# Structures Simplified

## Shear Lag in Steel Structures

September 11, 2020 Posted by Parishith Jayan
If we consider ourselves attending a design class in the undergraduate degree program, once the design philosophies are over, the first design lecture would be on "Design of Tension Members".

### Have you ever wondered, why we are learning this thing first?

You can check any standard textbooks, once the author finished talking about introductions, prerequisites, design philosophies, the first actual design would be "Design of Tension Members". In fact, if you check the codebooks, they too follow the same order. (Attached the content page of IS-800:2007 to support my argument)

### To answer the question, WHY?

To elaborate, it is easy to calculate the stress component in the case of tensile loading (Stress=Load/Cross-sectional Area), which is not so with transverse loading, since we need to calculate bending stresses.

To gain some insight regarding stresses and forces and why we calculate stress, read this article - Why do we calculate STRESS, when we have FORCES?

Regarding the limit states for tension members, the tension member can fail due to
1. Gross section yielding
2. Rupture of Net section
Kindly note, I have not listed the limit states for the failure of connections.

See, it is simple. Very few concepts and that is why it is placed first in the order.

Before moving to the topic, would like to ask one more question. What are the most widely used tension member sections?

As a designer, say, I am about to design a tension bracing member, the first section that comes to my mind would-be "Angle Sections" (Hoping, you too think of the same).

When considering angle members, there is one important behavior or phenomenon which should be addressed or taken into account while determining its tensile capacity. It is nothing but "Shear Lag", which is the main concentration for today's article.

## Contents

• What is meant by "Shear Lag effect"?
• Exercise to feel shear lag
• How do we incorporate "Shear Lag" in the calculation of tension capacity?
• As per AISC-360-10
• As per IS-800:2007
• Inference/Conclusion

## What is meant by "Shear Lag effect"?

The attached image shows the gusset plate connection of typical bracing members.

The most important thing to notice is the way in which they are connected. As we can clearly see, they are connected using bolts to the gusset plate, which in turn is welded to the main member.

"Only one of the leg is connected". So what?

You can say, that if the bolts are properly designed to take care of the design force, then the connection should work fine.

The problem is not with the connection, it is with the behavior of the member. Since only one leg is connected, the entire tensile force will get transferred to that leg, to flow through the connection. Which totally cancels out our limit state checks for the tension member.

We will be calculating yielding strength based on the entire area of the member (including both the legs) and rupture strength based on the net area (including both legs neglecting the bolt hole). Since at the location of connection, all the forces are getting transferred to the connected leg, the area of concentration of force reduces, so both of these limit states are not valid.

Usually, the force which is acting in one of the legs gets transferred to the entire cross-section in the form of shear. At the location of the connection, the stress distribution is non-uniform due to the concentration of force on the connected leg. This "lag" in stress distribution in the tension member is termed as "SHEAR LAG".

## Exercise to feel shear lag

Just try to make a replica of an angle section using paper which is around 50 to 75mm long. Let the sides of the angle be 10mm x 10mm.

Ask your friend to hold one end of the angle by grabbing both the legs of the paper angle, which you made. And on the other side, you too do the same and slightly apply a pull force. You can feel the stress in the entire cross-section of the paper (i.e. you can feel, both of the legs are experiencing stress due to the pull). Now, you just hold only one leg of the angle and apply a little pull. As you can evidently see, the outstanding leg of the angle at your end just casually drops off. You can touch and verify, that there is no stress on the outstanding leg at the location of the connection. But, as we move away from the connection, the angle still would experience stress on both the legs. This lagging in stress distribution is what we call, "Shear Lag".

## How do we incorporate "Shear Lag" in the calculation of tension capacity?

Due to the behavior of shear lag, the effective area of the tension member is getting reduces. So, this reduction in the area is applied in the form of "shear lag factor, 'U'" in AISC-360-10/16.

Herewith attached the snap from AISC-360-10 to determine the effective area and the table to determine the shear lag factor for different cross-sections.

The shear lag factor can be determined based on case 2 of the table. As we closely look, the factor depends on two things, firstly the centroidal distance from the plane of connection, secondly the length of the connection.

So, if we are about to increase the effective area (which would ultimately increase the tensile capacity of the section), we should try to reduce the value of 'x' (centroidal distance from connection plane) and increase the length of the connection.

Keeping the concepts that we learned so far in front, let me ask you a question.

### Obviously, the longer leg. Why do we do so? Because it will considerably reduce the value of "x" considering the short leg being connected.

In IS-800:2007, we do follow the following steps to determine the tension capacity of the section.
What is happening in the above formula, the capacity of the connected leg and outstanding leg are separately calculated. The assumption here is that the outstanding leg will reach yielding stress when the connected leg probably would have attained the ultimate stress.

