T-beam with overhang: max bending stress

Assumption: flange 30 cm × 10 cm on top, web 10 cm wide, overall depth 25 cm (web depth 15 cm). Self-weight ignored.

   4 kN/m over A to B              10 kN
 vvvvvvvvvvvvvvvvvvvvvvvvvv          |
 |========================|==========v
 A (hinge)           B (roller)      C
 |<-------- 6 m --------->|<- 2 m -->|

Reactions

Taking moments about A:

$$R_B \times 6 = 4\times 6\times 3 + 10\times 8 \Rightarrow R_B = 25.33\ \text{kN}$$

$R_A = 24 + 10 - 25.33 = 8.67$ kN (check: $\Sigma F_y = 0$).

Bending moments

• At B (hogging): $M_B = -10\times 2 = -20$ kNm.
• Zero shear at $x = R_A/4 = 2.167$ m from A, so the sagging moment is maximum there:

$$M_{max}^{+} = 8.67\times 2.167 - 4\times\frac{2.167^2}{2} = 9.39\ \text{kNm}$$

Section properties

Measure from the base (cm):

$$\bar y = \frac{30\times10\times20 + 10\times15\times7.5}{300+150} = 15.83\ \text{cm}$$ $$I = \frac{30\times10^3}{12} + 300(20-15.83)^2 + \frac{10\times15^3}{12} + 150(7.5-15.83)^2 = 20937.5\ \text{cm}^4$$

Distance to top fibre $= 25 - 15.83 = 9.17$ cm; to bottom fibre $= 15.83$ cm.

Stresses ($\sigma = My/I$)

$$\sigma = \frac{20\times10^6\times 158.3}{20937.5\times10^4} = 15.12\ \text{N/mm}^2$$

Answer: maximum bending stress = 15.12 N/mm² (compressive) at the bottom fibre over support B; the maximum tensile stress is 8.76 N/mm² at the top fibre over B.