Understanding structural behaviour and stability in response to lateral loading

Stability

In part 2 of this tutorial series on plastic collapse analysis, we explore plastic analysis of portal frames through worked examples.

Plastic Analysis of Frames – A Complete Guide – Part 2 | EngineeringSkills.com

Build on what we learned in part 1 to determine the critical collapse load for portal frame structures

Plastic Analysis of Frames – A Complete Guide – Part 2

This tutorial focuses on plastic analysis methods and determining the collapse load factors for determinate and indeterminate structures.

Plastic Analysis and Plastic Collapse – A Complete Guide – Part 1 | EngineeringSkills.com

Use plastic analysis to determine the collapse load factors for determinate and indeterminate structures.

Plastic Analysis and Plastic Collapse – A Complete Guide – Part 1

In this tutorial we'll explore P-Delta analysis, a geometric non-linearity that can lead to large deflections in slender structures

P-Delta Analysis and Geometric Non-linearity | EngineeringSkills.com

Free Download - The Complete Python Code for this Tutorial

In this tutorial, we'll explore P-Delta analysis, a geometric non-linearity that can lead to large deflections in slender structures

P-Delta Analysis and Geometric Non-linearity


# Structural Analysis and Stability – Asymmetrically Propped Structures

In the [previous post in this two part series](/posts/structural-analysis-and-stability/), we considered a multi-storey structure in which the stability elements were placed symmetrically within the floor plan. If you haven’t read that post, I suggest going back and starting there. In that case, when considered on a 2D plan from above, the centre of mass for the structure (the point through which the resultant wind loading acts) is coincident with the centre of stiffness. So, the loading is being applied through both the centre of mass **and** the centre of stiffness. When force passes through the centre of stiffness of any structure, only translation of that structure will occur, i.e. no rotation occurs.

However, if the stabilising elements are not placed symmetrically (much more realistic!), the centre of stiffness may no longer be coincident with the centre of mass. As a result, twisting or rotation of the structure occurs, about the centre of stiffness. So, our analysis must determine forces transmitted to the stabilising elements due to:

- linear translation of the structure along the line of action of the external force
- rotation of the structure about the centre of stiffness. If we assume small magnitude deflections, these rotations can be resolved into linear translations (which we will discuss below).

To work our way through this analysis we will consider the following amendments to the stabilising elements in last post:

- Shear wall B shortened from the south by 5 m
- The core moved west by 4 m and north by 3 m.

These amendments can be seen below. For this analysis let’s assume the design wind loading is from the south.

<Image
	src="/images/posts/structural-analysis-lateral-stability/Asymmetrical_final-1024x591.png"
	alt="Structural Analysis and Stability - Asymmetric propping"
	width="768"
	caption="Fig 1. Asymmetrically stabilised structure"
	height="443"
/>
## Locating the centre of stiffness (CoS)

The fact that the line of action of the resultant wind load no-longer passes through the CoS, means a torque or moment is induced. To evaluate this moment we must first locate the CoS. This can be done by taking the sum of the ‘_moments of stiffness_‘ (analogous to taking the sum of the moments of a set of forces) about a convenient set of axes. For our calculations we will take the southern and western edges of the floor plate as our x and y axes respectively. We can start by summarising the relative stiffness of each element in both the x (east-west) and y (north-south) direction:

$$
\begin{align*}
\hat{I}_{swa,NS} &= 2.048\times 10^{14}\\
\hat{I}_{swb,NS} &= 1.08\times 10^{13}\\
\hat{I}_{core,NS} &= 7.719\times 10^{14}\\
\hat{I}_{swa,EW} &= 5.12\times 10^{11}\\
\hat{I}_{swb,EW} &= 1.92\times 10^{11}\\
\hat{I}_{core,EW} &= 2.66\times 10^{15}\\
\end{align*}
$$

### Taking moments about y-axis to determine the x-coordinate of the CoS

The x-coordinate of the CoS, $\overline{x}$ is obtained from the following equation:

$$
\overbrace{\Big(\sum \hat{I}_{NS}\Big)}^{\textup{Sum of all } \hat{I}_{NS} }\:\:\bar{x} = \overbrace{\sum_{i=1}^{n}\underbrace{\Big(\hat{I}_{i,NS}\:\:x_i\Big)}_{\textup{moment of stiffness}}}^{\textup{Sum of components}}
\tag{1}
$$

Putting this equation into action we can determine the x-coordinate of the CoS:

$$
\begin{align*}
(9.875\times 10^{14})\:\:\bar{x} &= (2.048\times 10^{14})(0.2)\\
&+(7.719\times 10^{14})(16)\\
&+(1.08\times 10^{13})(39.8)\\
\bar{x}&= 12.98 \:\textup{m (from eastern edge of floor plate)}
\end{align*}
$$

Intuitively this makes sense if we think about the stiffer shear wall A ‘_dragging_‘ the centre of stiffness to left of the core. The less stiff shear wall B is unable to fully balance the effect.

### Taking moments about x-axis to determine the y-coordinate of the CoS

Based on our previous discussion of the shear wall stiffness in the east-west direction we can see by inspection that they will have negligible influence in the CoS y-coordinate, so we would expect to see the CoS in the y-direction lie in the centre of the core, 15 m from the southern edge of the floor plate. We can confirm this as follows:

$$
\begin{align*}
(2.664\times 10^{15})\:\:\bar{y} &= (5.12\times 10^{11})(12)\\
&+(2.66\times 10^{15})(15)\\
&+(1.92\times 10^{11})(14.5)\\
\bar{y}&= 15 \:\textup{m (from southern edge of floor plate)}
\end{align*}
$$

A revised sketch of the floor plate helps visualise our progress so far,

<Image
	src="/images/posts/structural-analysis-lateral-stability/CoS_final-1024x699.png"
	alt="Structural Analysis and Stability - Centre of stiffness"
	width="768"
	height="524"
	caption="Fig 2. Floor plate showing individual element CoS and combined CoS for the structure with coordinates in square brackets."
/>

The asymmetrical arrangement of stability elements induces a moment, $M_{wind}$, given by:

$$
M_{wind} = F_{wind}\times e
\tag{2}
$$

This moment induces a twisting of the structure. The task now is to determine the corresponding force transmitted into each stability element by this twisting effect.

## Forces due to rotation

Consider a rotation through $\theta$ radians about some arbitrary CoS. At a radius $r$ from the CoS, the circumferential displacement is denoted by $s$.

<Image
	src="/images/posts/structural-analysis-lateral-stability/Rotation_to_force-300x135.png"
	alt="Structural Analysis and Stability - Rotation to translation"
	width="300"
	height="135"
/>
<Caption props="">
	Fig 3. Displacement resulting from rotation through an angle $\theta$
</Caption>

However, for small angles, the linear displacement $\Delta$, is approximately equal to $s$. Thus we have:

$$
r\:\theta = s \approx \Delta
\tag{3}
$$

We know from Hooke’s Law that the force $F$ that develops in an element $i$, is the product of the element’s stiffness $k_i$ and displacement, $\Delta_i$, therefore using equation (3), we have:

$$
F_i = k_i\:\Delta_i
\tag{4}
$$

$$
F_i =k_i\:r_i\:\theta
\tag{5}
$$

Now, the moment of resistance offered by $F_i$ in response to an externally applied moment is:

$$
\begin{align*}
M_i &= F_i\:r_i\\
M_i &=k_i\:r_i^2\:\theta
\end{align*}
$$

Since there must be moment equilibrium, i.e. the externally applied moment about the CoS, $M_{wind}$, must be balanced by the individual moments generated by the forces developed in each stability element, $M_i$, we have,

$$
\begin{align*}
M_{wind} &= \sum M_i\\
&=\theta\:\sum k_i\:r_i^2
\end{align*}
$$

Rearranging we have,

$$
\theta = \frac{M_{wind}}{\sum k_i\:r_i^2}
\tag{6}
$$

Substituting equation (6) into equation (5) and rearranging, we obtain the component of force developed in each stability element as a result of the eccentrically applied external wind force (and the moment that creates):

$$
F_i = M_{wind}\:\frac{k_i\:r_i}{\sum k_i\:r_i^2}
\tag{7}
$$

## Evaluating total force transmitted

Returning to our analysis, the calculation procedure hereafter is best summarised in a table, Tb. ??. Note that element stiffness, denoted as $k_i$ in equation (7) is replaced with our more familiar relative stiffness $\hat{I}_i$.

