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Internal Loadings
a. Stability of the body
External Loads
Mechanics of Materials
b. Intensity of internal forces
c. Intensity of the deformation
of the body
Type of External loads:
1. Surface forces: Due to direct contact either
a) Concentrated: External force act at a point
b) Distributed: load is distributed along a narrow area over
a specified length. ( To deal with distributed loads, the
resultant force is equivalent to the area under the
distributed loading curve and act through the centroid of
that area)
FR
FR= 0.5x 9 x 270=1215 N
Acting 1/3 the length
(Centroid of triangle)
2. Body Forces: No direct contact, like body weight.
3. Support Reactions: If the support prevents motion
(translation, rotation) in a given direction , then a force or a
moment must be developed on the member in that direction.
Type of Supports
*Equilibrium of a body:
ΣF=0 Æ ΣFx =0 , ΣFy =0 , ΣFz =0
ΣMο= 0 Æ ΣΜx=0 ΣMy = 0 ΣMz=0
For coplanar forces: There are three equations of
equilibrium
ΣFx =0 , ΣFy =0 ,ΣΜx=0
A useful definition
Free body diagram: A sketch of the outlines shape of the body
isolated from its surrounding. On this sketch all forces and couple
moments that the surrounding exert on the body together with any
support reactions must be shown correctly. Only then applying
equilibrium equations will be useful.
Internal loadings: These internal loading acting on a specific region
within the body can be attained by the Method of Section.
Method of Section: Imaginary cut is made through the body in the
region where the internal loading is to be determined. The two parts are
separated and a free body diagram of one of the parts is drawn. Only
then applying equilibrium would enable us to relate the resultant
internal force and moment to the external forces.
Point O is often chosen as
the centroid of the
sectioned area
Apply Equilibrium
at this stage
•Three Dimensional Loading:
Four types of internal loadings can be defined:
Normal force, N. This force act perpendicular to the area.
Shear Force, V. This force lies in the plane of the area (parallel)
Torsional Moment, T. This torque is developed when the external
loads tend to twist one segment of the body with respect to the other
Bending Moment, M. This moment is developed when the
external loads tend to bend the body.
• Coplanar Loading:
If the body is subjected to a coplanar system of forces
then only normal force N, shear force V, and bending
moment Mo components will exist at the section.
Procedure of Analysis
1. After sectioning, decide which segment of the body will be
studied. If this segment has a support or connection than a
free body diagram for the entire body must be done first to
calculate the reactions of these supports.
2. Pass an imaginary section through the body at the point
where the resultant internal loadings are to be determined and
put the three unknowns (V, Mo, N) at the cut section. Then
apply equilibrium.
Suggestion: take the summation of moment around a point on
the cut section (V and N will not appear in this equation) and
solve directly for Mo)
Example 1.1
Determine the internal loading at C
A cut will be made through C and the right part will be studied
= 0.5x180x6
Fc = 6
270 9
Fc = 6/9 * 270 = 180N/m
Example 1.1 (Cont.)
If the cut was made at C and the left part was taken
First a free body diagram for the entire body is made and
equilibrium is applied to get the support reactions.
1215
=0.5x 270x9
270
A
c
3
6
Apply equilibrium and get support
reactions at A (N,V,M)
ΣFy = 0 ,
1215 –135-540 –Vc=0 -Æ Vc= 540 N
ΣMc = 0 Mc+ 540x1.5 + 135x2 – 1215x3 + 3645 = 0
Mc= - 1080 Nm
Example 1-2
A free Body
Diagram of the
entire body
Example 1-2 (Cont.)
A cut will be made at C and the left part will be studied
Example 1-3