Transverse loading

Assumptions

1. Elastic, homogeneous material; the flexure formula holds.
2. Shear stress is uniform across the width $b$ at the level considered, and acts parallel to the shear force.
3. The beam is prismatic and the load acts through the shear centre in a plane of symmetry.

Derivation

  M          M+dM
  |<--- dx --->|        cut at y1; shaded area A* above it
  +-------------+  --- top fibre
  |    A*       |
  |-------------|  y1   width b
  |      NA     |
  +-------------+

Take a small element of length $dx$ with bending moments $M$ and $M + dM$ at its faces. The bending stress at a distance $y$ from the NA on the left face is $\sigma = My/I$. Consider the part of the element above the level $y_1$, with area $A^*$.

Force on the left face: $F_1 = \int_{A^*}\dfrac{My}{I}\,dA$. Force on the right face: $F_2 = \int_{A^*}\dfrac{(M+dM)y}{I}\,dA$.

Net unbalanced horizontal force:

$$F_2 - F_1 = \frac{dM}{I}\int_{A^*}y\,dA = \frac{dM}{I}\,Q$$

where $Q = \int_{A^*}y\,dA = A^*\bar y^*$ is the first moment of the area above the level about the NA.

This force is balanced by the horizontal shear force on the cut face of area $b\,dx$:

$$\tau\,b\,dx = \frac{dM}{I}Q \;\Rightarrow\; \tau = \frac{Q}{I\,b}\,\frac{dM}{dx}$$

Since $dM/dx = V$:

$$\boxed{\tau = \frac{V\,Q}{I\,b}}$$

By complementary shear, the same stress acts on the vertical plane. The stress is zero at the top and bottom fibres (where $Q = 0$) and is usually greatest at the neutral axis.

Section property

Overall depth $240$ mm. Moment of inertia about the NA (flanges plus web):

$$I = \frac{150\times240^3 - 130\times200^3}{12} = 86.13\times10^6\ \text{mm}^4$$

(here 130 mm = $(150-20)$ is the total width of the two voids on either side of the web).

First moments of area

• Flange area above the web junction (junction at 100 mm from NA): $A_f = 150\times20 = 3000$ mm$^2$, centroid at $110$ mm:

$$Q_{junction} = 3000\times110 = 330\,000\ \text{mm}^3$$

• At the NA, add the web half-depth $100\times20 = 2000$ mm$^2$ at $50$ mm:

$$Q_{NA} = 330\,000 + 2000\times50 = 430\,000\ \text{mm}^3$$

Shear stresses $\tau = VQ/(Ib)$

The stress jumps from 2.55 to 19.16 MPa at the junction because the width changes from 150 to 20 mm.

 flange  |==|            0 - 2.6 MPa
         |  \
 web     |   )---> 19.2 (junction)
         |   )---> 25.0 (NA, maximum)
         |   )---> 19.2
 flange  |==|

Share of shear force

In the web the stress varies parabolically with $y$ (mm from the NA): $\tau(y) = \dfrac{V}{I\,b}\left[330\,000 + \dfrac{b\,(100^2-y^2)}{2}\right]$. Integrating $\tau\,b\,dy$ over the web (100 mm each side of the NA):

$$V_{web} = \int \tau\,b\,dy = 92.1\ \text{kN} \approx 92\ \%\ \text{of}\ V$$

Check with the simple estimate $\tau_{avg} = V/(\text{web area}) = 100000/(20\times240) = 20.8$ MPa, close to the web values.

Answer: $\tau_{NA} = 24.96$ MPa (maximum), $\tau_{web\,top} = 19.16$ MPa, $\tau_{flange} = 2.55$ MPa; the web carries about 92 % of $V$. The flanges carry the bending moment and the web carries almost all the shear.

Rectangular section ($b\times h$)

At a level $y$ above the NA the area above is $b\left(\frac{h}{2}-y\right)$ with centroid at $\frac12\left(\frac{h}{2}+y\right)$:

$$Q = b\left(\frac h2 - y\right)\cdot\frac12\left(\frac h2+y\right) = \frac b2\left(\frac{h^2}{4}-y^2\right)$$

With $I = bh^3/12$ and width $b$:

$$\tau = \frac{VQ}{Ib} = \frac{6V}{bh^3}\left(\frac{h^2}{4} - y^2\right)$$

This is a parabola: $\tau = 0$ at $y = \pm h/2$ and maximum at $y = 0$:

$$\tau_{max} = \frac{6V}{bh^3}\cdot\frac{h^2}{4} = \frac{3V}{2bh} = 1.5\,\tau_{avg}, \qquad \tau_{avg} = \frac{V}{bh}$$

Solid circular section (diameter $d$, radius $r$)

At a level $y$ above the NA the chord has width $b = 2\sqrt{r^2-y^2}$. The first moment of the segment above $y$ is

$$Q = \int_y^r 2\sqrt{r^2-y_1^2}\;y_1\,dy_1 = \frac23\,(r^2-y^2)^{3/2}$$

With $I = \pi r^4/4$:

$$\tau = \frac{VQ}{Ib} = \frac{V\cdot\frac23(r^2-y^2)^{3/2}}{\frac{\pi r^4}{4}\cdot2\sqrt{r^2-y^2}} = \frac{4V}{3\pi r^4}(r^2-y^2)$$

Maximum at $y = 0$:

$$\tau_{max} = \frac{4V}{3\pi r^2} = \frac43\cdot\frac{V}{\pi r^2} = \frac43\,\tau_{avg}$$

Distributions

  Rectangle            Circle
  |-----|              |  ___  |
  |  \   \             | /   \ |
  |   )   ) max=1.5avg | )   ) | max=1.33avg
  |  /   /             | \___/ |
  |-----|              |       |

Both are parabolic, zero at the top and bottom edges and maximum at the neutral axis.