<Image
	src="/images/posts/structural-analysis-lateral-stability/table.png"
	alt="Structural Analysis and Stability - Table of forces"
	width="557"
	height="207"
	caption="Fig 4. Table to determine the distribution of forces due to rotation of the structure."
/>

The force distributed into each stability element is calculated as the combination of force due to translation $F_t$ and rotation $F_r$. For all elements, $F_t$ will act in the direction of the external wind load. We will call this sense of force direction positive. However the sense of $F_r$ will depend on the overall sense of rotation and the position of each element relative to the CoS. So, the core and shear wall B experiences a positive force while shear wall A experiences a negative force. Noting that $M_{wind} = 1152\: \textup{kN}\times 7.02\:\textup{m} = 8087\:\textup{kNm}$, the total force on each element is calculated as follows.

$$
\begin{align*}
F_{core} &= \overbrace{1152\Bigg(\frac{7.719\times 10^{14}}{9.875\times 10^{14}}\Bigg)}^{F_t} +\: \overbrace{8087\Bigg(\frac{2.331\times 10^{15}}{4.827\times 10^{16}}\Bigg)}^{F_r}\\
F_{core} &= 900.5 + 390.5\\
F_{core} &= 1291\:\textup{kN (to the north)}
\end{align*}
$$

$$
\begin{align*}
F_{swa} &= 1152\Bigg(\frac{2.048\times 10^{14}}{9.875\times 10^{14}}\Bigg)-\: 8087\Bigg(\frac{2.617\times 10^{15}}{4.827\times 10^{16}}\Bigg)\\
F_{swa} &= 238.9 - 438.4\\
F_{swa} &= -199.5\:\textup{kN (to the south)}
\end{align*}
$$

$$
\begin{align*}
F_{swb} &= 1152\Bigg(\frac{1.08\times 10^{13}}{9.875\times 10^{14}}\Bigg)+\: 8087\Bigg(\frac{2.9\times 10^{14}}{4.827\times 10^{16}}\Bigg)\\
F_{swb} &= 12.6+48.6\\
F_{swb} &= 61.2\:\textup{kN (to the north)}
\end{align*}
$$

These force components are represented graphically below. Pay particular attention to the orientation of the rotational force components.

<Image
	src="/images/posts/structural-analysis-lateral-stability/Forces-1024x638.png"
	alt="Structural Analysis and Stability - Wind induced forces"
	width="768"
	height="479"
	caption="Fig 5. Forces transmitted into stabilising elements due to translation and rotation of the structure – wind from the south."
/>

For completeness we can repeat the procedure, this time considering wind approaching from the east. We note that the total wind loading acting on the eastern facade is 691.2 kN. The resultant of this force will act at the mid-point of the eastern facade. As a result, the lever arm about the CoS is 3 m, producing a clockwise moment of $M_{wind} = 2073.6$ kNm. Remember that the shear walls provide no resistance to loading in this direction, so the full wind load is transmitted into the core. The components of force transmitted due to rotation are calculated as follows:

$$
\begin{align*}
F_{core} &= 2073.6\Bigg(\frac{2.331\times 10^{15}}{4.827\times 10^{16}}\Bigg)\\
F_{core} &= 100.1\:\textup{kN (to the south)}
\end{align*}
$$

$$
\begin{align*}
F_{swa} &= 2073.6\Bigg(\frac{2.617\times 10^{15}}{4.827\times 10^{16}}\Bigg)\\
F_{swa} &= 112.4 \:\textup{kN (to the north)}
\end{align*}
$$

$$
\begin{align*}
F_{swb} &= 2073.6\Bigg(\frac{2.9\times 10^{14}}{4.827\times 10^{16}}\Bigg)\\
F_{swb} &=12.46\:\textup{kN (to the south)}
\end{align*}
$$

These force components can again be represented graphically,

<Image
	src="/images/posts/structural-analysis-lateral-stability/Forces_east_1000.png"
	alt="Asymmetrically Propped Structures - Force from East | EngineeringSkills.com"
	width="750"
	height="432"
	caption="Fig 6. Forces transmitted into stabilising elements due to translation and rotation of the structure – wind from the east."
/>
## Wall Stresses and Model Validation

We can evaluate the normal stresses that develop in each stability element following the procedure outlined previously using the engineer’s bending equation. To conclude our discussion we will evaluate the stresses that develop in shear wall B as a result of wind from the south. Finally, we will compare the results from our simplified hand analysis with those derived from a finite element computer model of the structure to gauge their accuracy.

### Normal stresses due to bending of shear wall B

Our analysis has indicated that shear wall B resists 61.2 kN of the external wind loading. By following the same procedure outlined previously, we can determine that the lateral load transmitted by a typical floor, $P = 11.13$ kN. The load transmitted by the roof slab is therefore $\approx 5.6$ kN. Thus the bending moment at the base of the wall evaluates to 802.2 kNm. If we assume that the wall bends about a neutral axis that passes through the centroid of the wall cross section, then $y = 1500$ mm and $I$ can be easily evaluated as:

$$
\begin{align*}
I &= \frac{400 \times 3000^3}{12}\\
I &= 9 \times 10^{11}\: \textup{mm}^4
\end{align*}
$$

The maximum values of axial or normal stress that develop at the base of the wall can therefore be estimated as:

$$
\begin{align*}
\sigma_{max} &= \pm \frac{M_{base}\:y}{I}\\
&=\pm \frac{802.2\times 10^{6}\times 1500}{9\times 10^{11}}\\
&=\pm 1.34\:\textup{N/mm}^2
\end{align*}
$$

This value is significantly higher than we observed for a symmetrical arrangement of stability elements. This is due to the fact that our structure now experiences torsion which induces additional forces in the wall. Shear wall B has also been reduced in depth from 8 m to 3 m. This means that for a given magnitude of applied loading, it will develop larger stresses than a deeper wall subjected to the same force.

### Finite Element Analysis

The same task completed above can be carried out using an alternative method of analysis known as finite element analysis. The details of this analysis method are beyond the scope of this post, however we can summarise by saying it is an analysis technique easily implemented using computer algorithms. As such, it is synonymous with analysis software. A well constructed FE model will often give more accurate results in a shorter time and be easily modified allowing for iterative analysis and design.

A word of caution at this point; FE analysis and computer analysis in general is not a substitute for having a solid understanding of the physics and mechanics of a given problem. A good engineer should always have the ability to reduce any problem to a simplified model suitable for hand analysis methods. That said, in a commercial environment, engineers must also be comfortable with computerised methods of analysis and design.

The following analysis was performed in [Autodesk Fusion 360](https://www.autodesk.co.uk/campaigns/fusion-360-for-hobbyists). This is a cloud based 3D modelling and analysis package. If you’re a student you likely have access to this software free of charge. Autodesk’s own tutorial material is excellent.

The image below shows a set of results indicating the qualitative deflection of the structure when wind loading is applied from the south as in our discussion above. The magnitude of deflection has been exaggerated to make it more visible in the deflected shape diagrams, (c) and (d). The heat map indicates the relative magnitude of deflection across the structure from dark blue (no deflection) to red (maximum deflection). As we would expect, deflection at the base of the walls is zero and this increases as we move vertically up through the structure (a-c). The counter-clockwise rotation of the structure is also evident, particularly when viewed from above (d).

<Image
	src="/images/posts/structural-analysis-lateral-stability/disp.jpg"
	alt="Structural Analysis and Stability - Finite Element Analysis"
	width="1000"
	caption="Fig 7. (a & b) Isometric views of the undeformed structure with relative deflection indicated by a heat map. (c) Isometric view showing a heap map on top of the deformed structure. (d) Plan view of the deformed structure with heat map emphasising the rotation of the structure about a CoS."
	height="707"
/>
Is should be noted that in order to determine an accurate estimate of the absolute
deflection of the structure, we would need to provide the software with an accurate
estimate of the *secant modulus* for the concrete. This is approximately analogous
to Young’s modulus and is quite difficult to accurately estimate. For this reason
we are not going to focus on the actual magnitude of the deflection in this analysis.
In a real world design you certainly would want to estimate the deflection.

In the image below we can see how the stresses are distributed within the structure. Image (a) shows the full stress range mapped onto the structure. As expected the largest stresses occur in the vertical stability elements. Imaged (b-d) show successively larger magnitudes of stress removed from the heat-map, leaving only higher magnitudes of stress indicated by the heat map. This allows us to emphasis the areas where the stress is at its highest, at the base of the walls.

<Image
	src="/images/posts/structural-analysis-lateral-stability/Stress_qual.jpg"
	alt="Structural Analysis and Stability - Qualitative stress distribution"
	width="1000"
	caption="Fig 8. (a) heat map indicating magnitudes of Von Mises stress across the structure. (b-d) heat map with successively larger magnitudes of stress removed for clarity."
	height="770"
/>
The stress depicted in the images above is called Von Mises stress. All we’re concerned
with in this visualisation is the qualitative information it provides, i.e. the relative
magnitude and location of stresses in the structure. We can clearly see that the
core and shear wall A are undergoing the most stress. Shear wall A in particular
is developing relatively high levels of stress. As mentioned above, this is due to
its smaller depth and its relatively large distance away from the CoS. This means
it undergoes large translations as a result of the rotation of the structure.