The net area is calculated for the connected leg to determine its capacity. The area of the outstanding leg is reduced with the reduction factor "Î²", which is our shear lag factor. It depends on the factors like the length (L) of the connection and shear lag width (bs).

Alternatively, a simpler approach is also provided in IS-800:2007 as shown in the snap below.

Here, the net sectional area is multiplied with the reduction factor "Î±" to account for the shear lag based on the number of bolts used in the connection (Length of the connection).

## Inference/Conclusion

As we have seen the concept of shear lag and the factors affecting it, the following are my inference
1. The thickness of the member does not play a role in the determination of shear lag factor or it does not affect the tensile capacity of the section.
2. If the distance between the plane of connection and the centroidal axis of the outstanding leg increases, it reduces the tension capacity of the section.
3. From the experimental studies, it is proved that if the length of the connection is increased up to 4 bolts, the tensile capacity increases. And if we further increase the length it does not have a significant effect. (Kulak and Wu, 1997).

## Lateral Torsional Buckling

September 02, 2020 Posted by Parishith Jayan

## What happens to a member, when it is subjected to transverse loading?

A member subjected to transverse loading will undergo bending and start to deflect along the plane in which it is loaded. It seems so simple, right?

Now, what will happen if we keep on increasing the load?

It will further bend until it reaches its maximum moment capacity and fails. This is what we call a bending behavior of the laterally supported beam. If a beam needs to behave this way, there are certain conditions that the beam should satisfy. They are,
1. There should not be any local buckling in the elements.
2. The compression flange of the member should be restrained (i.e.) The beam should be restrained in the lateral direction so that it cannot move laterally.
What happens when the first condition fails? The beam would fail due to local buckling of the web or flange even before attaining its full moment capacity.

What if the second condition fails? The beam will fall under the laterally unsupported beam category. And when the load is increased, the beam will displace laterally (to be more precise, the compression flange moves laterally) and the failure would occur due to the combination of lateral displacement and bending. It considerably reduces the moment capacity of the section. This phenomenon is called "Lateral Torsional Buckling". For our ease, let's call it LTB.

## Whether all laterally unsupported beam exhibits LTB?

No. When the beam is too short, even though the beam is laterally unsupported, the beam won't buckle laterally. It will fail after the attainment of full moment capacity like a laterally supported beam.

## The behavior of Lateral Torsional Buckling

As we see, the short beam won't exhibit LTB, what about long beams?

Let's consider a simply supported (but laterally unsupported) beam with a concentrated load at the midspan. If we keep on increasing the load. Initially, the beam will undergo bending and deflects vertically. Because of this, compressive stress develops in the top flange and tensile stress develops in the bottom flange.

When the load keeps on increasing, the stress developed will the beam section also increases. In order to relieve the compressive stress generated, the compression flange needs to get elongated (The stress created can be relieved in the form of displacement). Since the beam is oriented in such a way that its major axis taking up the loading, it can't elongate in the axis of loading. So, now the compression flange makes use of the lateral direction, which is unrestrained. It tries to displace in that direction, causing the whole section to twist about the axis of loading.

The key point to note in this behavior is that the cross-sectional shape won't change. The beam will undergo twist alone. This twisting makes the compression flange move away from the actual line of the beam resembling the buckling behavior of the column sections.

## How does the position of the load affect LTB?

Based on the position of the load, the moment resistance of the beam is enhanced or decreased. Say for example, if the load is applied at the top flange of the section, the effect of LTB will be more and if the load is applied at the bottom flange of the section, then the effect of LTB will be less.

What is the myth behind this?

In the first case, the load is applied above the shear center of the beam, when the top flange starts to buckle laterally, the applied load will be at some eccentricity now, which would cause some additional twisting moment. This load is called the "Destabilizing Load"

Contrarily when the load is applied at the bottom flange, the moment generated tries to stabilize the beam, thereby reducing the torsional buckling. This load is called the "Stabilizing load".

## Whether LTB occurs when an I-beam is positioned in H-shape and the load is applied along the minor axis?

In this case, the beam would be positioned to look like "H" and when the load is applied. Initially, the beam will bend in its minor axis. Comparing the bending capacity of the minor and major axis, the minor axis holds very low bending capacity. So, in this case, the beam will reach its maximum bending capacity in the minor axis before, the section displaces laterally. LTB won't occur.

## Whether LTB can be neglected?

The lateral-torsional buckling can be neglected in certain scenarios, they are as follows.
• As we already discussed, when the beam is loaded along its minor axis. The beam will generate its full bending capacity even before LTB could initiate.
• In sections like a square hollow tube and circular hollow tube. They have the same moment of inertia value in both the axis.
• When the beam is too short (this limit will be provided in design codes), the member will fail by creating its full moment capacity rather than LTB.

