Let's illustrate how to solve the deflections of trusses using the real work method. We shall demonstrate how to solve the movement of a truss joint with an applied load.
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Let's learn how to use the real work method to solve deflections. This example presents a case of axial strains.

The solution presented is in SI. The author will update the post soon to reflect English units.

To start, we shall consider a simple truss. It has a \(20 kN\) point load at \(E\). The value inside the parenthesis () is the member's cross-sectional area \(A\) in square centimeters. 

Simple truss example

Say that we are interested in finding the vertical translation \(\Delta_{E_v}\) at \(E\). Because there is a point load at said point, we solve for the vertical deflection using the general real work equation. We can use it on other points if there's a load. 

For this problem, we'll assume that all members are made of the same material so that the modulus of elasticity \(E\) is constant.

Main Solution

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Formulate S, A, and E Equations.

We start by identifying \(S\), \(L\), \(A\), and \(E\) per truss member. If there are variations in said variables, one must express these in terms of \(x\) (member's local axis). Our example has constant \(AE\) for all members. So, expressing it in terms of \(x\) is unnecessary.

Afterward, using the method of joints or sections, we solve for internal bar forces \(S\) of each due to the applied loads.

Bar forces S for real structure

We summarise our results for all bar forces \(S\), \(L\), \(A\), and \(E\) in the table. Positive bar forces are in tension, while negative-sensed bar forces are in compression.

Apply Real Work Equations

The only thing to do is apply the general real work equation. Using the same table, we create a column for \(S^2L/AE\). Afterward, we add the values of this column to obtain the required translation:

\(20^{k N} \times \Delta_{E_v}=\frac{675}{2 E}+\frac{1250}{E}+\frac{1350}{E}+\frac{675}{2 E}+\frac{3125}{E}+\frac{1600}{E}\)

\(\Delta_{E_v}=\frac{400}{E}\)

The positive direction indicates that the vertical translation at \(E\) matches the direction of the \(20 kN\) force, which is downward.

Created On
June 5, 2023
Updated On
February 23, 2024
Contributors
Edgar Christian Dirige
Founder
References

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Revision
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