
### 1.0 Question

Consider the reinforced concrete podium structure shown in Fig. 1 (a) below. It consists of a cantilever column and deck, supporting a lightweight steel frame structure. The column cross-section is also shown. The characteristic permanent, variable and wind loading is shown in Fig 1 (b).

If the maximum compressive stress in the column should not exceed $15\:N/mm^2$, confirm that the column satisfies the STR limit state.

<Image
	src="/images/courses/fundamentals-of-reinforced-concrete-design-to-eurocode-2/worked-example-2-determining-uls-actions/img1.jpg"
	alt="(a) R.C. podium supporting a lightweight steel frame structure, (b) characteristic loading applied to the R.C. structure. | EngineeringSkills.com"
	width="1000"
	height="1311"
	caption="Fig 1. (a) R.C. podium supporting a lightweight steel frame structure, (b) characteristic loading applied to the R.C. structure."
/>

### 2.0 Solution

This problem requires us to determine the most onerous combination of all three loads and to determine the maximum compressive stress in the column. The compressive stress, $\sigma_{max}$, will arise due to direct compression and bending. However, different combinations of actions will maximise the axial force and bending moment in the column. So, we must consider different combinations and check the maximum stress developed from each one.

If the axial force in the column is $N_{Ed}$ and the bending moment is $M_{Ed}$, we can calculate the maximum compressive stress from,

$$
\sigma_{max} = \frac{N_{Ed}}{A} + \frac{M_{Ed}\:y}{I}
\tag{1}
$$

where $A$ is the cross-sectional area of the column, $y$ is the distance from the section’s neutral axis to the extreme fibre of the cross-section, $y=500\:mm$ in this case, and $I$ is the relevant second moment of area of the section.

#### 2.1 Case 1: maximising bending

We'll start by identifying the design loads that maximise the bending moment in the column and, therefore, the normal stress due to bending. This is achieved by applying

- the maximum loading on the right outstand

- the minimum loading on the left outstand

- the maximum wind loading acting to the right

Referring to the table of partial factors (repeated below again), the maximum vertical loading is calculated as,

$$
\begin{align*}
w_{max} &= \gamma_G\:g_k + \gamma_Q\:q_k\\\\
w_{max} &= 1.35\times 25 + 1.5\times 12.5\\\\
w_{max} &= 52.5\:kN/m
\end{align*}
$$

The minimum vertical loading is calculated as,

$$
\begin{align*}
w_{min} &= \gamma_G\:g_k + \gamma_Q\:q_k\\\\
w_{min} &= 1.35\times 25 + 0\times 12.5\\\\
w_{min} &= 33.75\:kN/m
\end{align*}
$$

<Image
	src="/images/courses/fundamentals-of-reinforced-concrete-design-to-eurocode-2/worked-example-2-determining-uls-actions/img2.jpg"
	alt="Partial factors and combinations for ultimate limit states EQU, STR and GEO (repeated from lecture 4). | EngineeringSkills.com"
	width="1000"
	height="313"
	caption="Fig 2. Partial factors and combinations for ultimate limit states EQU, STR and GEO (repeated from lecture 4)."
/>

Similarly to the previous example, note the use of the single source principle again, forcing a value of $\gamma_G=1.35$ rather than $\gamma_G=1$, as we might initially expect from the table of factors.

As the variable wind loading, $q_{k,wind}$, is an accompanying variable action (we have selected $q_k$ as the leading variable action in this case), we must apply the combination factor, $\psi_0$ from the table in [lecture 6](/course/fundamentals-of-reinforced-concrete-design-to-eurocode-2/actions-on-structures), repeated below. Since this is a wind load, $\psi_0=0.5$.

$$
\begin{align*}
w_{max,w} &= \gamma_Q\:\psi_0\:q_{k,wind}\\\\
w_{max,w} &= 1.5\times 0.5\times 6\\\\
w_{max,w} &= 4.5\:kN/m
\end{align*}
$$