## Conclusion

It is evident that the behavior of lateral-torsional buckling largely influences the bending capacity of the laterally unsupported beam. Neglecting this would result in over-estimating the capacity of the section.

In order to reduce the impact of LTB, the long beams should be braced laterally at appropriate locations, to reduce the effective length of the member in a lateral direction, which would considerably increase the capacity.

## Bracing System - Structural arrangement that ensures stability in Longitudinal Direction

August 28, 2020 Posted by Parishith Jayan

## What happens to a building when it is subjected to wind loads?

Any building or structures, in general, must ensure stability in two directions (Lateral & Longitudinal) to safely transfer loads from the location of application to the ground.

Considering a typical steel warehouse building something similar to the following image, when it is subjected to wind load along the lateral direction, stability is ensured by the portal frame action.

Lateral Direction - Along width of the building

Longitudinal Direction - Along Length of the building

The column and the rafter connected using a rigid joint act as a portal to sustain the lateral loads that act on the building.

So, the building is fine in the lateral direction. What if the wind blows in the longitudinal direction?

## How longitudinal force gets transferred through the system?

In the longitudinal direction, when the force acts on the gable ends of the building, the first component to interact with the load is the cladding materials (Sheeting). Through which the force gets transferred to the next component, Girts. Obviously, we would have designed this member to withstand the wind force, so that member will be strong enough to take the load and transfer it to the supporting member.

Girts are in turn connected to wind columns. Now, the load reaches the wind column. This is where it gets interesting.

Mostly the wind columns will be pinned supported at the base as well as at the top (at rafter location). When wind load acts on these columns, there will be reactions at the ends (top and bottom) of the column. At the bottom end, the force is transferred to the ground as shear through anchor rods.

There arises a question.

## What happens at the top?

Whether we have to design the gable end frames with these wind reactions acting in a longitudinal direction?

No need. Actually, the reaction force which generates at the top of the wind column gets transferred through the strut tubes or purlins that we provide behind every wind column location.

And there arises another question.

## If the strut tubes carry the load from the wind column, then what is the role of providing bracing?

The actual purpose of the strut tube is to transfer the force from the gable end frame to the interior frames. But in order to transfer the force to the ground, we need a bracing system

The most commonly used bracing system in a steel building is “Diagonal Rod Bracing” (X-type bracing).

## How does Diagonal Rod bracing behave for the longitudinal load?

Once the force gets transferred to the bracing location, the brace rods won't let the force to continuously flow through the frames. Usually, in the diagonal rod bracing, one set of bracing rods will be tension and the others will be in compression. When the longitudinal load reaches the braced bay, the rod which is in compression buckles away and does not carry any force since it is slender. The other set of rods that are in tension will be active in carrying the force. A certain amount of lateral force will be absorbed by the bracing. These forces are transferred from the bracing system at the roof to the wall and get transferred to the ground as a shear force through anchor rods. (See below figure)

Note that, while providing diagonal rod bracing, only one set (rods in tension) of rods will be in action on transferring the force. So, you can ask me why we are providing two diagonals then?

The answer is so simple when the direction of force got reversed, the other set of diagonal members will be the one that transfers the force to the ground. That’s the reason why we are provide bracing in a pattern of X.

Since, these rods carry the longitudinal force through the tension action, while designing they have to be considered as a "Tension Only Members"

## What do we do, if we go for the Diagonal Angle or Tubular section as our bracing members?

If we are adopting Angle or Tubular sections as the bracing members. We should be aware that they are capable of transferring loads through compression as well as tension. So, the entire load transferring mechanism differs. Both the set of bracing members will be in action transferring the force. In that case, it is advisable to design those bracing members as a "Truss Members".

The provision of Rod, Angles, or Tubular sections as the bracing member depends on the magnitude of the lateral force that is acting on the building. One must calculate the lateral force that is intended to act on a structure and choose the bracing member accordingly.

Moving further, there arises the next question.

## How many braced bays are required for a building?

Let's look into this question with an example.

Consider a building whose length is around 64 m and we are dividing it to 8 nos. of 8m bay spacing. The minimum number of braced bays to be provided in this case is 3. You might wonder, from where did that magical number 3 came from..? Actually as per IS 800–2007, the maximum center to center distance between braced bay is 40 m. That means you cant go beyond 40 m, without providing a braced bay and which leaves us with 3 braced bays.

