Materials strength
Internal forces in frames
- Internal forces in frames - basic information
- Computational difficulties
- Problem with the interpretation of markings for a vertical rod
- Problem with an inclined rod
- Video - a course showing how to draw internal force diagrams
- Example - frame with an inclined rod
From this course, you will learn:
- how to determine support reactions in frames,
- how to verify the correctness of calculated reactions,
- how to draw diagrams of normal forces, shear forces, and bending moments in simple, inclined, and hinged frames,
- how to check the equilibrium of a joint (verification of the diagrams),
- you will learn to draw diagrams using two methods - by recording the functions of internal forces and without recording these functions,
all of this starting from very simple examples of frames, progressing to frames with more and varied loads, and ending with frames with inclined bars both with and without joints.
Internal forces in frames
In the field of material strength, understanding internal forces is a key element for the design and analysis of structures. Internal forces, such as normal force, shear force, and bending moment, are fundamental values that shape the behavior of materials under load.
Just like in beams, the same internal forces act in frames, affecting their strength and stability. Analyzing these forces allows engineers to better understand the behavior of frames under various loading conditions and to design them accordingly.
Frames are extremely versatile and are used in many different types of structures, from residential and commercial buildings to bridges, sports halls and infrastructure elements. Their application allows for the creation of durable, stable, and aesthetically pleasing structures that can withstand various loads and environmental conditions.
Computational difficulties
From the perspective of calculating internal forces, calculating frames is somewhat more difficult than calculating beams.
a) Problem with interpreting the labeling for a vertical bar
First, once we learn to use the labeling convention going from the left or right side of the beam, a problem arises with interpreting the labeling for a vertical bar.
It is probably easiest at first to draw the so-called "bottoms" on the labeling convention, meaning the assumed bottom of the bar. If we also mark the bottoms (on either side) when solving the frame on the vertical bar, it will be easier for us to use the labeling. We can always rotate the paper so that we look at the vertical bar as if it were a horizontal beam - I assume we have already mastered calculating beams.
b) Problem with an inclined bar
Another problem will arise if we have an inclined bar, because until now for a horizontal bar (for a beam) if the force acted axially (horizontally) it was a normal force, if it acted perpendicular to the beam (vertically), it was a shear force.
For a vertical bar on the other hand, vertical forces were axial forces, and horizontal ones were shear forces to the bar.
For an inclined bar - if we load it with a horizontal force, we must project that force onto the direction along the bar (the normal component to the bar) and onto the direction perpendicular to the bar (the shear component). And here comes another skill we need to demonstrate to correctly draw the diagrams - projecting forces onto components.
A nice verification of the correctness of the drawn internal force diagrams is checking the equilibrium of a joint which we will show in the example below.
Video - course showing how to draw internal force diagrams
Video course showing how to draw internal force diagrams for a hinged frame with an inclined bar by recording the functions of internal forces.
Example - frame with an inclined bar
Below is the solution with a description of projecting forces onto components for an inclined bar and checking the equilibrium of the joint at the end
Solution
1) Marking the bottoms, describing characteristic points, applying reactions at supports and calculating them
As mentioned in the introduction - we mark the bottoms on the frame.
For clarity, we label the characteristic points and must mark the reactions - we will proceed to calculate them.
Equations of static equilibrium
| \begin{aligned} &\sum M_{C}^L = 0\\ &-6 \cdot \frac{3}{2} + V_{A} \cdot 3 = 0\\ &V_{A} = \frac{6 \cdot \frac{3}{2}}{3} = 3 \, \text{kN}\\ \end{aligned} | \( \quad \quad \) | \begin{aligned} &\Sigma X = 0\\ &8 + H_{F} = 0\\ &H_{F} = -8 \, \text{kN}\\ \end{aligned} |
2) Calculation of internal forces
Having calculated the reactions, we move on to calculating the internal forces. Here we will show how to project the force onto the normal and shear directions to the bar.
