# Frame Analysis

## Contents

## Introduction

Frames are widely used in engineering practices around the world. Concepts of frames may be used from houses, to bridges, to buildings. Frame analysis is used to analyze simple frames. It is used to determine how the frame will act under external forces^{[1]}. The external forces impart forces on the internal members which will make them bend and could make them break. This results in shear and bending moment forces on the members. These shear and bending moment forces can be found and may be shown in the shear and bending moment diagrams. These can be constructed by analyzing each internal member individually.

## Shear Diagrams

Shear diagrams are used to help analyze the stresses within frames^{[2]}. This helps determine the type of structural beams that will need to be used in order to prevent failure within the beams.

### Sign Convention

In shear diagrams, you may either have positive or negative shear. In typical engineering practices, clockwise rotation is considered to be positive, and counter clockwise rotation is considered to be negative^{[2]}. When a frame is cut at the joints, we always assume positive shear. This means that the member will have a downwards arrow to the right of a cut, and an upwards arrow to the left to resemble shear (Figure 1)

^{[2]}.

### Analysis

Analysis of shear diagrams help determine the overall strength of the beams for design that will be needed^{[2]}. During this process, both the external and internal forces are taken into account. Once all the forces are found using the equations of equilibrium, we may produce a shear diagram. For frame analysis, the frame must be cut into each individual member^{[2]}. Once this is done, we may analyze the members due to the reaction forces and than create a shear force diagram.

## Bending Moment Diagrams

We use moment diagrams to determine how the members will bend over the distance of the entire member. The diagram shows which directions the members will bend and how much they will bend due to the forces that were found from solving the members with the equilibrium equations.

### Sign Convention

Sign convention for moment diagrams is different from shear diagrams.In bending moment diagrams, positive bending moment is denoted by two arrows pointing upwards. This will make the beam bend concave up. We always assume positive bending moment. Negative moment would conversely be denoted by the arrows pointing down and make the beam bend concave down (Figure 2).

### Analysis

Bending moment diagrams are used to determine how much a beam will bend when it is subjected to certain forces^{[2]}. The diagrams show how the beam reacts to the forces and how these forces affect the beam when taking into account the distance^{[3]}. This is very helpful because when we see how far the beam is going to bend we can design it to withstand that much force so it won't break. Bending moment diagrams also allow us to determine where the points of maximum moment will occur. This is the point where the most stress on the beam will be and it is always the most negative or positive point on the diagram. This is very helpful because when you know where you need to resist the most force you can design your frame with that in mind.

## Equations of Equilibrium

Below are the three equations of equilibrium we use in every problem to help solve for any unknowns^{[2]}.

$ \sum {F_x} = 0 $

$ \sum {F_y} = 0 $

$ \sum {M_} = 0 $

## Steps to Analyze Frames

Frame analysis is very similar to Beam Analysis because frames are made up of many beams. Each member of the frame is just analyzed like a single beam but it has to be done more than once.

- The first step for frame analysis is to determine if the structure is structurally sound. Please refer to Determinacy, Indeterminacy and Stability on how to determine this step.
- Next, perform the Calculation of Support Reactions. Once the support reactions are found you can move on to solving the whole frame.
- To solve the whole frame you must first break the frame into individual members and solve the reactions using the () (Figure 3).
- Once all the forces are determined, you may now construct your shear diagram.
- It is easiest to start at the left side and work your way to the right. The shear diagram starts at zero unless there is a reaction force at the support, which in most cases there is for frame analysis
^{[1]}. - Refer to the sign convention stated earlier in order to determine if the shear is negative or positive. Once this is determined, you may start constructing your diagram.
- Draw the reaction in the appropriate direction and move along the beam to the next reaction force.
- At the next reaction force, once again determine if it is positive shear or negative shear and add this number to the initial shear you determined. This is the next known point of shear. These points will be connected by drawing a horizontal line across to the next reaction force, than a straight line in the vertical direction will connect these points
^{[1]}. If it is a uniform distributed load over a section of the member, these points would have a linear connection where you basically connect the dots^{[1]}. - Once you go along the beam and determine all of the shears, the final shear at the end must be zero. If it is not zero, this means that you have made a mistake somewhere.