## P-Delta Effect on Structures

August 25, 2020 Posted by Parishith Jayan

P-Delta Analysis is a Non-linear second-order analysis which exhibits a considerable increase in the base moment when a structure is subjected to large lateral displacement due to wind or seismic forces.

Before moving further, there are two terms that need additional clarification to understand the behavior on the whole. They are "Non-linear" & "Second-order".

### What is Non-linear Analysis?

To put it in a simple statement, "Stress-Strain relationship is not linear as in the case of linear analysis". Non-linearity may be classified as follows.
1. Geometric Non-linearity
2. Material Non-linearity
3. Boundary condition Non-linearity
P-Delta effect falls under the Geometric Non-linearity category. This particular non-linearity is due to the excessive deformation or deflection of the material or structure, even though they are in the elastic limit.

### Why P-Delta analysis is called "Second-Order" Analysis?

For easy understanding, when a structure is loaded, it will deflect or deform to relieve the stress. This deflection is said to be the first-order effect. Without any additional loading, if any stresses or adverse effects induced in a structure due to the first-order deflection, it is called a second-order effect.

P-Delta analysis is called a second-order analysis because it depends on the initial deflection of the structure. If the initial first-order deflection is more, then obviously the P-Delta effect will be more in the structure.

### P-Delta Analysis

Usually, we used to analyze a structure in a static position with loads being applied to it and determine their responses. P-Delta analysis is nothing but analyzing a structure by applying loads on the deflected form of a structure. A deflected structure may encounter significant moments because the ends of the members have changed their position.

Consider a column of length 'h', which is fixed at the bottom and free at the top. It is encountering an axial load P and lateral load H as shown in the following image.

Notice that, in the case of linear static analysis, the lateral deflection depends on the lateral load, H. The deflection d1 due to the applied lateral load H, is the first-order deflection of the column. The moment at the base of the column, M0 will be "Hxh", as shown in the figure.

Now, this is where it gets interesting. There will be this axial load P which is a result of gravity loads acting on the column. If the initial deflection d1 is large, then geometric non-linearity would occur and this axial load P combines with the initial deflection d1 results in an additional moment "Pxd1". Now, the moment at the base would be "(Hxh) + (Pxd1)".

Now, what happens next? wouldn't the column deflect even more?

Yes, it will. And the new deflection will be d2 and the moment will be increased to "(Hxh) + (Pxd2)". This goes up to several iterations until the increase in deflection becomes negligible. This iterative analysis process is termed as "P-Delta" Analysis.

It is obvious that a P-Delta effect is named after the secondary moment, Pxd.

There are two types of P-Delta effects. Namely,

1. P-Large Delta effect
2. P-Small Delta effect
What we discussed so far, comes under P-Large Delta effect. It deals with the entire structure. Whereas, P-Small Delta effect is elementary level. It is a second-order deflection in the particular member due to applied axial load and initial lateral deflection of the member. In most of the cases, P-Small delta won't have a role, since it is elementary level, it requires huge deflection values to get effective.

### When P-Delta Analysis is necessary?

• P-Delta effect usually becomes prevalent in a tall structure that is experiencing gravity loads and large lateral displacement due to wind or other forces.
• If the lateral displacement or the vertical axial loads are significant, P-Delta analysis should be performed. In many cases, a linear static analysis can severely underestimate displacement in comparison with the P-Delta effect.

### Conclusion

P-Delta effect is a second-order effect experienced by any structure when subjected to lateral loads like earthquake or wind loads and is originated by an additional destabilizing moment generated due to the gravity load acting on the laterally deflected member further displacing it.

To be more precise, when a slender structure is subjected to lateral loads like wind or earthquake loads, it undergoes lateral displacement or sway. When this lateral displacement is reasonably large, gravity loads start to act with an eccentricity equal to the magnitude of elastic deflection causing an additional overturning moment. Due to which, the structure is pushed even further developing a second-order deflection. This second-order effect experienced is conveniently termed as P-Delta effect.

## Why "Prying Force" is important?

August 21, 2020 Posted by Parishith Jayan
Our previous post is about structural steel connections, emphasizing its importance and complexity. This post is sort of continuation of the previous post, which is going to emphasize on the additional tension force that is developed in the bolted connections or to make it simple, it is all about "Prying Force".

Before moving into the force aspect, we should know the behavior of prying.

### So, what is prying action?

Let's consider a steel plate that is bolted to the table (similar to the following image). Now we are about to apply the force, F on one end A of the steel plate. What would be the reactive force that is offered by the bolt to hold the steel plate in its position? Is it "F"?

No, the reactive force required is greater than F, since the other end B of the steel plate is pushing against the table. It will develop force Q at the end B.

