# AS 5100.2-2017 Clause 18.3 - Worked Example

Tags: as5100

## Problem statement

Calculate the restraint force and moment in a 10m x 1.5m, 40 MPa concrete slab in Canberra, soffit within 8m of ground.

### Material and section properties

Property Description Value Notes
$$E$$ Youngâ€™s modulus $$32800\textrm{ MPa}$$ AS 5100.5-2017 Table 3.1.2
$$\alpha$$ Thermal expansion coefficient $$10 \times 10^{-6} /^{o}C$$ AS 5100.5-2017 Cl 3.1.6
$$T$$ Temperature $$20\ ^{\textrm{o}}\textrm{C}$$ AS 5100.2-2017 Figure 18.3, Note 1
$$b$$ Section breadth $$10000\textrm{ mm}$$
$$d$$ Section depth $$1500\textrm{ mm}$$

### Temperature profile

$\Delta T(y) = \begin{cases} T(1 - \frac{y}{1200})^5 & y \leq 1200 \\ 0 & 1200 < y < 1300 \\ \frac{5}{200}y - 32.5 & y \geq 1300 \end{cases}$

### Restraint stresses

$\sigma_{res}(y) = -E \alpha \Delta T(y)$

### Restraint forces

\begin{align} F_{res} &= \int \sigma_{res} dA \\ &= -E \alpha b \int \Delta T(y) dy \\ &= E \alpha b \Bigg[ \int_{0}^{1200} T(1 - \frac{y}{1200})^{5}\,dy + \int_{1300}^{1500} \frac{5}{200}y - 32.5\,dy \bigg] \\ &= -32800 \times 10.0 \times 10^{-6} \times 10000 \Bigg[ \int_{0}^{1200} 20\bigg(1 - \frac{y}{1200}\bigg)^{5}\, dy + \int_{1300}^{1500} \frac{5}{200}y - 32.5\, dy \Bigg] \\ &= -14.76 \times 10^{6}\textrm{ N} \end{align}

### Restraint moments

$M_{res} = \int \sigma_{res} z \, dA$

where $$z = (750\textrm{ mm} - y)$$ is the distance above the centroid.

\begin{align} M_{res} &= -E \alpha b \Bigg[ \int_{0}^{1200} (750-y)T(1 - \frac{y}{1200})^{5}\,dy + \int_{1300}^{1500} (750-y)\bigg(\frac{5}{200}y - 32.5\bigg)\,dy \bigg] \\ &= -3280 \Bigg[ \int_{0}^{1200} (750 - y)20\bigg(1 - \frac{y}{1200}\bigg)^{5}\, dy + \int_{1300}^{1500} (750 - y)\bigg(\frac{5}{200}y - 32.5\bigg)\, dy \Bigg]\\ &= 6.47 \times 10^{9}\textrm{ N mm} \end{align}