Structural Analysis: Determining Support Reactions & Axial Forces

This report provides a detailed structural analysis of a statically determinate truss, focusing on the calculation of support reactions and member forces. The report begins by verifying the static determinacy of the structure using the formula m + r = 2j. Subsequently, it calculates the support reactions at points A and B using the equations of static equilibrium. The report then employs the method of joints to determine the axial forces in members AC and AF, utilizing a free body diagram of joint A. Finally, the method of sections is applied to calculate the force in member GH, dividing the structure into two sections and analyzing the equilibrium of one section. The report includes free body diagrams and detailed calculations to illustrate each step of the analysis.

Structural Analysis 1
STRUCTURAL ANALYSIS
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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):

Structural Analysis 4
∑ Fy=0
→ RA + AF sin 45° = 0
RA + 0.707107AF = 0; 120 kN = -0.7071AF
AF = -169.7 kN → AF ≈ 170 kN (compression).
∑ Fx=0
→ AC + AF cos 45° = 0
AC + 0.7071AF = 0 (substituting AF)
AC + 0.7071(-170) = 0
AC – 120 kN = 0
AC = 120 kN (tension)
Part e: Method of sections
This is the fastest method of determining member forces of a structure and it starts by dividing
the structure into two individual sections and analyzing each one separately (3). Since the
member force being determined is GH, a section x-x is taken by cutting the structure vertically
through members GH, DH and DE as shown in Figure 2 below

Structural Analysis 5
Figure 2: Section X-X of the structure
The two sections in Figure 2 above are in equilibrium hence the axial force in GH can be
determined by solving any of the sections. In this case, the section on the left hand side is used.
The free body diagram of the left hand section is as shown in Figure 3 below
Figure 3: Free body diagram of the left hand section of the structure
The forces of the left hand side section are determined using equations of static equilibrium. The
main forces of focus are FGH, FDH and FDE.

Structural Analysis 6
The simplest way of determining FGH is taking moments at point D. When this is done, it means
that the only unknown in the moment equation will be FGH. The reason for this is because forces
FDE, FDH and the point force of 35 kN will not have a moment arm hence will produce zero
moments at joint D. The moments at D are calculated as follows:
∑ MD=0
(120 kN x 2 m) – (90 kN x 1 m) + (FGH x 1 m) = 0
→ 240 km – 90 kNm + FGHm = 0
150 kNm + FGHm = 0
FGH = -150 kN
Thus GH = 150 kN (compression).

Structural Analysis 7
References
(1) Mississippi State University. Analysis of Statistically Determinate Trusses. [Online].
Mississippi State University, (n.d.). [27 February 2019]. Available from:
http://www.ae.msstate.edu/vlsm/truss/
(2) Erochko, J. The Method of Joints. [Online]. Learn About Structures, 2016. [27 February
2019]. Available from: http://www.learnaboutstructures.com/Method-of-Joints
(3) Moore, J. The Method of Sections. [Online]. Adaptive Map, (n.d.). [27 February 2019].
Available from: http://adaptivemap.ma.psu.edu/websites/structures/method_of_sections/
methodofsections.html