The image below shows us another stress heat map. However the stress indicated is the normal stress in the vertical direction. This is the stress that we have approximated in our hand calculation. We can see that our FE analysis indicates a maximum normal stress at the base of shear wall A of $\approx$1.6 N/mm$^2$. This is in relatively good agreement with our approximation of 1.34 N/mm$^2$. The negative stress on the right hand side of the wall is indicating compression with the positive stress indicating compression where we would expect to see it.

When we consider that both estimates of maximum stress have been determined using very different methods, the level of agreement is quite good. In reality, having an FE model is helpful as it allows for relatively rapid design iterations to be performed, compared to running our hand analysis for every iteration. Having said that, it is good to see that straightforward hand analysis techniques can provide quite accurate results, even if a little more time-consuming to complete.

<Image
	src="/images/posts/structural-analysis-lateral-stability/Stress_quant-1024x649.png"
	alt="Fig 9. Structural Analysis and Stability - Quantitative stress distribution"
	width="768"
	height="487"
/>
<Caption props="">
	Heat map showing normal stress in the vertical direction with maximum stress
	magnitude equal to $1.594 N/mm$.
</Caption>


In this tutorial we’ll consider the lateral stability of an asymmetrically propped multi-storey structure and compare our results to a finite element model

Structural Analysis and Stability – Asymmetrically Propped Structures | EngineeringSkills.com

Lateral stability of asymmetrically propped multi-storey structures with comparison to a finite element model

Structural Analysis and Stability – Asymmetrically Propped Structures


All structures typically experience some form of lateral loading during their design life. Typical sources of lateral loading include forces due to wind blowing against the structure, hydrostatic forces due to groundwater (acting against basement walls for example) or inertia forces due to ground motion (earthquakes).

The designer must consider how to transmit these lateral loads safely back into the foundations of the structure. In the context of building structures, there are several well established structural schemes for facilitating this load transfer. In this, the first of a two-part series on structural stability, we will introduce these schemes before diving into some numerical examples in this and the next post.

## Structural Stability of Braced Structures

Consider a basic structural frame subject to a single lateral load, $P$,

<Image
	src="/images/posts/structural-analysis-and-stability/Bracing1-1024x423.png"
	alt="Structural Analysis and Stability - An unbraced structures"
	width="482"
	height="198"
/>
<Caption props="">
	Fig 1. (Left) Simple frame subject to a lateral load $P$. (Right) Sway
	deflection $\Delta$ resulting from $P$.
</Caption>

Under the action of the load, the frame will deflect to the right by an amount
denoted by $\Delta$. This deflection occurs due to:

- rotation of the structural members relative to each other at joints B and C.
- bending of the vertical members.

One of the simplest ways to reduce the degree to which the frame deflects is to provide a diagonal member, ‘triangulating the frame’. Depending on the orientation of the diagonal bracing member (relative to the direction of the external loading), it may develop a compression force (referred to as a strut) or a tensile force referred to as a tie, So, for any lateral deflection to occur, the diagonal member must get longer or shorter depending on its orientation.

<Image
	src="/images/posts/structural-analysis-and-stability/Bracing2-1024x395.png"
	alt="Structural Analysis and Stability - Strut and Tie bracing"
	width="514"
	caption="Fig 2. (Left) Braced frame with compression developing in the bracing strut. (Right) Braced frame with tension developing in the bracing tie."
	height="198"
/>
The design of a (compression) strut involves the additional consideration of buckling
and so if all else is equal, structural (tension) ties tend to be more efficient.
The preference for ties over struts is apparent when we see ‘cross-bracing’ consisting
of slender structural ties.

<Image
	src="/images/posts/structural-analysis-and-stability/Bracing3-1024x629.png"
	alt="Structural Analysis and Stability - multi-storey tie bracing"
	width="504"
	caption="Fig 3. Cross-bracing in a multi-storey structure utilising only tension members."
	height="309"
/>
In such cases, depending on the direction of loading, one member will be ‘active’
and in tension while the other member is dormant not providing any resistance to
lateral loading. If the external load direction reverses as seen on the right in
the image above, the ties switch roles; the dormant member becoming active and vice-versa.

## Structural Stability of moment or portal frames

When considering our simple frame structure above, for a deflection to occur, we said that the angles between the vertical members and the horizontal member at joints B and C must change. Another way to reduce the amount of lateral frame deflection is to reduce the amount of relative rotation at these joints. In other words, by stiffening the joints and increasing their resistance to rotation, we reduce the overall lateral deflection of the frame.

<Image
	src="/images/posts/structural-analysis-and-stability/Portal1.png"
	alt="Structural Analysis and Stability - portal frame"
	width="360"
	height="330"
	caption="Fig 4. Schematic view of a portal frame with stiffened joints indicated in red"
/>
A structural frame in which the resistance to lateral loading is provided by the
beam to column connections is called a portal frame. Joint stiffness is enhanced
by providing ‘haunches’ that serve to increase the lever arm available at the joint,
(below). Haunches also reduce the effective span of the members between joints.

<Image
	src="/images/posts/structural-analysis-and-stability/Portal2.png"
	alt=""
	width="452"
	caption="Fig 5. Typical symmetrical portal frame."
	height="194"
/>

Portal frame structures are most often encountered
as steel frame structures, e.g. large warehouses, agricultural buildings, hangers.
Portal frames are characterised by their ability to span large distances efficiently.
In doing so they typically make use of plastic hinge formation. Lateral deflection
of portal frames is largely dependent on the stiffness of structural joints and also
the flexural rigidity of vertical (column) members. The result of this is that portal
frames generally experience larger lateral (sway) deflections than braced frames.

## Stability of Propped Structures

The third stabilising mechanism and the one we will explore in more detail in this series of posts is the so-called propped structure scheme, regularly employed in multi-storey frame structures, e.g. reinforced concrete (RC) frame buildings. In multi-storey RC frame buildings it’s common for concrete lift and stair shafts (also known as cores) to act like vertical cantilevers, providing the building’s resistance to lateral loads such as wind.

<Image
	src="/images/posts/structural-analysis-and-stability/mainBraced-1024x910.png"
	alt=""
	width="568"
	caption="Fig 6. Cantilevering concrete cores providing lateral stability in an RC frame structure."
	height="501"
/>

In this case the combined stiffness of the cores is significantly greater than the
lateral stiffness provided by the column-beam connections. As a result the RC walls
of the cores ‘attract’ or resist the majority of the lateral force imposed by wind
for example. Such a building is said to be propped by the cantilevering cores. Vertical
shear walls serve the same function.

Look around any large site with multi-storey structures and there is a strong possibility you will see this stabilising mechanism. Also note that the RC cores are constructed ahead of the floors and tower over the structure below. This is because they provide the stiff points within the structure with all other structural elements (beams and floor slabs) being propped against them.

In order to utilise the RC cores as stabilising elements, the lateral forces (say the wind blowing on the facade) must be transmitted back into the core. This is achieved by utilising the floor slabs (also known as floor plates) as stiff diaphragms. This can be seen in the diagram, above, with the red arrows indicating compression force transmission through the floor slabs back to the cores. Along the line of action of the horizontal compression forces (from the facade), the floor plates are exceptionally stiff and provide an ideal load path back into the cores.

## Stability of a Symmetrically Propped Structure

To develop our understanding of this stability scheme, we will consider the case study structure, pictured below. This RC frame consists of a ground floor (not shown) and 6 floor slabs (shown in green). Indicative finished floor levels (FFL) are also indicated. The stabilising elements are shown in red and consist of two shear walls and a central core. The columns provide for vertical load transmission only and do not contribute to lateral stability. The bottom image also shows a typical floor plan with dimensions indicated. We will assume a design wind loading of 1.2 $kN/m^2$ from the south. Our task is to estimate the stresses induced at the base of the stabilising elements. As you will see, this can be accomplished by utilising quite rudimentary analysis techniques. We will assume all RC walls are 400 mm thick and the core is located centrally on the floor plan.

<Image
	src="/images/posts/structural-analysis-and-stability/RC1_final1-839x1024.png"
	alt="Structural Analysis and Stability - symmetrically propped structure example"
	width="502"
	caption="Fig 7. (Top) A multi-storey reinforced concrete frame building stabilised by shear walls and central core (red). Stiff floor plates (green) provide a load path from facade back into vertical stability elements. (Bottom) Typical floor plan."
	height="612"
/>{" "}

Our first task is to determine the total wind loading acting on the building facade:

$$
\begin{align*}
F_{wind} &= 40\:\textup{m} \times 24\:\textup{m} \times 1.2\:\textup{kN/m}^2\\
&=1152 \:\textup{kN}
\end{align*}
$$

This force must be distributed between the three stabilising elements in proportion to their stiffness or flexural rigidity, $EI$. Since all members are made of the same material, we can focus on their second moment of area:

$$
I=\frac{bd^3}{12}
\tag{1}
$$

Since it is only the relative stiffness that is important we can evaluate only the numerator since the denominator is common to all. This relative stiffness will be denoted as $\hat{I}$. Evaluating the shear wall relative stiffness:

$$
\begin{align*}
\hat{I}_{swa} &= \hat{I}_{swb} = 400\times8000^3\\
&=2.048\times10^{14}\:\textup{mm}^4
\end{align*}
$$

Note that the shear walls act to resist lateral loading in the north-south direction only. In the east-west direction the shear walls have negligible stiffness (in comparison to their stiffness in the north-south direction).

