Design of Beams and shafts

A prismatic beam has a constant cross-section along its length. Design means choosing a section that keeps the maximum stresses within allowable limits and, where needed, the deflection within the permitted value.

Basic considerations

1. Bending governs most designs. The required section modulus is

$$Z_{req} = \frac{M_{max}}{\sigma_{all}}$$

2. Shear must be checked, especially for short, heavily loaded beams, and for thin webs: $\tau_{max} \le \tau_{all}$.
3. Principal stresses may be critical at points where both $\sigma$ and $\tau$ are large, such as the web-flange junction of an I-beam.
4. Deflection limits (for example span/325 to span/250 for many structural members) and stability against lateral buckling may govern.
5. Economy: choose the section with the least area satisfying $Z \ge Z_{req}$, with depth $>$ width so that the material is placed away from the neutral axis.

Ductile versus brittle materials

Short versus long beams

• Long beams ($L/d$ greater than about 10 to 15): bending governs; the shear stress is small.
• Short beams ($L/d$ small): the shear force is large relative to the moment; the shear stress, bearing at supports and web crippling may govern.

Design steps

1. Find reactions, draw the shear force and bending moment diagrams, get $M_{max}$ and $V_{max}$.
2. Compute $Z_{req} = M_{max}/\sigma_{all}$.
3. Select a section (standard section from tables or fix the proportions $h = 2b$ for timber).
4. Check shear stress at the NA, principal stresses at critical points, and deflection.
5. Revise the section if any check fails.

Reactions and maximum bending moment

          12 kN
  6 kN/m   |
 _vvvvvvvvvvvvvv_
 A                B     L = 5 m

Total load $= 6\times5 + 12 = 42$ kN, so $R_A = R_B = 21$ kN. Maximum shear force $V_{max} = 21$ kN at the supports. Maximum bending moment at mid-span:

$$M_{max} = \frac{wL^2}{8} + \frac{PL}{4} = \frac{6\times25}{8} + \frac{12\times5}{4} = 18.75 + 15 = 33.75\ \text{kN m}$$

Design for bending

$$Z_{req} = \frac{M_{max}}{\sigma_{all}} = \frac{33.75\times10^6}{10} = 3.375\times10^6\ \text{mm}^3$$

With $h = 2b$:

$$Z = \frac{bh^2}{6} = \frac{b(2b)^2}{6} = \frac{2b^3}{3}$$ $$\frac{2b^3}{3} = 3.375\times10^6 \;\Rightarrow\; b^3 = 5.0625\times10^6 \;\Rightarrow\; b = 171.7\ \text{mm}$$

Adopt $b = 175$ mm and $h = 350$ mm.

Check bending stress with the adopted section:

$$Z = \frac{175\times350^2}{6} = 3.573\times10^6\ \text{mm}^3, \qquad \sigma = \frac{33.75\times10^6}{3.573\times10^6} = 9.45\ \text{MPa} < 10\ \text{MPa}\ \checkmark$$

Check for shear

$$\tau_{max} = \frac32\,\frac{V}{bh} = \frac32\times\frac{21\times10^3}{175\times350} = 0.514\ \text{MPa} < 1.0\ \text{MPa}\ \checkmark$$

The shear stress is only about half the allowable value, so bending governs the design (the beam is long, $L/h = 14$).

Answer: provide a 175 mm $\times$ 350 mm section ($\sigma_b = 9.45$ MPa, $\tau_{max} = 0.51$ MPa). Deflection should also be checked.

The point is at a distance $y = 100$ mm from the neutral axis (the junction of web and flange).

Bending stress

$$\sigma = \frac{My}{I} = \frac{60\times10^6\times100}{86.13\times10^6} = 69.7\ \text{MPa (compression at the top, for a sagging moment)}$$

Shear stress

$Q$ of the flange about the NA: $150\times20\times110 = 330\,000\ \text{mm}^3$. Width at this level (web) $b = 20$ mm.

$$\tau = \frac{VQ}{Ib} = \frac{100\times10^3\times330\,000}{86.13\times10^6\times20} = 19.2\ \text{MPa}$$

Principal stresses

Take $\sigma_x = -69.7$ MPa (axial direction of the beam), $\sigma_y = 0$, $\tau_{xy} = 19.2$ MPa.

$$\frac{\sigma_x}{2} = -34.8\ \text{MPa}, \qquad R = \sqrt{34.8^2 + 19.2^2} = 39.75\ \text{MPa}$$ $$\sigma_1 = -34.8 + 39.75 = +4.9\ \text{MPa (tension)}$$ $$\sigma_2 = -34.8 - 39.75 = -74.6\ \text{MPa (compression)}$$

Maximum shear stress $\tau_{max} = R = 39.75\ \text{MPa}$.

Direction: $\tan2\theta = \dfrac{2\tau}{|\sigma_x|} = 0.551$, so $2\theta = 28.8^\circ$ and the compressive principal stress acts at $\theta = 14.4^\circ$ to the beam axis.

Comparison with the extreme fibre

At the extreme fibre ($y = 120$ mm) there is no shear stress and

$$\sigma_{max} = \frac{60\times10^6\times120}{86.13\times10^6} = 83.6\ \text{MPa}$$

The principal stress at the junction ($74.6$ MPa) is smaller than the extreme-fibre stress ($83.6$ MPa), so for this loading the extreme fibre governs. For deep plate girders, or where a large shear force and large moment act together, the junction can govern, so it must be checked.

Answer: $\sigma = 69.7$ MPa, $\tau = 19.2$ MPa; $\sigma_1 = 4.9$ MPa, $\sigma_2 = -74.6$ MPa; $\tau_{max} = 39.8$ MPa; $\sigma_{max,fibre} = 83.6$ MPa.