FBD, Free body diagram, static equilibrium

1. Free Body Diagram
V K Jadon
V K Jadon, Professor, Mechanical Engineering
𝑑
𝐷
𝑙 𝑓

2. Maximum induced stress at any point in a loaded machine member <= Design Stress
How many design equations are needed for one component to fix one dimension?
Design equation for Strength (Static Load)
Induced stress
Calculation of Load Effect of Load
V K Jadon, Professor, Mechanical Engineering
?

3. Free body diagram is used to evaluate the forces
(or/and reactions) acting on a member under the action
of some external known or unknown forces.
It is the diagram of (i) an assembly isolated from all the
supports or (ii) a member(s) of assembly disconnected
from the other members from its supports or (iii) portion
of the member showing all the external forces acting on it
and replacing supports/connections by appropriate
reactions.
125 250
375
2225 𝑁
The FBD of a member is in the state of static
equilibrium under the forces and reactions shown
on it.
A body is in static equilibrium, it must satisfy all
equations of static equilibrium.
Sometimes, a single FBD may not enable us to
find the required force(s) in the member.
In such cases, free body diagrams of different
members are analyzed.
𝐹𝑥 = 0; 𝐹𝑦 = 0; 𝑀𝑧 = 0
Free body diagram
V K Jadon, Professor, Mechanical Engineering

4. 1. Isolate the member (assembly or members together or portion of
member) from the supports.
2. Replace the supports with appropriate reactions
3. Show all external forces acting on the member(s) considered for FBD only.
4. Apply all equations of equilibrium for each FBD separately and solve for
the unknowns. 125 250
375
2225 𝑁
Roller Support Hinge/Pin Support
Fix Support
𝑅 𝑦 𝑅 𝑅 𝑦
𝑅 𝑥
𝑅 𝑦
𝑅 𝑥
𝑀𝑧
Steps to draw FDB
V K Jadon, Professor, Mechanical Engineering

5. FBD of assembly
1. Remove the supports at A (hinge) and B (hinge).
2. Replace support by reactions (𝐴 𝑥, 𝐴 𝑦) and (𝐵𝑥, 𝐵𝑦).
3. Show 2225 N at point C.
125 250
375
2225 𝑁
𝐴 𝑦
𝐴 𝑥 𝐵𝑥
𝐵𝑦
𝐹𝑥 = 0; 𝐹𝑦 = 0; 𝑀𝑧 = 0
FBD is in static equilibrium and hence should satisfy
equations of static equilibrium.
Important Notes:
Do not try to visualize the number and directions of the reactions.
It does not follow any rule when the point considered first time. If the same
point is repeating in subsequent FBD on different member, it must obey
Newton’s third law and the directions would be opposite.
If the point is on the same member in subsequent FBD, the direction should
not be changed.
How many maximum unknown we can find from one FBD ?
𝐶 𝑦
𝐶 𝑥
𝐴 𝑦
𝐴 𝑥
𝐵𝑥
𝐵𝑦
𝐶 𝑥
𝐶 𝑦
2225 𝑁
Example
125 250
375
2225 𝑁
?
V K Jadon, Professor, Mechanical Engineering

6. 125 250
375
2225 𝑁
𝐴 𝑦
𝐴 𝑥 𝐵𝑥
𝐵𝑦
𝐵𝑥
𝐵𝑦
𝐶 𝑥
𝐶 𝑦
2225 𝑁
𝐶 𝑦
𝐶 𝑥
𝐴 𝑦
𝐴 𝑥
(𝐼) (𝐼𝐼) (𝐼𝐼𝐼)
125 250
375
2225 𝑁
𝐴 𝑦
𝐵𝑦
(𝐼𝑉)
𝐹𝑥 = 0; 𝐹𝑦 = 0; 𝑀𝑧 = 0
𝑭𝑩𝑫 (𝑰𝑰)
𝑀𝐴 = 0 gives 𝐶 𝑥 = 0
𝐹𝑥 = 0 gives 𝐴 𝑥 = 0
𝐹𝑦 = 0 gives 𝐴 𝑦 = −𝐶 𝑦
𝐹𝑥 = 0; 𝐹𝑦 = 0; 𝑀𝑧 = 0
𝑭𝑩𝑫 (𝑰𝑰𝑰)
𝐹𝑥 = 0 gives 𝐵𝑥 = 0
𝑀 𝐵 = 0 gives 2225 375 + (𝐶 𝑦) 250 = 0
𝐶 𝑦 = −3337.5 𝑁
𝐹𝑦 = 0 gives 𝐵𝑦 = 3337.5 𝑁
V K Jadon, Professor, Mechanical Engineering

7. References
V K Jadon, Professor, Mechanical Engineering