So, the bolt requires a reactive force, which is equal to F+Q to hold the steel plate to the table, this particular behavior is called "Prying Action" and the additional tensile force, Q is termed as "Prying Force".

### Why it is important?

While determining the tension force of a bolted connection, it is hard to separate the discussion of bolt tension from the connecting element. So, the behavior of connecting elements needs to be incorporated in determining the bolt forces. Since they can incorporate additional forces due to the prying action.

As discussed in the above example, every bolted connection, when subjected to tensile load eccentrically, will try to separate the connecting part from the connected surface, due to the eccentricity in the applied tensile load and bolt placement, addition prying forces will be created which has to be resisted by the bolts.

If the bolts are not checked for this additional force, eventually it may lead to the failure of the connection bolt.

### Different Modes of Prying Action

Normally, the concept of prying action will be presented in terms of T-stub in textbooks and codal provisions.

As we see in the figure, the tensile load is not concentric and acting at some eccentricity to the center of the bolt which may induce prying force on the bolt. The behavior of the T-stub can be analyzed in three different modes.

#### Mode 1: Complete Flange Yielding (Thin Plate & Strong Bolt)

This mode occurs with a flexible connection flange (thin plate) and a strong bolt. What will happen in this situation, on the application of tensile load, the flexible flange tries to elongate. Whereas the bolt is rigid and won't undergo much of an elongation, holding the connecting flange tight to the surface (resulting in the double curvature). This makes the counterpart of the connecting plate to induce a very strong prying force (refer the attached image below).

#### Mode 2: Bolt Failure with Flange Yielding

In this mode, the flange will be relatively rigid compared to the bolt. In this case, when a tensile load is applied, the plate elongates. Since the bolts are flexible, they indeed elongate with the flange and allows the flange to form a single curvature (refer the attached image below). Here, the bolt is not resisting much, so the prying force created is little compared to the previous mode.

#### Mode 3: Bolt Failure (Thick Plate & Weak Bolt)

This occurs with thick plates and weak bolts. As the tensile load is applied, the plates which are thick enough won't elongate that easily. Meanwhile, since the bolts are weak, they fail even before the plate elongates resulting in no prying force.

### Conclusion

• The key take away from the post is that "the prying force if neglected could result in failure of the connection".
• Based on the connection geometry, the prying force can result in 40% of the tension in the bolt.
• The calculation of prying force is not as simple as it seems, it is complex and involves a lot of variables.
• The prying force will be maximum with thin plates and strong bolts.
• Thick plates and weak bolts won't form a prying action.

#### References

1. SCI-P207 - Joints in Steel Construction - Moment Connections
2. IS 800-2007 - General Construction in Steel - Code of Practice

## The most CRITICAL structural part in a steel structure..

August 16, 2020 Posted by Parishith Jayan

What are the structural elements does a steel structure holds?

As similar to every other building, the major load-carrying members in a steel structure are columns and beams.

Let us consider an industrial steel building to analyze its structural components.

From the above image, the major structural elements are rafters, columns, mezzanine joist (secondary beams), mezzanine beams, purlins & girts (for cladding support), and staircase stringers.

The interesting part is, we can easily design all these major elements using the STAAD package or similar software package. The only thing we should know is how loads get transferred from one element to another (i.e. what are the loads to be considered while designing each of these elements).

Understanding the load path and basic operation of any structural design software would suffice.

Then what is the most critical part?

“THE CONNECTIONS”

They are the most critical portion of any steel structure. The structural member may have sufficient strength, but if the connection does not meet the requirement, the member has no use at all.

Understanding how a connection behaves, what are the forces contributing to the connection, what are the connection parts to be verified makes it very crucial to design.

Say, for example, if you are designing a structural element, a mezzanine beam.

what are the checks we have to do?

1. The flexural capacity check
2. The shear capacity check
3. Check for Lateral Torsional Buckling
4. Slenderness check
5. Check for deflection

So, roughly 5 checks are enough to design a beam, which can be done easily with a basic understanding of all the above checks, and moreover, we have clear guidelines in the design code (IS 800–2007) to work with.

Now let us think about the connection of column and mezzanine beam Extended end plate connection. (Something similar to the following image )

What are the checks to be considered in order to design this connection?

There are around 15 checks to be done to properly design each and every component of this connection.

Herewith attached the checklist from “Joints in Steel Connections - Moment Connection”[3]

Without understanding how the components in the connection behave, and the force to consider to work with this design checks, we cannot reach a proper design solution. This is where most of the designers commit a mistake.

This makes the steel connections, the most critical portion in a structural steel building.

