How Do We Use Frames?

Post #10 of the Engineering Mechanics Series

Hello friends, 👋👋

Another week, another static system we use in structural engineering. Today we’ll break down how frames work, where we use them and why.

So far, we’ve explained the following static systems in detail:

• Simply supported beam

• Cantilever beam

• Trusses and

• Arches

You can find all previous posts → here ←

Here’s what we’ll cover:

• What are frames, and where are they used?

• Different static systems of frames

• Examples structures of frames

• Example hand calculation of internal forces

So let’s get into it. 🚀🚀

What is a Frame?

A frame is a static and structural system that consists of beams and columns. The main characteristic of a frame is that it’s a stable structure without depending on floors and walls (in-plane). In the next picture, you can see that the horizontal loads travel through the frame and to the foundation. Frames are built in all 3 main materials - timber, steel and reinforced concrete. The static system of the frame can have different support and connection conditions, which are explained further in the following.

The Static System of a Frame

The frame structure has a variety of different static systems due to different support and connection conditions.

Here are the most common static systems you can choose from: 👇👇

1. Rigid frame (Fixed-fixed frame)

The rigid frame is categorized by having 2 fixed supports and momentstiff connections. Therefore, the static system is indeterminate. The fixed supports (a) and (b) take up

• a vertical reaction force V

• a horizontal reaction force H and

• a moment reaction M

2. Pinned-pinned frame

The pinned-pinned frame is characterized by having 2 pin supports and momentstiff connections. This static system is also statically indeterminate.

The pinned supports (a) and (b) take up

• a vertical reaction force V

• a horizontal reaction force H and

3. Pinned-roller frame

The pinned-roller frame is characterized by having 1 pin support, 1 roller support, and momentstiff connections. This static system is also statically determinate.

The pinned support (a) takes up

• a vertical reaction force V

• a horizontal reaction force H and

The roller support (b) takes up

• a vertical reaction force V

4. Frames with hinges

You can add 1 a hinge connection to a pinned-pinned frame or 2 hinge connections to a fixed-fixed frame. But you can’t introduce a hinge to a pinned-roller frame because that would make the static system instable.

Example Structures of Frames

Frames are commonly used in warehouses, parking garages, carports, industrial production facilities, sports halls to mention a few. Frames are even used for small span bridges (roughly 10 meter span).

Frames are lightweight and quick to construct.

The most common material that is used for frames is steel, but as mentioned earlier timber and concrete are also often used.

Let’s look at some more specific examples

1. Warehouses

I designed a steel frame warehouse for my bachelor thesis. 👇👇

As the static system, I used a fixed-fixed frame and the frame extension (beam and column) where connected with a hinge to the column of the rigid frame.

2. Timber frame as a support for a folding door

This is a structure I designed some time ago. The only purpose of the frame was to resist the vertical loads of the self-weight of a folding door inside an office building.

So, I choose the static system of a pinned-pinned frame with momentstiff connections in the corners.

Now, I had one challenge with this frame: I couldn’t connect the frame to any of the existing structure which meant that I had to make the frame also stable out-of-plane for potential horizontal loads like internal wind pressure or a person hitting the door.

So I designed the support out-of-plane as a fixed support. The frame out-of-plane was then a cantilever beam/column.

3. Frame bridge

Bridges with smaller spans are often frames. You can see an example → here ←.

The beam is the bridge deck and the columns are the abutment walls. I designed a bunch of these types of bridges in a earlier job as a BIM modeller. The amount of reinforcement that went into these structures was massive. It was very hard to read the reinforcement plans. Even for me, who created them. So we decided to also provide our 3D models to the construction companies.

• Snippets from 3D reinforcement models

• Snippets from bridge reinforcement plan

Example hand calculation of internal forces of a pinned-roller frame

Let’s define the dimensions and loading situation of the frame. 👇👇

1. The first thing we always calculate for determinate structures are the reaction forces/moment. In our case, that is Va, Ha at support (a) and Vb at support (b) due to the equilibrium conditions.

Horizontal equilibrium:

\(\sum H = 0: 3 kN - H_a = 0\)

\(H_a = 3 kN\)

Now, to get the vertical reaction forces we’ll take the moments in point (b):

\(\sum M_b=0: V_a \cdot 5m - 3kN \cdot 3m = 0 \)

We now solve this equation for Va:

\(V_a = \frac{3 kN \cdot 3m}{5m} = 1.8 kN\)

We can now find Vb with the vertical equilibrium:

\(\sum V = 0: V_a + V_b = 0\)

\(V_b = - V_a = -1.8 kN\)

2. Calculation of the shear, normal and moment distribution along the frame due to the reaction forces and external load. The parameter x is introduced as the length between point a and any point on the left column.

\(\sum V = 0: V_a + N_x = 0 \rightarrow N_x = - 1.8 kN\)

\(\sum H = 0: H_a + V_x = 0 \rightarrow V_x = - 3 kN\)

\(\sum M_x = 0: M_x + H_a \cdot x = 0 \rightarrow M_x = - H_a \cdot x\)

We can see from these results that the shear and normal forces in the left column are constant, but the bending moment is linear (increases/decreases with an increase of x).

Let’s set x = 3 m:

\(M(x=3m) = - H_a \cdot 3m = - 9 kNm\)

At the support M(x=0m) = 0 which makes sense when we have a pin support.

3. Calculation of the shear, normal and moment distribution along the beam.

\(\sum V = 0: V_a - V_x = 0 \rightarrow V_x = 1.8 kN\)

\(\sum H = 0: H_a + N_x = 0 \rightarrow N_x = - 3 kN\)

\(\sum M_x = 0: M_x + H_a \cdot 3m - V_a \cdot x = 0 \rightarrow M_x = V_a \cdot x - H_a \cdot 3m\)

Let’s first set x=0m:

\(M_x = V_a \cdot 0 - H_a \cdot 3m = - 9 kNm\)

It’s the same value as for the column when we set x=3m.

And now, x=5m:

\(M_x = V_a \cdot 5m - H_a \cdot 3m = 0 kNm\)

With some experience, we already know that the bending moment in the right column is 0. Let’s run through the calculation anyways.

4. Calculation of the shear, normal and moment distribution along the right column.

\(\sum V = 0: V_b + N_x = 0 \rightarrow N_x = 1.8 kN\)

\(\sum H = 0: V_x = 0\)

\(\sum M_x = 0: M_x = 0 \)

5. Bending moment, normal force and shear force diagrams

We can plot the diagrams by hand.

Normal force diagram

Shear force diagram

Bending moment diagram

Conclusion

With the 10th edition of the engineering mechanics series, we have now covered simply supported beams, cantilever beams, arches, frames and trusses.

What will come next week then? 🤔🤔

Tune in on Wednesday again to find out. See you there. ✌️✌️

Have a great rest of the week.

Cheers,

Laurin.

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