Horizontal shear stress variation, SF 50 kN

Section (in cm): an inverted T. Flange 5 cm wide × 2 cm thick at the bottom; stem 1 cm wide × 5 cm high above it; overall height 7 cm. $F = 50$ kN.

     +-+   <- top of stem
     | |
     | |  5
     | |
 +---+-+---+
 |         |  2
 +---------+
     5

Neutral axis (from base)

$$\bar y = \frac{5\times2\times1 + 1\times5\times4.5}{10+5} = \frac{32.5}{15} = 2.167\ \text{cm}$$

Moment of inertia

$$I = \frac{5\times2^3}{12} + 10(1-2.167)^2 + \frac{1\times5^3}{12} + 5(4.5-2.167)^2 = 54.58\ \text{cm}^4$$

Shear stress $\tau = \dfrac{F\,A\bar y}{I\,b}$

The top of the stem is 4.833 cm above the NA. The junction of stem and flange (2 cm from the base) lies 0.167 cm below the NA, since the NA is at 2.167 cm.

Example: $\tau_{NA} = \dfrac{50\times10^3\times11.68\times10^3}{54.58\times10^4\times10} = 107.0\ \text{N/mm}^2$.

Variation

• In the stem the stress varies parabolically from 0 at the top to the maximum at the NA, then stays almost constant down to the junction.
• At the junction it drops suddenly from 106.9 to 21.4 N/mm² because the width changes from 1 cm to 5 cm.
• In the flange it falls parabolically to zero at the bottom.

Answer: $\tau_{max} = 107.0\ \text{N/mm}^2$ at the neutral axis; average stress on the stem $= 100\ \text{N/mm}^2$, so $\tau_{max}/\tau_{avg} \approx 1.07$.