Question

A simply supported wood beam with a span of L = 15 ft supports a uniformly distributed load of w0 = 270 lb/ft. The allowable bending stress of the wood is 1.95 ksi. If the aspect ratio of the solid rectangular wood beam is specified as h/b = 1.75, calculate the minimum width b that can be used for the beam.

Answer 1

Answer:

The minimum width is $b = 4.32 \ in$

Explanation:

The free body diagram is shown on the first uploaded image

The Summation of the moment about A  is equal to 0 from the diagram we can see its wedged

     i.e $\sum M_A = 0$

        $C_y *L = 270 \frac{2L}{3} *\frac{L}{3}$

       $C_y *L -270 \frac{2L}{3} *\frac{L}{3} = 0$

Given from the question that L = 15 ft

           $C_y *15 -270 * \frac{2*15}{3}*\frac{15}{3} =0$

          $C_y = 900lb$

        => $C_y =900lb \ (upward)$

The net upward force experienced by the beam = 0

i.e          $\sum F_y = 0$   because of the canceling downward force

              $A_y +C_y = 270 * \frac{2L}{3}$

                            $A_y + 900 = 270 * 10$

                             $A_y = 1800lb$

The maximum bending always occurs where the shear force is zero

           $A_y - 270*x =0$

           $1800 - 270 * x = 0$

           $x = \frac{1800}{270} =6.67ft$

To obtain the maximum bending moment

     $M_{max} = A_y *6.67 - 270 *6.67 *\frac{6.67}{2}$

     $M_{max} = 1800 *6.67 - 270*6.67*\frac{6.67}{2}$

               $=6000\ lb.ft$

To obtain the width

         $\frac{M}{I} =\frac{\sigma}{y}$

The equation above is the bending equation

Where M is the bending moment  $= 6000(12) \ lb in$

Note: the multiplication by 12 is to convert the value to inches  

I is the moment of inertia $=\frac{bh^3}{12}$

Note: the division by 12 is to convert the value to inches

and $\sigma$ is the bending stress = 1.95 ×1000 ksi

Then y is the distance from natural axis = $=\frac{h}{2}$

Substituting this into the formula we have

         $\frac{6000(12)}{\frac{bh^3}{12} } = \frac{1.95(1000)}{\frac{h}{2} }$

     $\frac{6000(12)}{\frac{bh^2}{6} } =1.95 * 1000$

  given that $\frac{h}{b} = 1.75$

                =>h = 1.75 b

     Substituting into the equation

             $\frac{6000(12)}{\frac{b(1.75b)^2}{6} } = 1.95(1000)$

         $\frac{6000(12)}{1} \frac{6}{3.063b^3} = 1750$

          $\frac{432000}{3.063b^3} =1750$

         $3.06b^3 =\frac{432000}{1750}$

         $3.063b^3 = 246.86$

         $b^3 = 80.593$

        $b = \sqrt[3]{80.593}$

         $b = 4.32 \ in$