The relative stiffness of the core can be determined by first calculating $\hat{I}$ for the outer enclosing rectangle and then subtracting $\hat{I}$ for the two voids:

$$
\begin{align*}
\hat{I}_{core} &= \hat{I}_{out} - \hat{I}_{in}\\
&=(8000\times6000^3)-2(3400\times5200^3)\\
&=7.719\times10^{14}\:\textup{mm}^4
\end{align*}
$$

After determining the relative stiffness of each active element, we can distribute the lateral load, $F_{wind}$, accordingly:

$$
\begin{align*}
F_{swa} = F_{swb} &= \frac{\hat{I}_{sw}}{\sum \hat{I}}\times F_{wind}\\\\
&=\frac{2.048\times10^{14}}{11.815\times 10^{14}}\times 1152\\\\
&=199.7\:\textup{kN}\approx 200\:\textup{kN}\:\: \textup{say}
\end{align*}
$$

Therefore, $F_{core} = 1152\:\textup{kN} - 2(200) = 752\:\textup{kN}$. Perhaps not surprisingly, the central core is resisting the largest proportion of the external wind loading in the north-south direction.

We can repeat the same exercise now assuming wind acting in the east-west direction. As mentioned above, in their current orientation, the shear walls provide negligible resistance to external loading in this direction. This is confirmed by evaluating the relative stiffnesses:

$$
\begin{align*}
\hat{I}_{swa} = \hat{I}_{swb} &= 8000\times400^3\\&=5.12\times 10^{11}\:\textup{mm}^4\\
\hat{I}_{core}&=6000\times 8000^3 - 2(5200\times 3400^3)\\
&=2.663\times10^{15}\:\textup{mm}^4
\end{align*}
$$

By inspection we can see that the core will resist in excess of 99% of the external loading due to its much larger relative stiffness. By visualising the shear walls in isolation and considering them as simple cantilevers we can easily see the importance of the axis of bending.

<Image
	src="/images/posts/structural-analysis-and-stability/shear_wall_axes_final.png"
	alt="Structural Analysis and Stability - shear wall axes"
	width="513"
	caption="Fig 8. The structure on the left (A), is said to be bending about its minor or weak axis, while the structure on the right is bending about its major or strong axis. Intuitively we would expect (B) to provide significantly more resistance to the load, $P$, than (A). This is what we have seen in the previous example."
	height="154"
/>

Now that we have a model
of the lateral load distribution in the structure, it is a relatively straightforward
exercise to estimate the stresses that develop at the base of the shear walls and
core. By way of demonstration, we will consider the stress that develops at the base
of one of the shear walls in response to wind loading from the south.

Our previous analysis suggests one shear wall resists approximately 17.3% of the wind load or $200 kN$. This is transmitted back into the wall via 6 floor plates, with the top floor (roof) transmitting half the magnitude of any of the lower floors, thus we can say, $5.5\:P=200\:kN$. Therefore $P=36.4\:kN$ and the load distribution can be modelled as shown below.

<Image
	src="/images/posts/structural-analysis-and-stability/shear_wall_moment-751x1024.png"
	alt="Structural Analysis and Stability - load distribution from floors"
	width="358"
	height="486"
	caption="Fig 9. Distribution of floor loading into one shear wall"
/>

By modelling the load distribution onto the wall, we can determine the bending moment generated at the base of the wall, $M_{base}$. It is then a trivial exercise to use the engineer’s bending equation to evaluate the corresponding stress. We can determine the bending moment at the base as the sum of the individual bending moments generated by each force, so;

$$
\begin{align*}
M_{base} &= 36.4\times 4 \\
&+ 36.4\times 8 \\
&+ 36.4\times 12 \\
&+ 36.4\times 16 \\
&+ 36.4\times 20 \\
&+ 18.2\times 24 \\\\
M_{base} &=2620.8\:kNm
\end{align*}
$$

We can evaluate the second moment of area for the wall’s cross section as:

$$
\begin{align*}
I&=\frac{400\times 8000^3}{12}\\
&=1.707\times 10^{13}\: \textup{mm}^4
\end{align*}
$$

Assuming the wall’s neutral axis is coincident with its centroid, the maximum tension and compression stress will occur at a distance of $y=4000\:$mm from the neutral axis, therefore the maximum axial stress that develops at the base of the wall is given by:

$$
\begin{align*}
\sigma_{max} &= \pm \frac{M_{base}\:y}{I}\\
&=\pm \frac{2620.8\times 10^{6}\times 4000}{1.707\times 10^{13}}\\
&=\pm 0.61\:\textup{N/mm}^2
\end{align*}
$$

So our analysis suggests that the maximum tensile stress that will develop is $0.61\: \textup{N/mm}^2$. If we assume the walls are constructed with concrete with a characteristic cylinder strength $f_{ck}=30\:\textup{N/mm}^2$, an approximate estimate of the tensile capacity of the concrete would be $2.5-3.0 \textup{N/mm}^2$. Therefore we would not expect the wind loading to cause tension cracks at the base of the wall. Furthermore, when we factor in the self-weight of the wall and any additional compression stresses due to floor loadings, the tensile stress developed due to wind loading is easily accommodated. A similar exercise can be completed to confirm the acceptability of stresses developed in the core walls.

The [the next post](/posts/structural-analysis-lateral-stability/) we’ll consider how this analysis plays out for asymmetrically propped structures and confirm the results of our hand analysis with a simple FE model analysis.


In this tutorial we'll introduce common lateral stability structural schemes before diving into some numerical examples in this and the next post.

Structural Analysis and Stability – Symmetrical Structures | EngineeringSkills.com

An introduction to common lateral stability structural schemes with numerical examples

Structural Analysis and Stability – Symmetrical Structures


In the [first post in this series](/posts/column-buckling-structures-and-stability/) on Column Buckling, we introduced the fundamental ideas of equilibrium and stability. In the [second post](/posts/column-buckling-equations/) we expanded on these ideas to look at more realistic column structures with various support conditions and distributed stiffness.

In this final post in this series on Column Buckling, we’ll look at more realistic buckling behaviour you’re likely to observe in reality. In particular we’ll explore the behaviour of columns subject to eccentric axial load and columns with an initial deformation, i.e. columns that don’t start out straight.

## 1.0 Columns with eccentric axial load

So far we have assumed that all compression forces on a column are axial forces that have their line of action along the column’s longitudinal axis. As a result of this assumption the column will potentially experience the three states of equilibrium previously discussed:

- Stable equilibrium
- Neutral equilibrium
- Unstable equilibrium

However, in reality a column is not likely to experience such ideal conditions. Compression forces are often not axial forces but forces applied at some eccentricity from the column’s longitudinal axis. This results in quite different failure behaviour that we will investigate here.

### 1.1 Lateral deflection

We’ll set the analysis up by considering a column, pinned at both ends, but with an ‘outstand’ or cantilever protruding from each end. The cantilever is somewhat unrealistic but it simply serves here as a way of applying an eccentric compression force. The compression forces are applied at the ends of these cantilevers as shown below at an eccentricity, $e$:

<Image
	src="/images/posts/column-buckling-realistic-buckling-behaviour/img-1.png"
	alt="Column Buckling with eccentric compression force"
	width="700"
	height="606"
	caption="Fig 1. Column Buckling with eccentric compression force"
/>

The first thing we note is that the load $P$ applied an an eccentricity $e$, is the same as simultaneously applying an axial load $P$ and a moment, $M_o$,

$$
M_o=P\times e
$$

We can start by following the same procedure as before. We need to determine the differential equation of the deflection curve. We do this by evaluating the internal bending moment, $M_x$ at some distance $x$ along the height of the deflected column as shown in the right hand diagram above.

$$
M_x = M_o + P(-v)
$$

$$
M_x = M_o-Pv
$$

Now substituting this expression for $M_x$ into the differential equation of the deflection curve yields,

$$
EI\frac{\mathrm{d}^2v}{\mathrm{d}x^2} = M_o-Pv = Pe-Pv
$$

If we make the following substitutions,

$$
k^2 = \frac{P}{EI} \:\:\: and\:\:\:\frac{\mathrm{d}^2v}{\mathrm{d}x^2} = \ddot{v}
$$

We can express our differential equation as,

$$
\ddot{v} + k^2v = k^2e
$$

The solution of this equation follows very closely that of the fixed-free column discussed in the previous post. The general solution is given by,

$$
v=C_1 \sin(kx) + C_2 \cos(kx) + e
$$

We can apply the following boundary conditions:

$$
v(x=0) = 0 \:\:\: and \:\:\: v(x=L) = 0
$$

Using these boundary conditions we find that,

$$
C_2 = -e
$$

and,

$$
C_1 = \frac{-e(1-\cos(kL))}{\sin(kL)}
$$

Which simplifies to,

$$
C_1 = -e\tan \left(\frac{kL}{2}\right)
$$

The solution to the differential equation of the deflection curve is therefore,

$$
v=-e\left(\tan\left(\frac{kL}{2}\right)\sin(kx) + \cos(kx)-1\right)
$$

We note that for a given value of $e$ and $P$, the deflection is always defined. In the previous axially loaded column analyses we considered, we could only determine a buckling mode shape, but the maximum deflection, when $P=P_{cr}$, was undefined. This corresponded to a state of neutral equilibrium.

