2. Forces
Compression
A body being squeezed
Tension
A body being stretched
3. Truss
A truss is composed of slender members
joined together at their end points.
– They are usually joined by welds or gusset
plates.
4. Simple Truss
A simple truss is composed of triangles, which
will retain their shape even when removed
from supports.
5. Pinned and Roller Supports
A pinned support can
support a structure in two
dimensions.
A roller support can
support a structure in
only one dimension.
6. Solving Truss Forces
Assumptions:
All members are perfectly straight.
All loads are applied at the joints.
All joints are pinned and frictionless.
Each member has no weight.
Members can only experience tension or
compression forces.
What risks might these assumptions pose if
we were designing an actual bridge?
7. Static Determinacy
A statically determinate structure is one that
can be mathematically solved.
2J = M + R
J = Number of Joints
M = Number of Members
R = Number of Reactions
8. Statically Indeterminate
B
Did you notice
the two pinned
connections?
A C
D
FD = 500 lb
A truss is considered statically indeterminate when the
static equilibrium equations are not sufficient to find the
reactions on that structure. There are simply too many
unknowns.
Try It
2J = M + R
9. Statically Determinate
B Is the truss
statically
determinate
now?
A C
D
FD = 500 lb
A truss is considered statically determinate when the
static equilibrium equations can be used to find the
reactions on that structure.
Try It 2J = M + R
10. Static Determinacy Example
Each side of the main street bridge in Brockport, NY has 19
joints, 35 members, and three reaction forces (pin and roller),
making it a statically determinate truss.
2J M R What if these
numbers were
2 19 35 3 different?
38 38
12. Equilibrium Equations
Fx 0
The sum of the forces in the x-direction is
zero.
Do you remember the Cartesian coordinate system? A
vector that acts to the right is positive, and a vector that
acts to the left is negative.
13. Equilibrium Equations
Fy 0
The sum of the forces in the y-direction
is zero.
A vector that acts up is positive, and a vector that
acts down is negative.
14. Using Moments to Find RCY
A force that causes a B
clockwise moment is
negative.
A C
RAx
D
A force that causes a 3.0 ft 7.0 ft
counterclockwise R Ay RCy
moment is positive. 500 lb
FD contributes a negative MA 0
moment because it causes FD(3.0 ft) RCy(10.0 ft) 0
a clockwise moment.
500lb (3.0 ft) RCy(10.0 ft) 0
RCy contributes a positive 1500lb ft RCy(10.0 ft) 0
moment because it causes
RCy (10.0 ft) 1500lb ft
a counterclockwise
moment. RCy 150lb
15. Sum the y Forces to Find RAy
We know two out of the B
three forces acting in the
y-direction. By simply A C
summing those forces RAx
D
together, we can find the 150. lb
R Ay
unknown reaction at 500. lb
point A.
Fy 0
FD RCy RAy 0
Please note that FD is 500. lb 150.00 lb RAy 0
shown as a negative
350. lb RAy 0
because of its direction.
RAy 350. lb
16. Sum the x Forces to Find Ax
Because joint A is pinned, it is B
capable of reacting to a force
applied in the x-direction. A C
RAx
D
350. lb 150. lb
However, since the only load 500. lb
applied to this truss (FD) has no
x-component, RAx must be zero.
Fx 0
Ax 0
17. Method of Joints
Use cosine and sine to determine x and y vector
components.
Assume all members to be in tension. A positive answer will
mean the member is in tension, and a negative number will mean the
B
member is in compression.
As forces are solved, update free body diagrams. Use
correct magnitude and sense for subsequent joint free body diagrams.
19. Method of Joints
Using Truss Dimensions to Find Angles
B
tan opp
1
adj
4.0 ft
4.0 ft
A θ1 tan 4.0 ft
θ 2 C
1
D 3.0 ft
3.0 ft 7.0 ft
tan 4.0
1
1
3.0
1
53.130
20. Method of Joints
Using Truss Dimensions to Find Angles
tan 1 opp B
adj
4.0 ft 4.0 ft
4.0 ft
tan 1 A θ θ2 C
7.0 ft 1
D
3.0 ft 7.0 ft
tan 4.0
1
1
7.0
1
29.745
21. Method of Joints
Draw a free body diagram of each pin.
B
A 53.130 29.745 C
RAx
D
RAy 500lb
RCy
Every member is assumed to be in tension. A positive
answer indicates the member is in tension, and a negative
answer indicates the member is in compression.
22. Method of Joints
Where to Begin
Choose the joint that has the least number of unknowns.
Reaction forces at joints A and C are both good choices to
begin our calculations.
B
BD
A C
RAx 0 AD CD
D
RAy
350lb 500lb RCy
150lb
23. Method of Joints
FY 0
RAy AB y 0
437.50 lb 350lb AB sin 53.130 0
AB
AB sin 53.130 350lb
A 53.130
AD
350lb
AB
350 lb sin 53.130
AB 438 lb
24. Method of Joints
Update the all force diagrams based on AB
being under compression.
B
BD
A C
RAx= 0 AD CD
D
RAy= 350lb 500lb RCy= 150lb
25. Method of Joints
FX 0
ABx AD 0
437.50 lb cos53.130 AD 0
AB 437.50 lb
AD 437.50 lb cos53.130
A 53.130
AD AD 262.50 lb
262.50 lb
350 lb
26. Method of Joints
FY 0
RCy BC y 0
302.33 lb 150 lb BC sin 29.745 0
BC
29.745 C BC sin 29.745 150 lb
CD
150lb
BC
150 lb sin 29.745
BC 302 lb
27. Method of Joints
Update the all force diagrams based on BC
being under compression.
B
BD
A C
RAx= 0 AD CD
D
RAy= 350lb 500lb RCy= 150lb
28. Method of Joints
FX 0
BCx CD 0
302.33 lb cos29.745 CD 0
BC 302.33 lb
29.745 C CD 302.33 lb cos29.745
CD
262.50 lb CD 262.50 lb
150 lb
29. Method of Joints
500lb
BD FY 0
D
BD FD 0
BD 500lb 0
BD 500lb
500lb
