Example of Statical Determinate Pin-Jointed Frame #1

In this post we will learn how to calculate the member forces and joint deflection in Statical Determinate Pin-Jointed Frame.

Find the explanation of the theory from previous post.

EXAMPLE

1. Find the vertical and horizontal deflection at joint D!

    - The cross sectional area for all members , A = 1200 m^2

    - The modulus of elasticity for all members, E = 205 kN/mm^2

ANSWER

1. Member forces

Therefore the summary of member forces is :

2. Horizontal deflection at point D
Based on previous post, the member forces coefficient due to imaginer force will be :

Therefore the summary of member forces is :

Use this formulae to compute the deflection : 

Where :
u = Member's coefficient due to imaginer force
L = Member's length
P = Member's axial force
E = Member's modulus of elasticity
A = Member's cross section area

Thus the member's horizontal deflection at point D will be 2.746 mm (positive value means the deflection direction is equal to the imaginer force).

3. Vertical deflection at point D

The member's force coefficient due to imaginer force will be :

Therefore the summary of member forces is :

Use this formulae to compute the deflection :  
 
Where :
u = Member's coefficient due to imaginer force
L = Member's length
P = Member's axial force
E = Member's modulus of elasticity
A = Member's cross section area

Thus the member's vertical deflection at point D will be 1.179 mm (positive value means the deflection direction is equal to the imaginer force).