Projecting reaction VA onto the normal and shear directions to the bar
It is a good idea to make a drawing where we mark the vertical direction, horizontal direction, the axis of the bar (normal direction), and perpendicular to the axis of the bar (shear direction).
Angle alpha is taken by looking at the frame between the level and the axis of the bar.
Additionally, one can mark the direction of the positive normal force and the positive shear force, which applies when analyzing the inclined bar from point A.
In such a drawing, every other angle will be angle alpha.
Now, when I want to project reaction \( V_A=3 \ kN \) onto the normal and shear directions to the bar, I draw it on this "web" above.
The two closest components are normal and shear, and most importantly to remember is that: the component where there is angle alpha is multiplied by cosine of that angle, and the other component by sine alpha. See the drawing below.
So basically, we know everything, we write down the functions of internal forces in all intervals:
Interval AB (0 ≤ x ≤ 1.5)
| \( \quad \quad \) | \begin{aligned} &Q(x) = 3 \cos (\alpha) = 1.8 \, \text{kN} \\ &M(x) = 3x \\ &M(0) = 0 \\ &M(1.5) = 4.5 \, \text{kNm} \\ &N(x) = 3 \sin (\alpha) = -2.4 \, \text{kN} \end{aligned} |
Interval BC (1.5 ≤ x ≤ 3)
| \( \quad \quad \) | \begin{aligned} &Q(x) = 3 \cos (\alpha) - 6 \cos (\alpha) = -1.8 \, \text{kN} \\ &M(x) = 3x - 6 \cdot(x - 1.5) \\ &M(1.5) = 4.5 \, \text{kNm} \\ &M(3) = 0 \\ &N(x) = 3 \sin (\alpha) + 6 \sin (\alpha) = -2.4 \, \text{kN} \end{aligned} |
Interval CD (0 ≤ x ≤ 4)
| \( \quad \quad \) | \begin{aligned} & \mathrm{N}(\mathrm{x})=-8 \ \mathrm{kN} \\ & \mathrm{Q}(\mathrm{x})=3-6-3 \cdot \mathrm{x}=-3 \cdot \mathrm{x}-3 \\ & \mathrm{Q}(0)=-3 \ \mathrm{kN} \\ & \mathrm{Q}(4)=-15 \ \mathrm{kN} \\ & \mathrm{M}(\mathrm{x})=3 \cdot(3+\mathrm{x})-6 \cdot(1.5+\mathrm{x})-3 \cdot \mathrm{x} \cdot \frac{\mathrm{x}}{2} \\ & \mathrm{M}(0)=0 \ \mathrm{kNm} \\ & \mathrm{M}(4)=-36 \ \mathrm{kNm} \end{aligned} |
Interval FE (0 ≤ x ≤ 2)
| \( \quad \quad \) | \begin{aligned} & \mathrm{N}(\mathrm{x})=-15 \ \mathrm{kN} \\ & \mathrm{Q}(\mathrm{x})=8 \ \mathrm{kN} \\ & \mathrm{M}(\mathrm{x})=-8 \cdot \mathrm{x} \\ & \mathrm{M}(0)=0 \\ & \mathrm{M}(2)=-16 \ \mathrm{kNm} \end{aligned} |
Interval ED (2 ≤ x ≤ 4)
| \( \quad \quad \) | \begin{aligned} & \mathrm{N}(\mathrm{x})=-15 \ \mathrm{kN} \\ & \mathrm{Q}(\mathrm{x})=8 \ \mathrm{kN} \\ & \mathrm{M}(\mathrm{x})=-8 \cdot \mathrm{x}-4 \\ & \mathrm{M}(2)=-20 \ \mathrm{kNm} \\ & \mathrm{M}(4)=-36 \ \mathrm{kNm} \end{aligned} |
Based on the calculated values of internal forces, we draw the diagrams
3) Diagrams of internal forces
4) Checking the equilibrium of a joint
How to check the equilibrium of a joint is shown in a separate theoretical introduction - SEE HERE.
Below you will also find links to numerous examples of solutions for both simple frames and those with inclined bar(s), as well as frames with hinges.