FREE-BODY DIAGRAM (FBD) BASICS
FBDs are sketches of a body (or a segment of a body), for example, a shaft or a beam or a car, with the supports
replaced by the applied and internal forces, moments, or torque symbols. The top-level FBDs show the entire body.
Sub-level FBDs show the applied and internal forces, moments, and torques for a segment of a body. The location and
direction of the applied forces, moments, and torques are shown in both FBDs. Cut plane locations are where imaginary
planes are passed through the body to reveal internal forces, moments, and torques.
Axially Loaded Bodies
Bar with Load
Top-Level FBD
Cut Plane
C
A
RA
B
A
C
B
Support replaced by reaction force
Always shown in the plus x direction with a plus sign.
Slash indicates reaction force.
Forces signs are plus if directed in the plus axis direction.
Equilibrium Equation
∑Fx = RA - P = 0
RA = P
RA has the correct assumed direction.
Next Level FBD at Cut Plane to Find Internal Force
N1
Internal normal force, N1, is perpendicular to cut
surface and always assumed to be in tension. The
sign is plus if the force points in the positive axis
direction. If equilibrium equation solution results in
minus sign, force is compressive
Equilibrium Equation
∑Fx = - N1 - P = 0
N1 = - P
So N1 is a compressive force and the
assumed direction is opposite.
1
Torsionally Loaded Bodies
Shaft with Loads
Top-Level FBD
Cut Plane
TA
B
A
Support replaced by reaction torque
Always shown in the plus x direction with a plus sign
by the Right Hand Rule.
Slash indicates reaction force.
Equilibrium Equation
∑Tz = TA + T1 – T2 = 0
TA = T2 – T1
Torque signs are + = CW and – = CCW with respect to the plus z direction
Whether TA has a plus or minus sign depends on magnitudes of T2 and T1
Next Level FBD at Cut Plane B to Find Internal Torque at B
TB
B
Equilibrium Equation
∑Tz = TB – T2 = 0
TB = T2
Internal torques are always shown in the positive direction.
TB has a plus sign; the assumed direction is correct.
CCW (-)
CW (+)
Arrows and arcs equivalency
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Beams with Concentrated Loads
Beam with Concentrated Loads
Top-Level FBD
d
N1
N1
NA
A
B
C
RA
Cut Plane a-a
Equilibrium Equations
∑Fx = NA – N1= 0, NA = N1
∑Fy = RA – P = 0, RA= P
∑MA = -MA – d*P = 0. MA = -d*P
Applied forces and moments are shown positive if directed in the positive axis direction
Reaction forces, and reaction moments are always shown in their positive direction.
RA is plus; the assumed direction is correct.
MA is negative; the assumed direction is correct.
Next Level Cut Plane a-a to Find Shear Force and Bending Moment at B
NA
A
N1
RA
Two FBDs (Left Side and Right Side) result at Cut-Plane a-a.
Either FBD can be used to find the internal force and moment. Identical equilibrium equations result.
Applied forces are shown in their given directions. Internal force, NB, is shown as tensile.
Its sign depends on its direction. Moment, MA, is shown as positive.
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Left Side FBD
x
NA
A
RA
Equilibrium Equations
∑Fx =NA + NB = 0, NB = -NA
∑Fy = RA- VB = 0, VB = RA
∑MB = MA + MB – x*RA= 0, MB = -MA + x*RA
Either FBD can be used to find the internal force and moment.
Applied forces are shown in their given directions. Internal force, NB, is shown as tensile.
Its sign depends on its direction. Moment, MA, is shown as positive.
NB is negative. The correct direction is opposite from the assumed direction.
Mx is negative so the correct direction is opposite from the assumed direction.
Shear Force and Bending Moment Diagrams
The shear force and bending moment diagrams are shown below. Positive shear and bending
moments are plotted in the positive axis direction. Just to the left of C, the shear force is P. Just
to the right of C, the shear force is 0 because the applied load removes the shear force (V C = P).
y
VA = P
VC- = P
VC+ = 0
Shear Force
x
y
x
Moment
-MA
Slope = VB = RA = P = dMx/dx
Mx = -MA + P*x
0
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