<Image
	src="/images/courses/fundamentals-of-reinforced-concrete-design-to-eurocode-2/worked-example-2-determining-uls-actions/img3.jpg"
	alt="Values of $\psi$ factors for buildings. | EngineeringSkills.com"
	width="1000"
	height="694"
/>

<Caption title="Values of $\psi$ factors for buildings">
	Fig 3. Values of $\psi$ factors for buildings.
</Caption>

Finally, we can't neglect the self-weight of the column itself. If we allow $25\:kN/m^3$ for reinforced concrete, the load imposed by the column self-weight is,

$$
\begin{align*}
P &= 1\times 0.6\times 10 \times 25\\\\
P &= 150\:kN
\end{align*}
$$

Applying $\gamma_G=1.35$, we obtain a ULS force of $P_{max} = 202.5\:kN$. The loading for this case is summarised in Fig. 4.

<Image
	src="/images/courses/fundamentals-of-reinforced-concrete-design-to-eurocode-2/worked-example-2-determining-uls-actions/img4.jpg"
	alt="Ultimate limit state loading for STR load case 1. | EngineeringSkills.com"
	width="1000"
	height="622"
	caption="Fig 4. Ultimate limit state loading for STR load case 1."
/>

The axial load at the base of the column can now be calculated as,

$$
\begin{align*}
N_{Ed} &= (33.75\times 4) + (52.5\times 7) + 202.5\\\\
N_{Ed} &= 705\:kN
\end{align*}
$$

Similarly, the bending moment is,

$$
\begin{align*}
M_{Ed} &= (52.5\times 7\times 3.5) + (4.5\times 10\times 5) -(33.75\times 4\times 2)\\\\
M_{Ed} &= 1241.25\:kNm
\end{align*}
$$

The second moment of area about the bending axis is,

$$
\begin{align*}
I &= \frac{600\times1000^3}{12}\\\\
I &=5\times 10^{10}
\end{align*}
$$

Using Eqn. 1, the maximum compressive stress at the base of the column is therefore,

$$
\begin{align*}
\sigma_{max} &= \frac{705\times 10^3}{1000\times 600} + \frac{1241.25\times10^6 \times 500}{5\times 10^{10}}\\\\
\sigma_{max} &= 1.18 + 12.41 \:N/mm^2\\\\
\sigma_{max} &= 13.59\:N/mm^2\:(<15\:N/mm^2)
\end{align*}
$$

So, load case 1 satisfies the STR ultimate limit state. We can also see that bending has, by far, the bigger impact on the compressive stress, so this load case that sought to maximise the bending moment is likely to be the worst case. However, just to be sure, we will examine a second load case that seeks to maximise the axial load.

#### 2.2 Case 2: maximising direct compression

To generate the loading, we simply need to apply the maximum loading across both cantilever outstands. We can also maintain the maximum wind loading already calculated. The axial load is therefore,

$$
\begin{align*}
N_{Ed} &= 52.5\times 11 + 202.5\\\\
N_{Ed} &= 780\:kN
\end{align*}
$$

The bending moment is,

$$
\begin{align*}
M_{Ed} &= (52.5\times 11 \times 1.5) +(4.5\times 10\times 5) \\\\
M_{Ed} &= 1091.25\:kNm
\end{align*}
$$

Finally, the stress is calculated as,

$$
\begin{align*}
\sigma_{max} &= \frac{780\times 10^3}{1000\times 600} + \frac{1091.25\times10^6 \times 500}{5\times 10^{10}}\\\\
\sigma_{max} &= 1.3 + 10.91 \:N/mm^2\\\\
\sigma_{max} &= 12.21\:N/mm^2\:(<15\:N/mm^2)
\end{align*}
$$

As suspected, case 1 was more onerous than case 2 and both were within the limit specified, so, _based on this simple stress check_, the column satisfies the STR ultimate limit state.


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