<Callout type="info">
	📌 From this we conclude that a column with eccentric compression forces has no
	neutral equilibrium state and therefore does not exhibit sudden column buckling
	behaviour.
</Callout>

### 1.2 Maximum deflection

For the pinned-pinned column considered in this case, the maximum lateral deflection, $\delta$, occurs at the mid-height point, $x=L/2$,

$$
\delta=-v\left(\frac{L}{2}\right)
$$

So, evaluating our expression for $v$ at $x=L/2$, gives,

$$
\delta=e\left( \tan\left(\frac{kL}{2}\right)\sin\left(\frac{kL}{2}\right) + \cos\left(\frac{kL}{2}\right)-1\right)
$$

After simplifying the expression we get,

$$
\delta = e\left(\sec\left(\frac{kL}{2}\right)-1\right)
$$

Now, remembering that,

$$
k=\sqrt{\frac{P}{EI}}
$$

and that for a pinned-pinned column we know that,

$$
P_{cr} = \frac{\pi^2EI}{L^2}
$$

we can rewrite $k$,

$$
k=\frac{\pi}{L}\sqrt\frac{P}{P_{cr}}
$$

Therefore,

$$
kL = \pi\sqrt{\frac{P}{P_{cr}}}
$$

We can now substitute this expression back into our equation for the maximum deflection at the mid-height to yield,

$$
\delta=e\left[\sec\left(\frac{\pi}{2}\sqrt{\frac{P}{P_{cr}}}\right)-1\right]
$$

We can plot this equation to obtain a load versus deflection curve for various values of load eccentricity, $e$. This relationship is represented qualitatively below...

<Image
	src="/images/posts/column-buckling-realistic-buckling-behaviour/img-2.png"
	alt="Column Buckling with eccentric compression force; Load versus maximum lateral deflection"
	width="700"
	height="468"
	caption="Fig 2. Column Buckling with eccentric compression force; Load versus maximum lateral deflection"
/>

<Callout type="warning">
	📌 We can observe from this graph that the relationship between lateral
	deflection at mid-height and applied load is non-linear. This means we cannot
	use the principle of superposition to determine the influence of multiple
	simultaneously applied loads.
</Callout>

The vertical line in the graph above, represents the case when $e=0$. In this case we observe the equilibrium states discussed previously. For values of $e>0$, as $P$ approaches the critical load, the deflection increases and the horizontal line representing the value of critical load becomes an asymptote for the load-deflection curves.

Remember that the preceding derivation is based on the assumptions of small deflections. So in reality as the deflections increase, the observed behaviour of the column will deviate form the strict mathematically predicted behaviour visualised in the graph above.

> _💡A key takeaway from this discussion is that the behaviour of a real world column under realistic loading conditions is not likely to accord with the strict mathematical models we evaluated previously for perfectly axially loaded columns._

## 2.0 Columns with initial deformation

Now we’re going to consider the behaviour of a column that already has an initial lateral deformation. We will follow the same procedure as before to determine an equation that describes the shape of the column under axial load.

### 2.1 Governing differential equation

Consider the pinned-pinned column below that has an initial lateral deflection, $v_0$:

<Image
	src="/images/posts/column-buckling-realistic-buckling-behaviour/img-3.png"
	alt="Column Buckling: Column with an initial deformation"
	width="350"
	height="308"
	caption="Fig 3. Column Buckling: Column with an initial deformation"
/>

Note that the complete lateral deflection consists of the initial deflection and the buckled deflection:

$$
v(x) = \bar{v}(x) + v_o(x)
$$

By considering the equilibrium of a sub-section of the structure we can determine an equation for the internal moment of resistance at some position along the column length:

<Image
	src="/images/posts/column-buckling-realistic-buckling-behaviour/img-4.png"
	alt="Column Buckling: Column with an initial deformation; Internal bending moment"
	width="200"
	height="339"
	caption="Fig 4. Column Buckling: Column with an initial deformation; Internal bending moment"
/>

Note that for clarity, hereafter we’ll dispense with the $(x)$ as it’s clear that the lateral deflection is always a function of $x$. Taking moments about point A yields,

$$
M=-Pv
$$

$$
M=-P(\bar{v} + v_0)
$$

Substituting this expression for $M$ into the differential equation of the deflection curve yields,

$$
EI\frac{\mathrm{d}^2\bar{v}}{\mathrm{dx^2}}=-P\bar{v} - Pv_0
$$

Making the usual simplifying substitutions and letting,

$$
\frac{\mathrm{d}^2\bar{v}}{\mathrm{dx^2}}=\ddot{\bar{v}}
$$

and

$$
k^2=\frac{P}{EI}
$$

we have,

$$
\ddot{\bar{v}} + k^2\bar{v} = -k^2v_0
$$

To proceed with the derivation we need to assume some function of $x$ to describe the initial deformation of the column. For the purposes of this derivation we can assume that a simple sine function describes the initial deformation $v_0$. In this case we have the following differential equation of the deflection curve,

$$
\ddot{\bar{v}} + k^2\bar{v} = -k^2V_0\sin\left(\frac{\pi x}{L}\right)
$$

We’ve already seen that the complementary solution (i.e. the solution when the RHS equals zero) is given by,

$$
\bar{V}_c = C_1 \sin(kx) + C_2 \cos(kx)
$$

This just leaves the particular solution to be calculated. Because the right hand side of the differential equation contains a sinusoid, we can assume a general sinusoid for the particular solution, this gives us…

$$
\bar{v}_p=C_3\sin\left(\frac{\pi x}{L}\right)+C_4\cos\left(\frac{\pi x}{L}\right)
$$

Remember that the reason we assume a solution is so that we can differentiate it and sub it back into our governing differential equation, so differentiating yields,

$$
\dot{\bar{v}}_p = C_3\frac{\pi}{L}\cos\left(\frac{\pi x}{L}\right) - C_4\frac{\pi}{L}\sin\frac{\pi x}{L}
$$

$$
\ddot{\bar{v}}_p = -C_3\frac{\pi^2}{L^2}\sin\left(\frac{\pi x}{L}\right) - C_4\frac{\pi^2}{L^2}\cos\frac{\pi x}{L}
$$

Now substituting these expressions back into our governing differential equation yields,

$$
\frac{-\pi^2}{L^2}\left[C_3\sin\left(\frac{\pi x}{L}\right)+C_4\cos\left(\frac{\pi x}{L}\right)\right]+k^2\left[C_3\sin\left(\frac{\pi x}{L}\right) + C_4\cos\left(\frac{\pi x}{L}\right)\right]=-k^2V_0\sin\left(\frac{\pi x}{L}\right)
$$

Now, if we equate cosine terms on the left hand side with cosine terms on the right hand side, we get,

$$
C_4\left[\frac{-\pi^2}{L^2}\cos\left(\frac{\pi x}{L}\right)+k^2\cos\left(\frac{\pi x}{L}\right)\right]=0
$$

Now since we know the term in brackets doesn’t equal zero, we can deduce that $C_4=0$. Now equating sine terms on both sides gives us,

$$
C_3\left[\frac{-\pi^2}{L^2}\sin\left(\frac{\pi x}{L}\right) + k^2\sin\left(\frac{\pi x}{L}\right)\right]=-k^2 V_0\sin\left(\frac{\pi x}{L}\right)
$$

Rearranging this yields,

$$
C_3=\frac{-k^2V_0}{\left(\frac{-\pi ^2}{L^2}+k^2\right)}
$$

We can now state the particular solution to the equation as,

$$
\bar{v}_p =\frac{-k^2}{\left(\frac{-\pi ^2}{L^2}+k^2\right)}V_0\sin\left(\frac{\pi x}{L}\right)
$$