## Why web takes all the shear in an I- section?

August 15, 2020 Posted by Parishith Jayan
During the initial stage of one's career, every design engineer, especially a steel design engineer would have come across this statement "Bending force is carried by the flange of an I-Section, and shear force is taken care of by the web".
Like everyone, when I first heard this, just started to keep this as a thumb rule and practice it.

Never asked a question " WHY..?
Until someone else asked me, why we consider shear is taken by the web of an I-Section.

And this question can be answered in one simple sentence, as well as it can be discussed elaborately like a research article.

I just want to make it as simple as possible with sufficient explanation. With this in our mind, let's get into it.

#### Why webs carry all the shear?

Due to the applied transverse load on the beam, internal forces parallel and opposite to the applied load will be generated along the beam cross-section. This force is called "Shear Force", which has the effect of shearing away the cross-section.

The best example for a shear force is scissors, just imagine you are cutting cardboard with the help of scissors, the two blades of the scissors must employ force parallel and opposite to each other along the cross-section of the cardboard in order to tear it off.

As we already know, though we have shear force, we need shear stress to determine the shear capacity of the member. (This topic is already discussed in the blog)

Now, we are about to determine the shear stress distribution for an I-section. Once, we have the shear stress distribution, we can clearly understand why we consider the web is subjected to shear rather than the flange.

We could easily find a formula to determine the shear stress distribution along the cross-section from any strength of materials or statics book.

The shear stress, Ï„ is given as

Ï„ = VQ/ Ib

Where V = The shear force
Q = The first moment of area
I = Moment of Inertia of the cross-section
b = Width of the cross-section at the point of interest

If we observe the above equation, we could easily determine, which are the factors that influence the shear stress along the cross-section of a member.

The first parameter, V does not vary, since it is the shear force, it remains the same throughout the cross-section. Secondly, the "I" moment of inertia of the cross-section is also a constant for a particular cross-section.

The parameters, "Q" and "b", are the variables in the above equation, which determines the distribution of shear stress. Of these two, let's not worry about "Q" the first moment of area for a while. Let's concentrate on "b" the width of the cross-section.

From the cross-sectional image, consider we are determining the shear stress at point A, the width "b" is the entire breadth of the flange. Whereas, if we move to point B, it is just the thickness of the web.

Comparing the parameter "b" at point A and B, we could easily realize that at point A, the "b" value will be more. If we have a higher "b" value in the formula of shear stress, then obviously it would result in the least shear stress. And as the value of "b" is getting decreased, the shear stress increases.

Now, let's look into the "Q", the first moment of area. It is proved from the name of the parameter that as the area increases, the value of Q also increases. So, starting from the top and moving towards the neutral axis, the area is getting increased, which leads to an increase in Q value.

Plotting the value of shear stress distribution along the cross-section would result in the diagram as shown below.

As we can see, the shear stress is zero initially at the top, and as we move down the top flange, the "b" value will be constant, but the "Q" value will be increasing, resulting in the parabolic curve. As soon as we reach the web part, because of decrease in "b" value, there is a sudden increase in the shear stress and as we move further, the "b" value remains constant, the parabolic increase in the value of shear stress is due to the increase in the first moment of area "Q", reaching the maximum stress at the neutral axis.

#### CONCLUSION

As we can see visually from the shear stress distribution, the flange does have very little shear stress distributed compared to the web portion. For this reason, we designers follow the thumb rule that "web will carry all the shear".

## How to apply the effective length of a member in the design process?

August 11, 2020 Posted by Parishith Jayan

## What is the importance of employing Lx, Ly & Lz for a member during the design stage?

Every structural or civil engineer would be aware of column buckling behavior. Which is one of the predominant failure modes for the column, especially for a long column.

There will be this famous statement, "doubling the length of the column, reduces the load-carrying capacity of the column by the factor of 4".

Also, the effect of end conditions on the buckling strength will be strongly emphasized in each of our design classrooms.

The effective length factor (K) of the column varies with different boundary conditions, say for pinned-pinned boundary condition, it is "1". And for other boundary conditions, it varies respective to the distance between the point of inflection, approximating the pinned-pinned end condition. Hope you guys are aware of these details, if not kindly try to gain some insight as they are very basic concepts.

In general, everyone has some ideas on effective length and effective length factors. Now, let's move to the design aspect.

Say, I have decided to design a column of 5m length for a concentrated load of 100kN. Let's consider, after the design process, I am arriving at the ISMB300 section.

I am giving that section for approval, and the end-user rejects that section, saying they can't go for the ISMB300 section at that location, due to some site constraints, and they are asking to reduce the section depth.

## In that scenario, as a design engineer what would you do?

You have two choices,

1. Asking the client to shift the location of the column.
2. Reducing the section as per client request.
If the first option does not appear to be us. Then, we should be ready to reduce the depth of the section.

