An I-section of a beam has the following dimensions: flanges, 10 cm × 1 cm and web, 12 cm × 1 cm (Figure 6.20). If this section is subjected to a bending moment of 5 kN m and a shearing force of 5 kN, find out maximum tensile stress and maximum shearing stress induced in it.

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Accepted Answer

From symmetry, we can easily locate the centroid of the I-section and we can consider the neutral axis passing through that section. The area moment of inertia about that section is

[latex] I_x=\frac{10 imes 14^3}{12}-\frac{9 imes 12^3}{12}=990.76 cm ^4 [/latex]

(refer to Example 6.5 for further details).
It is already given that bending moment at the section is 5 kN m or 500 kN cm. Now maximum bending stress will occur at the topmost and lowermost fibres and its magnitude is

[latex] \sigma_{\max }=\frac{M y}{I}=\frac{500 imes 1000 imes 7}{990.67} N / cm ^2=35.33 MPa [/latex]

Obviously, for positive bending moment, the top fibre will be in compression and the bottom fibre will be subjected to tension. For determination of maximum shear stress, we must know how shear stress varies along depth. In case of I-section, the maximum shear stress occurring at neutral axis is given by

[latex] au_{\max }=\frac{V}{I t}\left[\frac{b}{2}\left\lgroup \frac{h^2}{4}-\frac{h_1^2}{4} ight group +\frac{t h_1^2}{8} ight] [/latex]

You must be guessing how do we get this equation. It is simple. We can use the same principle we have followed for derivation of shear stress equation for rectangular cross-section of beam. Let us try it here in brief.
Let us refer to Figure 6.21. If we consider the shaded portion and write the first moment of inertia about neutral axis or centroidal axis, what do we get?

Q is the first moment of inertia about neutral axis, given by

[latex] Q=b\left\lgroup \frac{h}{2}-\frac{h_1}{2} ight group \cdot \frac{1}{2}\left\lgroup \frac{h}{2}+\frac{h_1}{2} ight group +t\left\lgroup \frac{h_1}{2}-y_1 ight group \cdot \frac{1}{2}\left\lgroup \frac{h_1}{2}+y_1 ight group [/latex]

Putting it in the general equation of shear stress, Eq. (6.10), we get the shear stress at a level which is [latex] y_1 [/latex] distance away from NA as:

[latex] au_1=\frac{Q V}{I b}= au_2 \quad\left( ext { as } au_1= au_2 ight) [/latex] (6.10)

[latex] au=\frac{V}{I t}\left[\frac{b}{2}\left\lgroup \frac{h^2}{4}-\frac{h_1^2}{4} ight group +\frac{t}{2}\left\lgroup \frac{h_1^2}{4}-y_1^2 ight group ight] [/latex] (1)

This is the general equation for shear stress for I-section. If [latex] y_1=0 [/latex] , that is, at the neutral axis:

[latex] au= au_{\max }=\frac{V}{I t}\left[\frac{b}{2}\left\lgroup\frac{h^2}{4}-\frac{h_1^2}{4} ight group +\frac{t h_1^2}{8} ight] [/latex] (2)

For [latex] y_1=h_1 / 2 [/latex], we get minimum shear stress in the web at the junction with the flange portion as:

[latex] au_{\min }=\frac{V}{I t}\left[\frac{b}{2}\left\lgroup\frac{h^2}{4}-\frac{h_1^2}{4} ight group ight] [/latex] (3)

So, it is clear that from neutral axis to the web–flange junction that shear stress decreases as shown in Figure 6.21(b). A careful estimation reveals that the flanges contribute very little to the total shear force, and hence are not taken into consideration from practical point of view.

[latex] ext { Putting } V=5 kN , I=990.67 cm ^4, t=1 cm , h=14 cm , h_1=12 cm ext { and } b=10 cm [/latex]. into Eq. (6.15) we get [latex] au_{\max }=4.189 MPa [/latex]. So maximum bending stress is 35.33 MPa and maximum shear stress is 4.189 MPa.

Question 6.7: An I-section of a beam has the following dimensions: flanges...

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