Center. Once we know the Section Modulus, the Moment of Inertia can be computer using the Equation: Ix = Sx x c = 24 x 3 = 72 inch ^{ 4 } or the Equation Ix = bd ^{ 3 } / 12 may be used. or Sx = bd ^{ 2 } / 3 = 4 x 6 ^{ 2 } / 3 = 48 inch ^{ 3 } , for xx through the Bottom Edge. Assume that the Hollow Retangular Shape is a 4" (b) wide x 6" (d) high, with a 2" (b1) x 4" (d1) Hollow Center, therefore, to find the Section Modulus (Sx): Sx = (bd ^{ 3 } )(b1d1 ^{ 3 } ) / 6d = (4 x 6 ^{ 3 } )  (2 x 4 ^{ 3 } ) / 6 x 6 = 20.44 inch ^{ 3 } , for xx through the Center. Application of Section Modulus (Sx)The following are Pratical Examples of Loads and Calculations being applied to Retangular Members Uniformly Distributed Load (100 lb per ft) on a 2 x 10 Example of a Simple Supported Beam with a Point Concentrated Load with Equations and Solutions: Point Concentrated Load (600 lb) on a 2 x 12 Example of a Simple Supported Beam with a Point Concentrated Load with Equations and Solutions: Point Concentrated Load (1200 lb) on a 4 x 12 