Combining the particular and complimentary solutions gives us the general solution,

$$
\bar{v} = C_1\sin(kx) + C_2 \cos(kx) + \frac{k^2}{\left(\frac{\pi ^2}{L^2}-k^2\right)}V_0\sin\left(\frac{\pi x}{L}\right)
$$

### 2.2 General solution and boundary conditions

At this point we can apply boundary conditions to determine the remaining unknown constants of integration. At $x=0$ (base of the column at pin support), $v=0$ (the lateral deflection must equal zero). Imposing this condition on our general solution we find that $C_2=0$. And so our general solution simplifies to,

$$
\bar{v} = C_1\sin(kx) + \frac{k^2}{\left(\frac{\pi ^2}{L^2}-k^2\right)}V_0\sin\left(\frac{\pi x}{L}\right)
$$

The second boundary condition is that at $x=L$ (top of the column at pin support), $v=0$ (lateral deflection also zero). Again, imposing this condition yields,

$$
C_1\sin\left(\sqrt{\frac{P}{EI}}L\right)=0
$$

From this we can deduce that $C_1$ must equal zero as the sine term cannot equal zero for a non-trivial solution. Therefore the complete buckling deflection is given by,

$$
\bar{v} = \frac{\frac{P}{EI}}{\left(\frac{\pi^2}{L^2}-\frac{P}{EI}\right)}V_0\sin\left(\frac{\pi x}{L}\right)
$$

Now to make this a little easier to manage, remember that,

$$
P_E=\frac{\pi^2EI}{L^2}
$$

If we now define the ratio,

$$
\rho = \frac{P}{P_E}
$$

we can restate the buckling deflection as,

$$
\bar{v} = \left(\frac{\rho}{1-\rho}\right)V_0\sin\left(\frac{\pi x}{L}\right)
$$

Therefore the total deflection (remember this is the initial deflection plus the buckling deflection) is given by,

$$
v=V_0\sin\left(\frac{\pi x}{L}\right) + \frac{\rho}{1-\rho}V_0\sin\left(\frac{\pi x}{L}\right)
$$

We can simplify this to,

$$
v=\frac{V_0}{1-\rho}\sin\left(\frac{\pi x}{L}\right)
$$

or,

$$
v=\alpha V_0\sin\left(\frac{\pi x}{L}\right)
$$

where,

$$
\alpha = \frac{1}{1-\rho}
$$

### 2.3 Observations

We can see from the equation derived above that the factor $\alpha$ represents a magnification factor on the initial displacement. As the axial load, $P$ increases, the magnitude of lateral deflection increases but the deflected shape remains the same. Note that when $P=P_E$, the equation breaks down as the magnification factor $\alpha$ goes to infinity.

This is not likely to be a problem because $P$ is not likely to reach $P_E$ while the structure satisfies our small deflection assumption. We can plot the relationship between $\rho$ and $\alpha$ to visualise the behaviour of the column.

<Image
	src="/images/posts/column-buckling-realistic-buckling-behaviour/img-5.png"
	alt="Column Buckling: Column with an initial deformation: load versus deflection"
	width="700"
	height="449"
	caption="Fig 5. Column Buckling: Column with an initial deformation: load versus deflection"
/>

It’s important to recognise that for a column with an initial deformation, we do not observe the strict mathematical column buckling behaviour predicted for perfectly loaded perfectly straight columns. Even so, the Euler load is still an important quantity that has a role in predicted the lateral deflection of the column via the ratio $\rho$.

---


Real world columns do not always exhibit the strict mathematical column buckling behaviour predicted for perfectly loaded, perfectly straight columns.

Column Buckling - Realistic Buckling Behaviour | EngineeringSkills.com

Download our Guide to Column Buckling - eBook

Real-world columns rarely exhibit the strict mathematical buckling behaviour predicted for perfectly loaded, perfectly straight columns

Column Buckling - Realistic Buckling Behaviour


In the [previous post in this series](/posts/column-buckling-structures-and-stability/), we introduced some core concepts using a simplified idealised structure. In this post we’ll start to consider more realistic structures and determine the column buckling equations. In particular we’ll determine an expression for a critical load for an axially loaded column with pinned ends. Then we’ll explore some other support conditions.

We’ll cover quite a lot in this one, so here’s a table of contents…

1. **Columns pinned at both ends**
   - Establishing the differential equation
   - Solving the differential equation
   - Solving for the constants of integration
2. **Buckling modes and mode shapes**
   - Maximum deflection under critical load
   - Higher order buckling modes
3. **Columns with other support conditions**
   - Fixed-Fixed
   - Fixed-Free
   - Fixed-Pinned
4. **Axis of buckling**
5. **Critical stress**
6. **Column effective length**

When we refer to a pin support, this is a support that offers no resistance to rotation. Sometimes we come across pin supports in practice that very closely approximate this theoretical definition. Here are some examples:

<Image
	src="/images/posts/column-buckling-equations/img-1.png"
	alt="Column Buckling Behaviour - Pinned supports"
	width="800"
	height="719"
/>

For the purposes of our discussion here, we’ll assume we’re dealing with pin supports like these, and that they don’t inhibit the rotation of the structure.

## 1.0 Buckling equation for columns pinned at both ends

In order to analyse this structure, we’re going to use the differential equation of the defection curve in which $M$ is the internal bending moment at a given cross-section, $v$ represents the lateral deflection of the column and $x$ the distance along the length of the column. $EI$ is referred to as the flexural rigidity and is the product of Young’s modulus $(E)$ and the second moment of area of the cross-section $(I)$.

$$
EI\frac{\mathrm{d}^2v}{\mathrm{dx^2}}=M
$$

> _💡This equation simply models the deflected shape of the column for a given flexural rigidity._

You’re likely to have come across this equation elsewhere in your engineering studies. Our laterally deflected column behaves in much the same way as a beam. Remember buckling is a bending failure rather than a failure due to direct compression (crushing). Our task now, is to use this equation to derive an expression for the critical axial load in a pinned-end column.

<Callout type="info">
	📌 Just as for the idealised structure discussed in the previous post, our
	derivation here will also assume all deflections are small by comparison to the
	size of the structure.
</Callout>

### Establishing the differential equation

First consider the structure, its deflected shape and the free-body diagram that results from cutting the structure at a distance $x$ from the bottom support…

<Image
	src="/images/posts/column-buckling-equations/img-2.png"
	alt="Column Buckling Behaviour - Pinned supports"
	width="700"
	height="535"
/>

Taking moments about point A yields:

$$
M=-Pv
$$

Substituting $M$ back into the differential equation of the deflection curve and rearranging slightly yields:

$$
\frac{\mathrm{d}^2v}{\mathrm{d}x^2} + \frac{P}{EI}v=0
$$

This is a linear, homogeneous, 2nd order differential equation with constant coefficients. It also happens to be an equation that models the deflected shape of our structure and contains $P$, the axial load. Following the same approach as for the idealised structure discussed previously, if we solve this equation we can determine $P_{cr}$, the value of applied load at which it is balanced by the column’s flexural bending resistance.

So we’re faced with trying to solve a differential equation. At this point, it’s helpful to be able to point to the general solution of this form of differential equation:

$$
v(x) = C_1 \sin\left(\sqrt{\frac{P}{EI}}x\right) + C_2\cos\left(\sqrt{\frac{P}{EI}}x\right)
$$

### Solving the differential equation

If you want to take a detour at this point to see how the differential equation is solved, watch the video below. Otherwise, we’ll continue working out the constants of integration in the next section.

<Youtube id="es8keQYFnws" />

### Solving for the constants of integration

Now that we have the general solution for this format of equation, we need to solve for the constants of integration using the boundary conditions (pinned-pinned) for our column.

**Boundary condition 1:**

At $x=0$ (base of the column at pin support), $v=0$ (the lateral deflection must equal zero. Imposing these conditions on our general solution yields:

$$
0=C_1\sin(0) + C_2\cos(0)
$$

Therefore

$$
C_2=0
$$

As a result our solution reduces to:

$$
v(x)=C_1\sin \left(\sqrt{\frac{P}{EI}}x\right)
$$

**Boundary condition 2:**

At $x=L$ (top of the column at pin support), $v=0$ (lateral deflection also zero). Again, imposing these boundary conditions yields,

$$
0=C_1\sin\left(\sqrt{\frac{P}{EI}}L\right)
$$

So, either $C_1$ equals zero or $\sin(...)$ equals zero. If $C_1$ equals zero, the equation is satisfied for any value of $P$. This doesn’t really move us forward…we can think of this solution as the trivial solution. Of more interest is the case where:

$$
\sin\left(\sqrt{\frac{P}{EI}}L\right)=0
$$

Knowing what we know about $\sin$ functions, this equation can only be true, when,

$$
\sqrt{\frac{P}{EI}}L = 0, \pi, 2\pi,...n\pi
$$

We can dismiss the case where,

$$
\sqrt{\frac{P}{EI}}L = 0
$$

In this scenario, $P$ would need to equal zero (as all other terms have non-zero values by definition) and we know this is not the case. So for a pinned-pinned column, our solution is:

$$
\sqrt{\frac{P}{EI}}L = n\pi
$$

where $n=1,2,3$… Finally, we can rearrange this equation to make $P$ the subject of the equation:

$$
P=\frac{n^2\pi^2EI}{L^2}
$$

For $n=1,2,3,$…

<Callout type="info">
	📌 This equation represents an infinite series of buckling loads. The lowest
	one $(n=1)$ is the critical buckling load, also known as the **Euler Buckling
	Load** $P_E$.
</Callout>

$$
P_E = \frac{\pi^2EI}{L^2}
$$

## 2.0 Buckling modes and mode shapes

So far, we have established that there is an infinite series of buckling loads and the lowest one is the critical one and called the **Euler Buckling load**. This raises the question of what do the larger buckling loads correspond to?

