Chapter 6, Problem 1RP

Determine the shape factor for the wide-flange beam.

R6–1

To determine

The shape factor for the wide-flange beam.

Answer to Problem 1RP

The shape factor for the wide-flange beam is $\underset{\_}{k=1.22}$.

Explanation of Solution

Determination of Elastic moment:

Moment of inertia:

Show the free-body diagram of the beam as in Figure 1.

Consider the beam as rectangular and determine the moment of inertia for rectangular section and deduct the moment of inertia of the two smaller rectangular portions.

Determine the moment of inertia of the I-section using the formula.

$I=\frac{B{D}^3}{12}-2\times \frac{b{d}^3}{12}$ (1)

Here, B is width of rectangular section, D is depth of large rectangular section, b is depth of small rectangular section, and d is depth of small rectangular section.

Substitute 180 mm for B, 220 mm for D, 75 mm for b, and 180 mm for d in Equation (1).

$\begin{array}{c}I=\frac{180\times {220}^3}{12}-2\times \frac{75\times {180}^3}{12}\\ =86,820,000{\text{mm}}^4\times \frac{1{\text{m}}^4}{{\left(1,000\text{mm}\right)}^4}\\ =86.82\times {10}^{-6}{\text{m}}^4\end{array}$

Elastic moment:

Determine the elastic moment using the equation.

$M_Y=\frac{{\sigma }_{Y}I}{c}$ (2)

Here, ${\sigma }_Y$ is allowable stress in the beam and c is the distance between the neutral axis and the extreme fibre in the beam.

Substitute $86.82\times {10}^{-6}{\text{m}}^4$ for I and 110 mm for c in Equation (2).

$\begin{array}{c}{M}_Y=\frac{{\sigma }_Y\times 86.82\times {10}^{-6}}{110\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}\\ =\left(0.78927\times {10}^{-3}\right){\sigma }_Y\end{array}$

Determination of plastic moment:

Show the free-body diagram of the profile view of the I-beam as in Figure 2.

Determine the plastic moment in the I-section using the equation.

$\begin{array}{c}{M}_p={\sigma }_Y{A}_1{c}_1+{\sigma }_Y{A}_2{c}_2\\ ={\sigma }_Y{b}_1{h}_1{c}_1+{\sigma }_Y{b}_2{h}_2{c}_2\end{array}$ (3)

Here, $A_1$ is area of flanges, $c_1$ is the distance between the centre to centre of flanges where the force is resultant, $A_2$ is the area of web, and $c_2$ is the distance between the centre to centre of web where the force is resultant.

Substitute 180 mm for $b_1$, 20 mm for $h_1$, 200 mm for $c_1$, 30 mm for $b_2$, 90 mm for $h_2$, and 90 mm for $c_2$ in Equation (3).

$\begin{array}{c}{M}_p=\left(\begin{array}{l}{\sigma }_Y\times 180\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\times 20\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\times 200\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\\ +{\sigma }_Y\times 30\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\times 90\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\times 90\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\end{array}\right)\\ =\left(0.963\times {10}^{-3}\right){\sigma }_Y\end{array}$

Shape factor:

Determine the shape factor for the I-section using the formula.

$k=\frac{M_p}{M_Y}$ (4)

Substitute $\left(0.963\times {10}^{-3}\right){\sigma }_Y$ for $M_p$ and $\left(0.78927\times {10}^{-3}\right){\sigma }_Y$ for $M_Y$ in Equation (4).

$\begin{array}{c}k=\frac{\left(0.963\times {10}^{-3}\right){\sigma }_Y}{\left(0.78927\times {10}^{-3}\right){\sigma }_Y}\\ =1.22\end{array}$

Thus, the shape factor for the wide-flange beam is $\underset{\_}{k=1.22}$.