ENSC 016: STATICS OF RIGID BODIES

LECTURE 6: STRUCTURAL ANALYSIS`

1ST SEMESTER | A.Y. 2023-2024
LECTURER: ENGR. MELVIN SINGAYAN

SIMPLE TRUSS PROCEDURE OF ANALYSIS

➢ A truss is a structure composed of slender ➢ Draw the free-body diagram of a joint having
members joined together at their end at least one known force at most two
points. unknown forces.
ASSUMPTION FOR DESIGN ➢ Assume force to be either in tension or
- All loads are applied at the joints. compression.
- The members are joined together by ➢ Orient the x and y axes such that the forces
smooth pins. on the free-body diagram can be easily
SIMPLE TRUSS: ASSUMPTION FOR DESIGN resolved into their x and y components and
then apply the two force equilibrium
equations: Fx=0 and Fy=0. Solve for the two
unknown member forces and verify their
correct sense.
➢ Using the calculated results, continue to
analyze each of the other joints. Be sure to
choose a joint having at most two unknowns
and at least one known force.

SAMPLE PROBLEM 6-1 Determine the force in each

member of the truss shown. Indicate whether the
- Because of these two assumptions, each members are in tension or compression.
truss member will act as a two-force
member, and therefore the force acting
at each end of the member will be
directed along the axis of the member. If
the force tends to elongate the member,
it is a tensile force (T); whereas if it tends
to shorten the member, it is a
compressive force (C).
- This method is based on the fact that if
the entire truss is in equilibrium, then
each of its joints is also in equilibrium.
- If the free-body diagram of each join is Support reactions:
drawn, the force equilibrium equations
+→ Σ𝐹𝑥 = 0; 600𝑁 − 𝐶𝑥 = 0
can then be used to obtain the member
force acting on each joint. 𝐶𝑥 = 600𝑁

+↶ Σ𝑀𝐶 = 0; −𝐴𝑦 (6𝑚) + 400𝑁(3𝑚) + 600𝑁(4𝑚) = 0

𝐴𝑦 = 600𝑁

+↑ Σ𝐹𝑦 = 0; 600𝑁 − 400𝑁 − 𝐶𝑦 = 0

𝐶𝑦 = 200𝑁
 ZERO FORCE MEMBERS
➢ Truss s analysis using the method of joints is
greatly simplified if we can first identify
those members which support no loading.
These zero-force members are used to
increase the stability of the truss during
construction and to provide added support if
the loading is changed.
➢ The zero-force members of a truss can
Joint A:
generally be found by inspection of each of
4 the joints.
+↑ Σ𝐹𝑦 = 0; 600𝑁 − 𝐹𝐴𝐵 = 0 ➢ If only two non-collinear members form a
5
truss joint and no external load pr support
𝐹𝐴𝐵 = 750𝑁 (𝐶) reaction is applied to the joint, the two
members must be zero-force members.
3
+→ Σ𝐹𝑥 = 0; 𝐹𝐴𝐷 − (750𝑁) = 0
5
𝐹𝐴𝐷 = 450𝑁 (𝑇)

➢ If three members form a truss joint in which

two of the members are collinear, the third
member is a zero-force member provided no
external force of support reaction has a
3 component that acts along this member.
+→ Σ𝐹𝑥 = 0; −450𝑁 + 𝐹𝐷𝐵 + 600𝑁 = 0
5
𝐹𝐷𝐵 = −250𝑁 𝑜𝑟 250𝑁 (𝑇)

4
+↑ Σ𝐹𝑦 = 0; −𝐹𝐷𝐶 − (−250𝑁) = 0
3
𝐹𝐷𝐶 = 200𝑁 (𝐶)

+→ Σ𝐹𝑥 = 0; 𝐹𝐶𝐵 − 600𝑁 = 0

𝐹𝐶𝐵 = 600𝑁 (𝐶) SAMPLE PROBLEM 6-2 Using the method of joints,
determine all the zero-force members of the Fink
To check: roof truss shown. Assume all joints are pin
connected.
+↑ Σ𝐹𝑦 = 0; 200𝑁 − 200𝑁 = 0

Joint G:

+→ Σ𝐹𝑥 = 0; 𝐹𝐺𝐶 = 0
Joint F:
EQUATIONS OF EQUILIBRIUM
➢ Moments should be summed about a point
that lied at the intersection of the lines of
action of two unknown forces, so that the
third unknown force can be determined
directly from the moment equation.
➢ If two of the unknown forces are parallel,
+↑ Σ𝐹𝑦 = 0; 𝐹𝐹𝐶 cos 𝜃 = 0 force may be summed perpendicular to the
𝑆𝑖𝑛𝑐𝑒 𝜃 ≠ 90°; 𝐹𝐹𝐶 = 0 direction of these unknowns to determine
directly the third unknown force.
Joint D:
SAMPLE PROBLEM 6-3 Determine the forces in
members GE, GC, and BC of the truss shown.
Indicate whether the members are in tension or
compression.