These are loads that correspond to higher modes of buckling. Each mode of buckling has a corresponding buckled shape. To explore this further, refer back to our general solution after we established that $C_2=0$:

$$
v(x)=C_1\sin \left(\sqrt{\frac{P}{EI}}x\right)
$$

Application of our second boundary condition resulted in us establishing the relationship:

$$
\sqrt{\frac{P}{EI}}L = n\pi
$$

Combining these two equations yields:

$$
v(x)=C_1\sin\left(\frac{n\pi x}{L} \right)
$$

for $n=1,2,3,$… This equation describes the deflected shape of the column for each value of the buckling load $(n=1,2,3,...)$ defined above. Only when the axial load has one of the values given by the equation (previously defined)…

$$
P=\frac{n^2\pi^2EI}{L^2}
$$

…can the column have a bent shape and be in equilibrium. For any other load, the column would be straight and in a state of STABLE equilibrium $(PP_{cr})$. We recall this concept from our previous discussion or idealised column structures in the last post:

<Image
	src="/images/posts/column-buckling-equations/img-3.png"
	alt="Column Buckling Behaviour - Pinned supports"
	width="700"
	height="389"
/>

### Maximum deflection under critical load

Our equation for the deflected or buckled shape of a column contains the parameter $C_1$, left over from our general solution to the differential equation. We now recognise that $C_1$ represents the magnitude of the sine wave or in this context, the maximum lateral deflection for the deflected/buckled deflection (defined above):

$$
v(x)=C_1\sin\left(\frac{n\pi x}{L} \right)
$$

If we now recall the states of equilibrium, when the column is in its buckled shape $(P=P_{cr})$ and in a state of neutral equilibrium, the maximum lateral deflection can have any value that still satisfies our small deflection assumption. We can represent the value of $C_1$ graphically…

<Image
	src="/images/posts/column-buckling-equations/img-4.png"
	alt="Column Buckling Behaviour - Pinned supports"
	width="700"
	height="473"
/>

The value of $C_1$ is therefore mathematically undefined, it can have any ‘small’ value. Fortunately this doesn’t cause us any practical difficulty, as mentioned previously, the state of neutral equilibrium is not something we observe in civil engineering practice. For all intents and purposed, to a civil or structural engineer, a column experiencing $P_{cr}$ is considered to have failed. This theoretical state of neutral equilibrium is also known as a bifurcation point.

### Higher order buckling modes

Imagine an axially loaded  pinned end column that is restrained laterally at its mid-height point (by a floor slab say). The column would be prevented from buckling under the first critical (Euler buckling) load due to the lateral restraint. The floor slab restraint literally holds the column and stops it from buckling.

As a result of this restraint, the column can carry more load, until it reaches the second buckling load $(n=2)$. As the column is not restrained against buckling in the second mode, it would now buckle (or theoretically enter into a state of neutral equilibrium).

> *💡Adequate lateral restraint increases a column’s resistance to buckling by ‘*closing off*‘ the possibility of buckling in lower modes and ‘*unlocking’_ higher buckling loads._

We can easily visualise the first three modes of buckling simply by evaluating the load and deflected shape equations for $n=1,2,3$.

<Image
	src="/images/posts/column-buckling-equations/img-5.png"
	alt="Column Buckling Behaviour - Pinned supports"
	width="700"
	height="453"
/>

**Mode**

$$
n=1
$$

$$
n=2
$$

$$
n=3
$$

**Critical Load**

$$
P_{cr} = \frac{\pi^2EI}{L^2}
$$

$$
P_{cr} = \frac{4\pi^2EI}{L^2}
$$

$$
P_{cr} = \frac{9\pi^2EI}{L^2}
$$

**Buckled Shape**

$$
v(x) = C_1\sin\left(\frac{\pi x}{L}\right)
$$

$$
v(x) = C_1\sin\left(\frac{2\pi x}{L}\right)
$$

$$
v(x) = C_1\sin\left(\frac{3\pi x}{L}\right)
$$

<Callout type="info">
	📌 For any given mode, $P_{cr}$ represents the practical upper limit on axial
	load. Natural imperfections, for example column ‘out of plumb’ or poor load
	alignment will almost certainly induce buckling when the critical load is
	exceeded.
</Callout>

## 3.0 Columns with other support conditions

So far we’ve looked at the behaviour of a column pinned at both end. This is the typical starting point. But the same process can be followed to determine the corresponding equations for columns with different types of support conditions. In the videos below, I’ll carry out the derivations to demonstrate the process for:

- Columns with a base fixed against rotation and free at the top (fixed-free)
- Columns fixed against rotation at both ends (fixed-fixed)
- Columns fixed against rotation at the base and pinned at the top (fixed-pinned)

### Buckling equation for columns restrained against rotation at both ends (fixed-fixed)

<Youtube id="0Yc4irFtsGE" />

### Buckling equation for columns restrained against rotation at the bottom and free at the top (fixed-free)

<Youtube id="DZnFuWCF7xU" />

### Buckling equation for columns restrained against rotation at the bottom and pinned at the top (fixed-pinned)

<Youtube id="MR8CjOw7g8I" />

### Support condition summary

<Image
	src="/images/posts/column-buckling-equations/img-6.png"
	alt="Column Buckling Behaviour - Pinned supports"
	width="500"
	height="831"
/>

## 4.0 Axis of Buckling

We recall from the equation for the buckling load that it is a function of I, the second moment of area of the cross-section:

$$
P=\frac{n^2\pi^2EI}{L^2}
$$

So for a given cross-section, a column will always buckle about the axis with the lower second moment of area, the ‘weaker’ axis. This assumes that both axes have equal restraint. Consider the case of a universal column (UC) section under compression:

<Image
	src="/images/posts/column-buckling-equations/img-7.png"
	alt="Column Buckling Behaviour - Pinned supports"
	width="500"
	height="456"
/>

Since the X-X axis, is the major principle axis (with the largest value of second moment of area – you might remember this from your study of Mohr’s circle), it is the stronger axis and so the column will buckle about the minor principle or Y-Y axis first.

When considering the buckling load for a column structure, the cross-section shape plays a key role, you should evaluate the major and minor principal axes to determine the critical axis for buckling.

## 5.0 Critical Stress

<Callout type="info">
	📌 The critical stress is the average axial stress in a cross-section under the
	critical load
</Callout>

$$
\sigma_{cr}=\frac{P_{cr}}{A}=\frac{\pi^2EI}{AL^2}
$$

We now define the **Radius of Gyration**, as:

$$
r=\sqrt{\frac{I}{A}}
$$

We can therefore write the critical stress as:

$$
\sigma_{cr}=\frac{\pi^2E}{\left(\frac{L}{r}\right)^2}
$$

Finally we define the **Slenderness Ratio** as:

$$
\lambda=\frac{L}{r}
$$

The slenderness ratio is a very useful measure of a column’s geometry and susceptibility to buckling. A high slenderness ratio indicates greater susceptibility to buckling. The slenderness ratio should be determined separately for each principle axis. We can get an intuition for the slenderness ratio by visualising both extremes,

<Image
	src="/images/posts/column-buckling-equations/img-8.png"
	alt="Column Buckling Behaviour - Pinned supports"
	width="500"
	height="465"
/>

The equation for critical stress can therefore be written as a function of the slenderness ratio as follows,

$$
\sigma_{cr}=\frac{\pi^2E}{\lambda^2}
$$

Plotting the critical stress versus slenderness ratio for a given value of Young’s modulus yields an **Euler Curve** showing the safe range of average axial stress for a given slenderness ratio (or the safe range of slenderness ratio values for a given axial stress),

<Image
	src="/images/posts/column-buckling-equations/img-9.png"
	alt="Column Buckling Behaviour - Pinned supports"
	width="500"
	height="382"
/>

<Callout type="info">
	📌 An Euler curve is only valid for critical stresses below the material yield
	stress
</Callout>

## 6.0 Column Effective Length

The final concept we will look at in this post is column effective length. The effective length of a column, $L_e$, is the length between points of inflection on the deflected curve/shape (even if the shape must be extended until a point of inflection is reached – discussed below). For a column pinned at both ends, the effective length is simply the full length of the column between pin restraints:

<Image
	src="/images/posts/column-buckling-equations/img-10.png"
	alt="Column Buckling Behaviour - Pinned supports"
	width="200"
	height="366"
/>

Consider a column with a fixed support at its base and no lateral restraint at the top (fixed-free). At the base, no rotation can occur. In order to obtain two inflection points on the deflected shape, it must be extended…

<Image
	src="/images/posts/column-buckling-equations/img-11.png"
	alt="Column Buckling Behaviour - Pinned supports"
	width="300"
	height="435"
/>

The ‘imaginary’ red portion of the deflected shape is the extension required to provide inflection point number 2. Thus the effective length for a fixed-free column is 2L.

