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
\(E\) Young’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.

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

Further reading