- It is easiest to start at the left side and work your way to the right. The shear diagram starts at zero unless there is a reaction force at the support, which in most cases there is for frame analysis
- Determine Moment diagram.
- It is easiest to start at the left side and work your way to the right. The moment diagram starts at zero unless there is a reaction moment at the support, which in most cases there is for frame analysis.
- Refer to the sign convention stated earlier in order to determine if the moment is negative or positive. Once this is determined, you may start constructing your diagram.
- Draw the moment in the appropriate direction and with the appropriate magnitude of the moment and move along the beam to the next moment that was found
^{[1]}. - At the next moment, once again determine if it is positive shear or negative moment and add this number to the initial moment you determined. This is the next known moment. These points will be connected by drawing a line across to the next moment, then a straight line in the vertical direction will connect these points to the zero line
^{[1]}. If there are negative and positive moments present then the line connecting the two points will cross over the zero line which is where the maximum shear force will occur. - The moments must add up to zero or you have made a mistake somewhere.

These images show what the shear and moment diagrams would look like according to the directions of forces in Figure 3.

## Frame Example Problem

In this problem, we will analyze the frame shown in Fig 4 below using the steps listed above.

### Step 1 (Checking Determinacy)

- $ m = 3, \; j = 4, \; r = 3 and \;ec = 0 $

- $ i=(3m+r)-(3j+e_{c}) $

- $ i=(3*3+3)-(3*4+0) $

- $ i=0 $

- Therefore, the frame is stable and statically determinate.

### Step 2 (Finding Support Reactions)

- $ \sum F_{x} = 0 $
- $ -A_{x}+ 25 = 0 $
- $ A_{x} = 25 kN $

- $ \sum M_{A} = 0 $
- $ D_{y}(20)-10(20)(10)-25(10)=0 $
- $ D_{y}=112.5 kN $

- $ \sum F_{y} = 0 $
- $ 112.5-10(20)+A_{y}=0 $
- $ A_{y} =87.5kN $

### Step 3 (Finding Member End Forces)

* Joint A:*
By applying the three equations of equilibrium we obtain

- $ AB_{x} = -25 kN $
- $ AB_{y} = 87.5 kN $

**Member AB**

- $ \sum F_{x} = 0 $
- $ B_{x}+ 25 -25 = 0 $
- $ B_{x} = 0 $

- $ \sum F_{y} = 0 $
- $ -B_{y}+87.5= 0 $
- $ B_{y} = -87.5 kN $

- $ \sum M_{B} = 0 $
- $ M_{B}+25(10)-25(20)=0 $
- $ M_{B}=250 kN.m $

* Joint B (considering member BC):*
By applying the three equations of equilibrium we obtain

- $ BA_{x} = 0 $
- $ BA_{y} = 87.5 kN $
- $ BA_{M} = -250 kN.m $

**Member BC**

- $ \sum F_{x} = 0 $
- $ C_{x} = 0 $

- $ \sum F_{y} = 0 $
- $ 87.5-10(20)+C_{y}= 0 $
- $ C_{y} = 112.5 kN $

- $ \sum M_{C} = 0 $
- $ -M_{C}-250+10(20)(10)-87.5(20)=0 $
- $ M_{C}=0 $

* Joint C (considering member CD):*
By applying the three equations of equilibrium we obtain

- $ CD_{x} = 0 $
- $ CD_{y} = -112.5 kN $
- $ CD_{M} = 0 $

**Checking computations(considering member CD):**

- $ \sum F_{x} = 0 $

- $ \sum F_{y} = 0 $

- $ 112.5-112.5= 0 $

- $ \sum M_{B} = 0 $

### Step 4 (Axial, Shear, and Bending Moment Diagrams)

## References

- ↑
^{1.0}^{1.1}^{1.2}^{1.3}^{1.4}^{1.5}Jeffery Erochko, Class Lecture, Frame Analysis "Analysis of Plane Frames." Carleton University, Ottawa, Ontario, Azrieli Theatre, September 23rd, 2013 - ↑
^{2.0}^{2.1}^{2.2}^{2.3}^{2.4}^{2.5}^{2.6}Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning - ↑ Hibbeler, R.C. (2010). Mechanics of Materials: 8th Edition. Upper Saddle River, NJ: Pearson Education