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ENGR-102
W EEK 2 L ECTURE – B RIDGE M ODULE
O VERVIEW OF W EEK 2 LAB ACTIVITIES
T HE M ETHOD OF J OINTS
F REE BODY DIAGRAMS
Pramod Abichandani, Ph.D.
Richard Primerano, Ph.D.

T HE T RUSS




A structural member composed of triangular sections
The members of a truss are connected at nodes
Each member is in either tension or compression
A technique called the method of joints can be used to determine the forces in all members under a given load.
2

E XAMPLE : F IND THE F ORCES IN THIS
S TRUCTURE


We know the applied loads, the types of supports, and the geometry of the structure
From this, we can determine the tension and compression forces in the members AB, BC, and AC
100lb
B
1ft 1ft
1ft
pin joint – prevents motion in the X and Y directions.
A C
roller joint – prevents motion in the Y direction.
3

A NALYZING F ORCES IN A T RUSS
T HE M ETHOD OF J OINTS
1.
2.
3.
4.
Determine the support reactions.
Draw a free body diagram for each joint.
Write an equilibrium equation for each joint.
Solve for the forces at each joint.

A good video tutorial on the method of joints
 http://www.youtube.com/watch?v=ZOtSRbzKCC8
4

S TEP 1: D ETERMINE S UPPORT R EACTIONS

Employ Newton’s law to find the reaction forces keeping the structure stationary
100lb
B
F cx
A
F ay
1ft
1ft
1ft
C
F cy
Σ forces in X direction = 0
𝐹 𝑎𝑥
= 0
Σ forces in Y direction = 0
𝐹 𝑎𝑦
+ 𝐹 𝑐𝑦
− 100𝑙𝑏 = 0
Applying Σ of moments about A will tell us that:
𝐹 𝑎𝑦
= 𝐹 𝑐𝑦
= 50𝑙𝑏
5

S TEPS 2&3: D RAW F REE B ODY D IAGRAMS


Draw all the forces on node A
Each force must be resolved into its X and Y components
F ab
• sin 60 º F ab
0
50
60 º
F ac
F ab
• cos 60 º
Sum of forces in X direction
𝐹 𝑎𝑐
𝐹 𝑎𝑐
+ 𝐹 𝑎𝑏
= −𝐹 𝑐𝑜𝑠60 = 0 𝑎𝑏 𝑐𝑜𝑠60
Sum of forces in Y direction
50 + 𝐹 𝑎𝑏
𝐹 𝑎𝑏 𝑠𝑖𝑛60 = 0
= −50/𝑠𝑖𝑛60
F ab
F ac
= - 50 / 0.87 = -57.7 lb
= 57.7 * 0.5 = 28.9 lb
6

B RIDGE D ESIGN M ODULE W EEK 2
S OFTWARE T OOLS FOR B RIDGE D ESIGN

We introduce one of the three design tools (West
Point Bridge Designer) that will be used in this module for bridge design and analysis.

AutoCAD and Visual Analysis are the others



WPBD is an educational package developed and distributed by the United States Military
Academy.
It allows you to quickly design and simulate bridges using a variety of common construction materials.
Using this, you can visualize the forces in a truss for given loads.
7

S OME B ASIC P HYSICS
S TATIC E QUILIBRIUM

When an object is in static equilibrium, all of the particles that make up the object are stationary and there is no net force acting on the object.


Basically, the object is at rest and not accelerating
All applied forces are balanced by reaction forces.
100lb
B
0 lb
A
1ft
1ft
1ft
C
50lb 50lb
8

E XTERNAL F ORCES A CTING ON A
S TRUCTURE

Applied forces (loads) – these are the forces that the structure is designed to carry: cars on a bridge, shingles on a roof, lift force on a wing

Dead loads – loads permanently acting on a structure: the weight of members, road decking, roofing material

Live loads – time varying loads acting on a structure: cars on a bridge, people on a floor, snow on a roof

The dead loads can usually be calculated with high accuracy.

The live loads are generally estimated based on experience of calculation. video
9

E XTERNAL F ORCES A CTING ON A
S TRUCTURE

Reaction forces – exerted by support structures and responsible for keeping a structure in static equilibrium

These forces “adjust themselves” as the applied forces change.
100lb
B
B
100lb
1ft 1ft 1ft 1ft
0 lb
A
1ft
C
100 lb
A
1ft
C
50lb 50lb 43lb 43lb
10

E XAMPLE : T UG OF W AR

Consider a 3-way tug of war.




