Section modulus of fixed beam at collapse

Given data

Fixed beam AB, $L = 9$ m, loads 60 kN at 3 m from A and 120 kN at 3 m from B (6 m from A). $\sigma_y = 250$ MPa, load factor 1.15.

Factored (collapse) loads: $W_1 = 1.15\times60 = 69$ kN at C (3 m from A), $W_2 = 1.15\times120 = 138$ kN at D (6 m from A).

The required plastic section modulus $Z_p$ follows from $M_p = \sigma_yZ_p$ (the section modulus needed for the beam to just collapse under the factored loads).

Possible mechanisms (hinges at the fixed ends and under one load)

The beam is indeterminate to the second degree (3 hinges for a mechanism).

Mechanism 1: hinges at A, C and B (under the 69 kN load). Part AC (3 m) rotates $\theta_1$, part CB (6 m) rotates $\theta_2$ with $3\theta_1 = 6\theta_2$, i.e. $\theta_1 = 2\theta_2$. The hinge rotations are A: $2\theta_2$, C: $3\theta_2$, B: $\theta_2$; the deflection at C is $6\theta_2$ and at D (3 m from B) it is $3\theta_2$.

• Internal work $= M_p(2 + 3 + 1)\theta_2 = 6M_p\theta_2$
• External work $= 69\times6\theta_2 + 138\times3\theta_2 = 828\theta_2$

$$M_p = \frac{828}{6} = 138\ \text{kN·m}$$

Mechanism 2: hinges at A, D and B (under the 138 kN load). Part AD (6 m) rotates $\theta_1$, part DB (3 m) rotates $\theta_2$ with $6\theta_1 = 3\theta_2$, i.e. $\theta_2 = 2\theta_1$. Hinge rotations: A $\theta_1$, D $3\theta_1$, B $2\theta_1$; deflection at D $= 6\theta_1$, at C $= 3\theta_1$.

• Internal work $= M_p(1 + 3 + 2)\theta_1 = 6M_p\theta_1$
• External work $= 69\times3\theta_1 + 138\times6\theta_1 = 1035\theta_1$

$$M_p = \frac{1035}{6} = 172.5\ \text{kN·m}$$

Collapse mechanism and $M_p$

The larger requirement governs (the beam must resist the worst mechanism): $M_p = 172.5$ kN·m, with hinges at A, D and B.

Check (equilibrium): free reactions: $R_A = (69\times6 + 138\times3)/9 = 92$ kN, $R_B = 115$ kN. Free moments: $M_C = 92\times3 = 276$, $M_D = 115\times3 = 345$ kN·m. With the end moments $-M_p$ at A and B, $M_D = 345 - M_p = +M_p$ gives $M_p = 172.5$, and $M_C = 276 - 172.5 = 103.5 \le M_p$, so no section exceeds $M_p$.

Section modulus

$$Z_p = \frac{M_p}{\sigma_y} = \frac{172.5\times10^6\ \text{N·mm}}{250\ \text{N/mm}^2} = 6.9\times10^5\ \text{mm}^3 = 690\ \text{cm}^3$$

Answer: required (plastic) section modulus $Z_p = 6.9\times10^5$ mm³ = 690 cm³ (corresponding to $M_p = 172.5$ kN·m). Since the shape factor of the chosen section is not given, this is the plastic modulus; if the elastic modulus is wanted, use $Z_e = Z_p/S$ for the section selected.