How do we do that?

- By introducing "Intermediate Restraints".

As we see already, the predominant mode of failure, in this case, is buckling. What if we reduce the effective length of the member by providing a brace or lateral restraint to prevent it from buckling at the center. The effective length of the member would be reduced to half, which will increase the load-carrying capacity by the factor of 4. Is not it?

So, the best solution is to brace our column. Then there pop-outs the next question.

## Which direction should we brace?

If we consider an I-section, there are two axes, namely the major axis and minor axis. As we know, the buckling load depends on the slenderness ratio of the member. We have to brace the axis which has the least slenderness ratio.

Providing the brace along the major axis is merely useful to nothing. So, we should go for providing intermediate restraint along the minor axis.

How do we incorporate this intermediate bracing in the design?

In the site condition, we can manage to provide a strut tube or secondary beam connecting the minor axis to another column or support. In a 2D design platform, to incorporate this effective length concept, there are parameters like Lx, Ly, and Lz in software packages like STAAD Pro.

They are nothing but the effective length of the member. Once we assign proper value for Lx, Ly, and Lz it replicates the actual condition by considering that length for the determination of column buckling load.

Herewith attached the sample snap of the STAAD environment, where I modeled the column with full effective length and half effective length. Just look at the column buckling load capacity.

 Fully Unsupported Column
 Braced at Mid Height
You can visibly see that the load-carrying capacity of the column has been enhanced by applying the appropriate effective length parameter and replicated the actual capacity of the column.
These things are OK.

Now let's talk about pre-engineered buildings and tips to determine the value of effective length for the design.

Just look into the following image,
 Typical Pre-Engineered Building Cross-Section

## How do we determine the effective length for the rafter and column sections?

First, let's consider columns, the major axis of the column is oriented along the frame direction, and there is no supporting arrangement provided in that direction. If we look into the longitudinal direction, girts will be running perpendicular to the column. That is they run perpendicular to the minor axis of the column member and in pre-engineered buildings, we have a habit of providing flange braces connecting the girt/purlin and the main member to brace the compression flange. Hence, they can be considered as a lateral restraint for the column.

So, the effective length of the column along the minor axis (Ly) can be considered as the maximum distance between the girts, or between the girt and finished floor level.

This is applicable for rafters also.

## Why the shape "TRIANGLE" is widely used in structural engineering?

August 09, 2020 Posted by Parishith Jayan
Most of the structures that we know comprise of this particular shape "triangle" in any one of its structural systems.

#### What is so important? or why "TRIANGLES"?

Just think of basic 2-dimensional geometrical shapes, what do we have?
Squares, Rectangles, Triangles, circles, pentagon, hexagon, and the list go on.

Of these shapes, why a triangle is called the most stable shape. Let us compare these shapes and understand them.

For our example, let's consider the square and the triangle as shown below.

Applying a force on one of the edges and studying the behavior would lead to a better understanding regarding the stability of the structure.

First, let’s try with the vertical force.
What will happen? The load applied over the square moves directly downward through the vertical member and stands still. The triangle when subjected to a vertical force through one of its edges, distributes that force evenly to either side and stands stable.

Now, let’s apply the horizontal force.

The square could not withstand this force without deformation, it tries to sway along the direction of application of the force and distort from its original shape. Whereas, applying the horizontal force tries to push one edge of the triangle.

It is not possible to move the edge without the elongation of one of its sides, this makes them stay stable for the applied force.

This simple comparison applies to all the other shapes too. Also, this can be easily verified by making a simple model with popsicles. I tried it myself, and recommend you to do so.

Let us look into a design example, where this particular behavior is utilized.

Most of us would have seen a bracing system in the steel buildings.

What is the function of bracings systems in the steel buildings?

The longitudinal forces that act on the structure are transferred to the ground with the help of bracing systems, right.

So, the longitudinal force is about to act on the edges should be transferred safely without much distortion or deflection of the system. Which is the shape that can perform this task? Obviously, the triangle.

The attached image shows several bracing arrangements that are employed in the steel structure.

#### CONCLUSION

This stable behavior of the triangles makes them the supreme geometry. And for this reason, they are extensively employed in the formation of trusses.

What we have discussed so far is also applied to the 3-dimensional shapes and this emphasis the use of triangle configuration in the construction of towers, electrical grids, truss bridges, and so on.

August 06, 2020 Posted by Parishith Jayan

## What happens to the load applied to a structure?

Whatever the load which acts on the structure ultimately tries to reach the ground from its point of application. Is not it?

This path is called “Load Path”.

Just imagine, you are departing from your workplace and commuting to your home.

