Structural Analysis 1

Structural Analysis 1
STRUCTURAL ANALYSIS
Name
Course
Professor
University
City/state
Date

Structural Analysis 2
Structural Analysis
Part c: Support reactions
The formula m + r = 2j (where m is the number of members, r is the number of support reactions
and j is the number of joints of the structure) is used to determine if the structure is statically
determinate or not.
In this case, m = 21; r = 3 and j = 12
Therefore 21 + 3 = 2(12) → 24 = 24, hence the structure is statically determinate. This implies
that the reactions at supports A and B can be determined by use of equations of static equilibrium
only (1).
Sum of vertical forces of the structure, ∑ Fy=0
→ RA – 90 kN – 35 kN – 20 kN – 90 kN + RB = 0
RA – 235 kN + RB = 0
RA + RB = 235 kN
Sum of moments at A, ∑ MA =0
→ (90 kN x 1 m) + (35 kN x 2 m) + (20 kN x 4 m) + (90 kN x 5 m) – (RB x 6 m) = 0
90 kNm + 70 kNm + 80 kNm + 450 kNm – 6RBm = 0
690 kNm = 6RBm
RB = 115 kN
But RA + RB = 235 kN

Structural Analysis 3
→ RA = 235 kN – RB = 235 kN – 115 kN = 120 kN
Therefore RA = 120 kN and RB = 115 kN
Part d: Method of joints
The axial forces in members AC and AF can be determined using joint A only. This is possible
because joint A has only two unknown axial member forces, which can be determined using
equations of static equilibrium. The free body diagram of joint A is as shown in Figure 1 below.
Figure 1: Free body diagram of Joint A
It is assumed that all members are in tension (pulling away from the joint) hence negative values
will show that the member is in compression. In solving this problem, upward forces and those
pointing towards the right hand side are taken to be positive whereas downward forces or those
pointing to the left hand side are taken to be negative.
The member forces AC and AF from the free body diagram in Figure 1 above are determine by
taking the sum of vertical forces and horizontal forces separately as follows (2):