Examples
• Internal forces in simple structures
• Example: the truss
• Example with friction
(Serway 12.1-12.3 – see p. 372)
Physics 1D03 - Lecture 18
Trusses are composed of triangles made of
rigid, massless beams joined with pins. They
rest on two support points attached to pins, only
one of which is fixed.
Pins or dowels
through holes in beams do
Massless beams are a
not allow any torque to be
good approximation
transmitted through the joint.
because the truss can
support a load many
times its own weight.
With only one fixed support,
all horizontal supporting Normal force
forces are collected at one only here
point.
Fixed support point Roller support point
at a pin at a pin
Physics 1D03 - Lecture 18
“Two-Force Members”
Member in tension
B D
B D
forces on member forces by member
Member in compression
B D
B D
forces on member forces by member
In general, a member is subject to bending as well as tension or
compression; the truss is a special structure in which all of the
members are stressed in tension or compression only.
Physics 1D03 - Lecture 18
Concept: To solve for all the internal and
external forces in a ideal truss, set SF = 0
and St = 0 for the external forces. Then set
SF = 0 at each pin.
1. The torques due to internal forces are not required,
because they are automatically zero when the internal
forces are parallel to the beams.
2. Since the compression or tension forces at the end of
any beam are equal and opposite, not all of these
equations are required to solve for every force.
Physics 1D03 - Lecture 18
The Truss
Examples: Steel bridges, hydro towers, crane booms, roof supports
in buildings, the supports for the walkway between ABB and JHE, ...
A stable truss is built from triangles.
Provided the loads are applied only
at the joints, each member exerts a
force parallel to itself on each joint.
Physics 1D03 - Lecture 18
Example 6 kN C
The horizontal members are D
B
1.00 m long, and the angles A E
are 30o, 60o, and 90o. Find
the force of compression or I H G
tension in each member. 6 kN
(For now: in member CH).
Method:
1) Find the external forces on the whole truss.
2) Draw a free-body diagram for each joint. The members exert
forces on the joints; the directions of these forces are known.
3) Write equilibrium equations at each joint and solve.
Physics 1D03 - Lecture 18
Questions: C
1) If the directions shown for the FBC FDC
forces are correct, which members
would be in tension, and which in FH
compression? C
Forces on joint C
2) How many unknowns can be found
just by using the equations of static
equilibrium for joint C?
C
B D
A E
I H G
Physics 1D03 - Lecture 18
6 kN C
Results: B D
External forces, A E
F y N A N E Fg 0 I H G
6 kN NE
NA
NA = NE = 6 kN
Joint E: just two unknown forces,
solve to get: FED
FED = -12 kN, FGE = 12cos(30) 30o
= 10.4 kN FGE
NE
The – sign just means we guessed
wrong about the direction of FED . (It is 12 kN
actually in compression.)
30o
Which joint should be done next? FGE
NE
Physics 1D03 - Lecture 18
6 kN C
Joint G: D
B
FGD = 6 kN (tension) A E
FGH = 10.4 kN (tension)
I H G
Joint D: NA 6 kN NE
FDH = 6 kN (compression)
FDC = 6 kN (compression)
Joint C:
FCH = 6 kN (tension)
FBC = 6 kN (compression)
Question: Can we now say we know the rest of the forces “by
symmetry”? For example, can we say that FAB = FDE, and FBI = FGD?
Physics 1D03 - Lecture 18
Quiz
A
What does the force diagram C
for the pin at A look like?
B D
A) B) C) D)
Physics 1D03 - Lecture 18
Example: rolling a spool.
P
Inner radius a,
Outer radius b = 2a
ms = 0.60 at both surfaces
If the spool has weight w = 100 N, a
what minimum force P is needed to C
b B
rotate the spool?
A
Plan: Assume the spool is not rotating (still in equilibrium) but is
about to slip (so fs = ms N at each contact). Use the conditions for
equilibrium, and the two friction equations, to solve for P.
Physics 1D03 - Lecture 18
1) Free-body diagram: b = 2a
y P
w = 100 N
x ms = 0.60
a NB
C
b B
w fB
fA
A
NA
Physics 1D03 - Lecture 18
1) Free-body diagram: b = 2a
P
w = 100 N
Are the friction forces in the
y ms = 0.60
right directions?
Torques about centre: x
aP = b(fA + fB ) , so a NB
C
P = (b/a)(fA + fB ) = 2(fA + fB ) b B
w fB
Forces: fA
x-components, fA = NB A
y-components, NA + fB + P = w
Friction (about to slip): NA
fA = ms NA
fB = ms NB
Five equations in 5 unknowns – no sweat! (but plan your solution
before starting the algebra).
Physics 1D03 - Lecture 18
Quiz
When the rope is pulled as shown, which way will the spool move?
A) roll to the left pull
B) roll to the right
C) it won’t rotate
D) depends on the angle
a
b
Physics 1D03 - Lecture 18
Summary
• Internal forces may be found by considering the forces
on one part of a structure.
• If only two forces act on a member in equilibrium, the
forces must be parallel to the member.
Practice problems: 49, 57 (both versions of book)
Physics 1D03 - Lecture 18