What will be the first thing you do; you find your mode of transportation (say bike) and tries to figure out the route to reach your destination, right? There may be several routes for you to consider.

Similarly, for a load to reach the ground there may be more than one load path to consider.

After which, what you do, will choose the quickest way to travel. Correct?

The loads which are applied to the structure behave exactly the same. It always tries to find the shortest path to reach the ground.

Let’s see this with an example. The beam which is shown in the figure below is a simply supported beam of 3m length resting over a column. A load of 10kN acts at 1m from its left support.

What is the possible load path for this simple structure?

The load can move towards its left and reach the beam-column joint through which it gets transferred to the column, then from the column, it can reach the ground.

A similar path on the right is also possible.

But which is the shortest path? Transferring through the left side right. Yes, that is what happens. A major amount of load gets transferred to the ground through left support. This can be realized by calculating the reaction at supports S1 and S2.

Applying equilibrium equations and determining support reaction yields the value of 6.66kN for support S1 and 3.33kN for support S2 respectively.

What is so important in this? Yeah, the load is taking the shortest path. Why we need to know about which way the load is moving?

1. It is important because each element along the load path must have sufficient strength and stiffness to safely transfer the load. This is where design gets in. So, without understanding the load path, an engineer could not design a structure.

2. Secondly, a load gets transfer from one member to another through a connection. The load path plays a major role in determining what are the forces to consider while designing the connection.

Let’s look into the following figure to understand the importance of a load path in designing.

It is a gusset plate connection, connecting the floor brace to the column.

How the load is getting transferred here? Or more precisely what is the load-path?

If we draw a simple line diagram of the above connection, it will look like a crosshair with one diagonal, and this leads to the misconception that the load-path will be directly from the bracing member to the column through connection.

But the actual load path is different from what we assume.

The load from bracing moves to the beam through a gusset plate connection, and from the beam it is transferred to the column.

If we didn’t consider this load while designing the beam, then it will lead to the failure of the section due to this unexpected load. This simple example portrays the importance of understanding the load-path to reach a safe design.

Understanding how load travels through a structure is the key to optimal design.

If you guys want to know more regarding the behavior of a structure. Let me know in the comment.

## Bend the "RULER"...

August 04, 2020 Posted by Parishith Jayan

Why it is easy to bend material in the middle rather than at the ends?

In order to understand the concept, one must clearly visualize the context.

I like to clarify two things here, firstly the bending capacity of a member. Secondly, the bending force applied.

For our easy understanding, Consider a 1m length ruler. Let's place it between the two tables and apply 1 kN load on the ruler. We are about to determine the bending capacity of the ruler as well as the bending moment developed due to the applied load.

BENDING CAPACITY:

It is nothing but the capacity or strength of the object to withstand the applied bending force.

The bending capacity is given by the simple bending equation as M/I = F/y

Where, M = Bending moment capacity of the section,

I = Moment of Inertia of the section

F = Yield strength of the material

y = Position of Neutral Axis

Investigating the above equation,

· The moment of inertia of the section depends upon its cross-sectional (width, depth & area of material) parameters.

· The yield strength of a material is a constant value for a particular material.

· The neutral axis depth for a symmetrical section will be at its half depth, considering the major axis.

From the above statements, we infer that whatever the length of the member, unless and until if the cross-section and the material don't vary, the moment capacity of the section remains the same.

So, it has nothing to do with the bending of the member.

BENDING MOMENT:

Let us consider 3 different cases for the determination of bending moment.

1. The load (1 kN) placed exactly at the center of the ruler.

2. The load (1 kN) placed 0.8m from one support (i.e. near one end of the ruler).

3. Let's move the tables close to each other, making the length of the ruler 0.8m and placing the load at the center.

CASE-1:

Length of the ruler = 1m; Load = 1 kN located at center

Bending Moment under the load = WL/4 = 1*1/4 = 0.25 kN.m

CASE-2:

Length of the ruler = 1m; Load = 1 kN located at 0.8m from one end.

Bending Moment under the load = W.a.b/L = 1*0.8*0.2/1 = 0.16 kN.m

CASE-3:

Length of the ruler = 0.8m; Load = 1 kN located at center

Bending Moment under the load = WL/4 = 1*0.8/4 = 0.2 kN.m

What do we infer from this?

Comparing case-1 and case-2: The ruler is likely to bend more when the load is applied in the middle rather at the ends, because, the load acting at the middle develops a larger bending moment.

Comparing case-1 and case-3: As the length of the ruler decreases, even though the load acts in the middle, the moment value reduces for the shorter ruler. So, if we bend two members of the same material and different lengths, the lengthy member will bend more. With an increase in the length of the member, the bending moment also increases.