+↙ Σ𝐹𝑥 = 0; 𝐹𝐷𝐹 = 0

THE METHOD OF SECTIONS

➢ When we need to find the force in only a few
members of a truss, we can analyze the truss
using the method of sections.
➢ It is based on the principle that if the truss is
in equilibrium, then any segment of the truss
is also in equilibrium.

+→ Σ𝐹𝑥 = 0; 400𝑁 − 𝐴𝑥 = 0

𝐴𝑥 = 400𝑁

↺ +Σ𝑀𝐴 = 0; −1200𝑁(8𝑚) − 400𝑁(3𝑚)

+ 𝐷𝑦 (12𝑚) = 0

𝐷𝑦 = 900𝑁

+↑ Σ𝐹𝑦 = 0; 𝐴𝑦 − 1200𝑁 + 900𝑁 = 0

PROCEDURE OF ANALYSIS
𝐴𝑦 = 300𝑁
➢ Decide on how to “cut” or section the truss
through the members where forces are to be
determined.
➢ Before isolating the appropriate section, it
may force be necessary to determine the
truss’s support reactions. If this is done, then
the three equilibrium equations will be
available to solve for member forces at the
section.
➢ Draw the free-body diagram of that segment
of the sectioned truss which has the least ↺ +Σ𝑀𝐺 = 0; −300𝑁(4𝑚) − 400𝑁(3𝑚)
number of forces acting on it. + 𝐹𝐵𝐶 (3𝑚) = 0
➢ Use one of the two methods described
above for establishing the sense of the 𝐹𝐵𝐶 = 800𝑁 (𝑇)
unknown member forces.
 ↺ +Σ𝑀𝐶 = 0; −399𝑁(6𝑚) + 𝐹𝐺𝐸 (3𝑚) = 0 free-body diagrams of its parts must be
drawn.
𝐹𝐺𝐸 = 800𝑁 (𝐶)
SAMPLE PROBLEM 6-5 For the frame shown,
draw the free-body diagram of (a) each member,
SAMPLE PROBLEM 6-4 Determine the force in (b) the pins at B and A, and (c) the two members
connected together.
member CF of the truss shown. Indicate whether
the member is in tension or compression. Assume
each member is pin connected.

PROCEDURE OF ANALYSIS
FREE-BODY DIAGRAM
➢ Draw the free-body diagram of the entire
frame or machine, a portion of it, or each of
its members. The choice should be made so
that it leads to the most direct solution of the
problem.
➢ Identify the two force members. Remember
4 6 that regardless of their shape, they have
= = 4𝑚 equal but opposite collinear forces acting at
4 + 𝑥 (8 + 𝑥)𝑥
their ends.
➢ When the free-body diagram of a group of
members of a frame or machine is drawn,
the forces between the connected parts of
this group are internal forces and are not
shown on the free-body diagram of the
group.