<Callout type="info">
	📌 Another way to think about effective length is that it’s the length of an
	equivalent pinned end column.
</Callout>

Before concluding our discussion of effective length, we can define a general expression for the critical load as:

$$
P_{cr}=\frac{\pi^2EI}{{L_e}^2}
$$

If we now introduce the **Effective Length Factor**, $K$, such that $L_e=Kl$, we can state the critical load as a function of the effective length factor:

$$
P_{cr}=\frac{\pi^2EI}{{(KL)}^2}
$$


In this post we'll determine column buckling equations for axially loaded column with different end conditions including pinned-pinned supports.

Column Buckling Equations | EngineeringSkills.com

Determine column buckling equations for axially loaded column with different end conditions

Column Buckling Equations


Long slender structural elements under the action of an axial load may fail due to buckling rather than direct compression. Buckling failure occurs when axial load induces a lateral deflection leading to a bending type failure. Buckling can also occur in plate and shell structures and is a relatively common cause of structural collapse. Depending on the geometry of the structural element, buckling can occur long before the material yields.

> 💡Columns rarely fail in direct compression, buckling usually occurs first

## The Idealised Structure

<Image
	src="/images/posts/column-buckling-structures-and-stability/img-1.png"
	alt="Column Buckling Structures and Stability - Pinned supports"
	width="400"
	height="487"
/>

As a starting point to understanding buckling behaviour – let’s consider an idealised (simplified) structure that we can use as the basis for mathematical modelling. Our structure consists of 2 rigid bars connected by a rotational spring…

**Note the following about the structure:**

- The rotational spring has a stiffness of $\beta_R$

- Both bars are perfectly aligned and the external load $P$ is applied directly through the longitudinal axis
- Under ideal conditions the system is in compression and the spring in not under any load

## Adding a small lateral displacement

<Image
	src="/images/posts/column-buckling-structures-and-stability/img-2.png"
	alt="Column Buckling Structures and Stability - Pinned supports"
	width="250"
	height="448"
/>

Think about what happens if we impose a small lateral displacement on the structure…

The rotational spring applies a restoring moment $M_B$, equal to the angle through which the spring rotates multiplied by the spring’s rotational stiffness:

$$
M_B = 2\theta\times \beta_R
$$

The restoring moment acts to decrease the lateral displacement while the axial load, $P$, tries to increase it.

At this point, two things **could** happen:

- If the axial load is ‘small’,  $M_B$ _wins_ and the structure straightens up returning to its original position/state

**💡The structure is said to be STABLE**

- If the axial load is ‘large’, P _wins_, and the lateral displacement continues until the structure collapses

**💡The structure is said to be UNSTABLE**

At some point we must cross a boundary between the stable and unstable states. This boundary is characterised by the critical load $P_{cr}$.

<Callout type="info">
	📌 The critical load, $P_{cr}$ is the axial load at which a structure under
	compression will move from a stable condition or state to an unstable one.
</Callout>

Much of the discussion that follows will now focus on determining what the critical load is for a variety of column end conditions.

## Determining the critical load for the idealised structure

To determine the critical load for our idealised structure, we can ‘cut’ the structure at the location of the spring and consider equilibrium of the top half, the resulting free body diagram reveals the moment action of the spring at the cut end of the structure…

<Image
	src="/images/posts/column-buckling-structures-and-stability/img-3.png"
	alt="Column Buckling Structures and Stability - Pinned supports"
	width="300"
	height="308"
/>

- Note that we have vertical force equilibrium as a result of the internal axial force revealed by the cut at B
- We also have moment equilibrium due to the couple creating by both P forces balancing $M_B$.
- If we assume that the displacement of the system is small, we can approximate the lateral deflection as:

$$
\delta \approx \frac{L}{2}\times \theta
$$

So, taking the sum of the moments about B and assuming clockwise moments are positive yields:

$$
\begin{align*}
P\times\delta-M_B&=0\\\\
\frac{PL\theta}{2}-2\theta\beta_R&=0\\\\
\theta\left(\frac{PL}{2}-2\beta_R\right) &=0\\
\end{align*}
$$

Now since we know, $\theta$ isn’t equal to zero, the expression in brackets must equal zero. So solving for $P$ yields:

$$
P_{cr} = \frac{4\beta_R}{L}
$$

This is the value of $P$ for which the external axial load is balanced by the restoring moment of the rotational spring, as such, from what we said above, this is the critical load for our idealised structure, i.e. the axial load at which it is on the boundary between being stable and unstable.

## Equilibrium and Stability

It’s important to establish a clear understanding of what we mean when using the terms equilibrium and stability, particularly in the context of a discussion on buckling.

### When $P

The idealised structure is straight and in equilibrium.

<Callout type="info">
	📌 The structure is in a state of equilibrium and that equilibrium is
	**STABLE**
</Callout>

If a small lateral deflection is imposed on the structure, it will return to the straight position. A good analogy for this is a ball sitting on a concave surface. The ball will always return to the same stable state of equilibrium when not disturbed.

### When $P>P_{cr}$

The idealised structure is straight and in a state of equilibrium

<Callout type="info">
	📌 The structure is in a state of equilibrium **_BUT_** that equilibrium is
	**UNSTABLE**
</Callout>

The structure is in a precarious (unstable) state of equilibrium. The smallest possible lateral deflection or disturbance will cause the structure to buckle. From an engineering perspective we must avoid this state.

The corresponding ball analogy is that of a ball precariously positioned at the top of a convex surface…the smallest disturbance will set the ball rolling.

<Image
	src="/images/posts/column-buckling-structures-and-stability/img-4.png"
	alt="Column Buckling Structures and Stability - Pinned supports"
	width="300"
	height="186"
/>

### When $P=P_{cr}$

💡What happens if P=$P_{cr}$? and we apply a small lateral deflection??

In this case the disturbing effect of the axial load is matched by the restoring effect of the moment $M_B$.

<Callout type="info">
	📌 When $P=P_{cr}$, for any ‘small’ angle, the structure remains in a state of
	**NEUTRAL EQUILIBRIUM** – how practical do you think this is?
</Callout>

This case is analogous to a ball sitting on a perfectly flat surface. We can visualise the various states of equilibrium in a graph of axial load versus angle of rotation…

<Image
	src="/images/posts/column-buckling-structures-and-stability/img-5.png"
	alt="Column Buckling Structures and Stability - Pinned supports"
	width="400"
	height="252"
/>

- All points identified in red, represent states of equilibrium
- Notice that ‘_in theory_‘, for an undefined range of $\theta$ (angles of rotation), when $P=P_{cr}$, we have a state of neutral equilibrium and the structure could keep deflecting.

From a practical point of view, we always aim to maintain axial loads well below the critical buckling load for a structure. The concept of neutral equilibrium can be thought of as a theoretical state that is highly unlikely to be observed in practice.

Up to this point we’ve discussed a simplified idealised structure that has all of its resistance to rotation concentrated at the midpoint in a rotational spring. This has allowed us to get a feel for some important concepts.

In [the next post](/posts/column-buckling-equations/), we’ll expand to start thinking about more realistic structures that have distributed flexural rigidity.

<Callout type="warning">
	⚠️ Remember, the assumption that underpins everything we’ve said so far, is
	that we’re dealing with small deflections. As the magnitude of the deflection
	increases relative to the size of the structure, our assumption (and therefore
	our model) breaks down.
</Callout>

---


Long slender structural elements under load will likely fail due to buckling rather than direct compression. In this post we'll explore column buckling

Column Buckling and Stability | EngineeringSkills.com

Long slender structural elements in compression will typically fail due to buckling rather than direct compression. In this post we'll explore column buckling

HUB / STABILITY

Understanding structural behaviour and stability in response to lateral loading

Stability Tutorials