Force A is exerted along the -X axis
Force B is exerted along the -Y axis
Force C is exerted at a 45˚ angle w.r.t. the +X axis
The knot is in static equilibrium

If for force exerted by player A is 150lb, find all remaining forces.
F
A
= 150lb
45˚
F
C
+Y
+X
F
B
11

S OME INITIAL OBSERVATIONS


F
A
 is resisted (balanced) by a component of F
C
F
B plays no role in balancing F
A
F
B
 is also balanced by a component of F
C
F
A plays no role in balancing F
B
F
A
= 150lb
45˚
F
C
+Y
+X
F
B
12

R ESOLVING A V ECTOR INTO ITS
C OMPONENTS

The force F
C can be considered as the sum of two component forces, one acting in the X direction and the other acting in the Y direction.
+Y
+X
F
CY
F
CX
= F
C
= F
C
•sin(θ)
•cos(θ)
F
C
F
CY
θ
F
CX
As F
C becomes closer to horizontal:
• F
CX
• F
CY
→
F
C
→
0
θ
F
C
F
CX
F
CY
13

R ESOLVING A V ECTOR INTO ITS
C OMPONENTS

Note that we usually draw both component vectors with their tails at the origin, not head to tail.
same thing
F
C
F
C
F
CY
F
CY
θ θ
F
CX
F
CX
+Y
+X
14

B ACK TO OUR P ROBLEM

F
CY
F
C
F
A
= 150lb
F
B
45˚
F
CX
+Y
+X
15

T HE C ONDITIONS N EEDED FOR S TATIC
E QUILIBRIUM

An object in static equilibrium is one that “isn’t moving”

Neither translating nor rotating

Assuming the ropes can only move in the plane of the screen, this means


All forces in the X direction must sum to zero
All forces in the Y direction must sum to zero

We will come back to the rotation part
16

T HE EQUATIONS OF STATIC EQUILIBRIUM


If it were non-zero, the object would be accelerating
F
A
= 150lb
+Y
+X
F
CY
F
B
45˚
F
C
F
CX
F
CY
F
CX
= F
C
•sin(45) = F
C
= F
C
•cos(45) = F
C
/ √2
/ √2
ΣF
X
F
CX
F
A
F
C
= 0
– F
= F
A
CX
= 0
= F
C
= 150 √2 lb
/ √2
ΣF
Y
F
CY
F
F
B
B
= 0
– F
B
= F
CY
= 0
= F
= 150 lb
C
/ √2
17

W HAT IS M OMENT OF F ORCE

Also called moment or torque.
5lb
5lb
1ft
5 ft-lbs of torque
5lb
1ft
45˚
0.707ft
3.54 ft-lbs of torque 0 ft-lbs of torque 18

C ALCULATING THE R EACTION F ORCES
A CTING ON A S TRUCTURE

The reaction forces keep the object fixed both transnationally (X and Y directions) and rotationally (about the Z axis)
B 100lb
1ft
F
AX
A
F
AY
2ft
C
F
CY
19

T HE R ELEVANT E QUATIONS


The object is not translating: ΣF
X
The object is not rotating:

Σ torques about any point = 0

Lets choose point A: ΣM
A
= 0
ΣF
X
F
AX
F
AX
= 0
– 100 = 0
= 100
B 100lb
= 0, ΣF
Y
= 0
1ft
ΣF
Y
F
AY
F
AY
= 0
+ F
= -F
CY
CY
= 0
F
AX
A
2ft
C
ΣM
1ft*100lb + 2ft*F
F
CY
F
AY
A
= 0
= -50 ft-lb
= 50 ft-lb
CY
= 0
F
AY
F
CY
20

A NALYZING F ORCES IN A T RUSS
T HE M ETHOD OF J OINTS
1.
2.
3.
4.
Determine the support reactions.
Draw a free body diagram for each joint.
Write an equilibrium equation for each joint.
Solve for the forces at each joint.

A good video tutorial on the method of joints
 http://www.youtube.com/watch?v=ZOtSRbzKCC8
21

F INDING THE S UPPORT R EACTIONS


Last time we had a symmetrical structure that was symmetrically loaded.
What observations can we make


Weight distributions between support reactions
Sign of each support reaction
50lb
100lb
F
AX
A
F
AY 5 ft 5 ft
C
10 ft
F
CY
22

F INDING THE S UPPORT R EACTIONS

Last time we had a symmetrical structure that was symmetrically loaded.
100lb
F
AX
A
F
AY 5 ft 5 ft
50lb
C
10 ft
F
CY
ΣF
X
F
AX
= 0
= 0
ΣF
Y
F
AY
= 0
+ F
CY
= 150lb
ΣM
A
20F
CY
F
CY
F
AY
=
= 0
– 500 – 250= 0
=
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HANK YOU