PROCEDURE
↺ +Σ𝑀𝐶 = 0; −𝐹𝐶𝐹 sin 45 (12𝑚) + 3𝑘𝑁(8𝑚) ➢ Forces common to two members which are
− 4.75𝑘𝑁(4𝑚) = 0 in contact act with equal magnitude but
opposite sense on the respective free-body
𝐹𝐶𝐹 = 0.589 𝑘𝑁 (𝐶)
diagrams of the members.
➢ In many cases it is possible to tell by
inspection the proper sense of the unknown
FRAMES AND MACHINES forces acting on a member; however, if this
➢ Frames and machines are two types of seems difficult, the sense can be assumed.
structures which are often composed of pin- ➢ Remember that once the free-body diagram
connected multi-force members, i.e., id drawn, a couple moment is a free vector
members that are subjected to more than and can act at any point on the diagram.
two forces. Frames are used to support Also, a force is a sliding vector and can act at
loads, whereas machines contain moving any point along its line of action.
parts and are designed to transmit and alter
the effect of forces. EQUATIONS OF EQUILIBRIUM
➢ In order to determine the forces acting at the ➢ Count the number of unknowns and
joints and supports of a frame or machine, compare it to the total number of
the structure must be disassembled, and the equilibrium equations that are available. In
two dimensions, there are three equilibrium
 equations that can be written for each Bx = 0
member. −8𝑘𝑁(1𝑚) + 𝐶𝑦 (2𝑚) = 0 ; 𝐶𝑦 = 4𝑘𝑁
➢ Sum moments of a point that lies at the 𝐵𝑦 − 8𝑘𝑁 + 𝐶𝑦 = 0; 𝐵𝑦 = 4𝑘𝑁
intersection of the lines of action of as many
of the unknown forces as possible. 3
𝐴𝑋 − (10𝑘𝑁) ( ) + 𝐵𝑥 = 0; 𝐵𝑥 = 6𝑘𝑁
➢ If the solution of a force or couple moment 5
magnitude is found to be negative, it means 4
𝑀𝐴 − (10𝑘𝑛) ( ) (2𝑚) − 𝐵𝑦 (4𝑚) = 0;
the sense of the force is the reverse of that 5
shown on the free-body diagram. 𝑀𝐴 = 32𝑘𝑁 ∙ 𝑚
4
𝐴𝑦 − (10𝑘𝑁) ( ) − 𝐵𝑦 = 0; 𝐴𝑦 = 12𝑘𝑁
SAMPLE PROBLEM 6-6 Determine the horizontal 5
and vertical components of force which the pin
at C exerts in member BC if the frame. SAMPLE PROBLEM 6-8 The two planks are
connected by cable BC and a smooth spacer DE.
Determine the reactions at the smooth supports
A and F, and find the force developed in the cale
and spacer.

↺ +Σ𝑀𝐶 = 0; 2000𝑁(2𝑚) − (𝐹𝐴𝐵 sin 60) (4𝑚) = 0

𝐹𝐴𝐵 = 1154.7𝑁
+→ Σ𝐹𝑥 = 0; 1154.7 cos 60 − 𝐶𝑋 = 0
𝐶𝑋 = 577𝑁
+↑ Σ𝐹𝑦 = 0; 1154.7 sin 60 − 2000𝑁 + 𝐶𝑦 = 0
𝐶𝑦 = 1000𝑁
↺ +Σ𝑀𝐴 = 0; 𝐹𝐷𝐸 (6𝑓𝑡) − 𝐹𝐵𝐶 (4𝑓𝑡) − 100𝑙𝑏(2𝑓𝑡) = 0
SAMPLE PROBLEM 6-7 The compound beam ↺ +Σ𝑀𝐹 = 0; 𝐹𝐷𝐸 (4𝑓𝑡) − 𝐹𝐵𝐶 (6𝑓𝑡) + 200𝑙𝑏(2𝑓𝑡) = 0
shown is pin connected at B. Determine the 𝐹𝐷𝐸 = 140𝑙𝑏 ; 𝐹𝐵𝐶 = 160𝑙𝑏
components of reaction at its supports. Neglect
its weight and thickness. +↑ Σ𝐹𝑦 = 0; 𝑁𝐴 + 140𝑙𝑏 − 160𝑙𝑏 − 100𝑙𝑏 = 0
𝑁𝐴 = 120𝑙𝑏
+→ Σ𝐹𝑥 = 0; 𝑁𝐹 + 160𝑙𝑏 − 140𝑙𝑏 − 200𝑙𝑏 = 0
𝑁𝐹 = 180𝑙𝑏

SAMPLE PROBLEM 6-9 The frame supports the

50-kg cylinder. Determine the horizontal and
vertical components of reaction at A and the
force at C.

Segment BC:
+← Σ𝐹𝑥 = 0 ↺ +Σ𝑀𝐵 = 0 +↑ Σ𝐹𝑦 = 0

Segment AB:
+→ Σ𝐹𝑥 = 0 ↺ +Σ𝑀𝐴 = 0 +↑ Σ𝐹𝑦 = 0
 +→ Σ𝐹𝑥 = 0; 𝐷𝑥 − 50(9.81) = 0
𝐷𝑥 = 490.5 𝑁
+↑ Σ𝐹𝑦 = 0; 𝐷𝑦 − 50(9.81) = 0
𝐷𝑦 = 490.5 𝑁

↺ +Σ𝑀𝐴 = 0; 𝐹𝐵𝐶 (0.6𝑚) + 490.5𝑁(0.9𝑚)

− 490.5𝑁(1.20𝑚) = 0

𝐹𝐵𝐶 = 245.45 𝑁

+→ Σ𝐹𝑥 = 0; 𝐴𝑥 − 245.25𝑁 − 290.5 = 0

𝐴𝑥 = 736𝑁
+↑ Σ𝐹𝑦 = 0; 𝐴𝑦 − 490.5𝑁 = 0`
𝐴𝑦 = 490.5𝑁