Internal Forces in Planar Structures
Steven&Vukazich
San&Jose&State&University
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Internal Forces in Planar Structures: Equilibrium Analysis and Free Body Diagrams, Study notes of Material Engineering
An explanation of how to find internal forces in planar structures using equilibrium analysis and free body diagrams. It includes examples of calculating support reactions, axial forces, shear forces, and bending moments for a beam subjected to various external forces. Students can use this information to understand the principles of structural mechanics and apply it to solve similar problems.
What you will learn
- How can you use the equations of equilibrium to find the internal forces in a free body diagram?
- How do you calculate the support reactions for a planar structure using equilibrium analysis?
- What is the process for finding the internal forces at a given point in a planar structure?
- What are the three internal forces developed in each section of a planar structure to ensure equilibrium?
- What is the difference between an axial force, shear force, and bending moment in a structural member?
Typology: Study notes
2021/2022
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Internal Forces in Planar Structures
Steven Vukazich
San Jose State University
Consider the Frame
B
A
C
P
w
Roller support
Pin support
Cut the Frame into Two Sections
B
A
C
P
w
A
x
C
x
Ay
q
q
Make a cut at an arbitrary
point Q between B and C
perpendicular to member BC
Q
Free Body Diagrams of the Sections of the Frame
B
A
C
P
w
A
x
C
x
Ay
q
q
q
q
Q
M Q V Q F Q Q M Q V Q F Q
Three internal forces are
developed to ensure
equilibrium of each section
F = Axial Force
V = Shear Force
M = Bending Moment
Can use our three
equations of
equilibrium to find
F , V , and M
w
Internal Force Example Problem
A beam is supported by a pin support at point A and extends over a
roller support at point D. The beam is and subjected to a uniformly
distributed load from A to B, a point moment at point C an inclined
point load at point E as shown.
Find the internal forces at points q , r , s , and t.
9 ft 2 ft 3 ft 4 ft 5 k E 23 k-ft 2 k/ft D
B C
A
3 4 6 ft 1 ft 1 ft 2 ft q r^ s t
Find All of the External Forces FBD of beam 9 ft 2 ft 3 ft 4 ft 5 k E 23 k-ft 2 k/ft D
B C
A
3 4
Ax
Ay Dy
2 k/ft 9 ft = 18 k 4.5 ft 5 k
= 3 k 5 k
= 4 k 5
+ =^0
Dy = 10 k
=^0
Ax = 4 k
Find All of the External Forces 9 ft 2 ft 3 ft 4 ft 5 k E 23 k-ft 2 k/ft D
B C
A
3 4
Ax
Ay Dy
2 k/ft 9 ft = 18 k 4.5 ft 5 k
= 3 k 5 k
= 4 k 5
External Forces FBD of beam showing all external forces 9 ft 2 ft 3 ft 4 ft 5 k E 23 k-ft 2 k/ft D B C
A
3 4
4 k
13 k
6 ft 8 k
1 ft 1 ft 2 ft q r^ s^ t
FBDs of Segments A q and q BCDE
M q V q 2 k/ft A
4 k
13 k 6 ft
q q 3 ft (^2) ft 3 ft 4 ft 5 k E 23 k-ft 2 k/ft D B C 3 4
8 k
1 ft 1 ft^2 ft r (^) s t q q Find Internal Forces at Point q F q F q M q V q
Can apply equations of equilibrium
to either FBD to find internal forces
FBD of Segments A q
Find Internal Forces at Point q 2 k/ft 2 k/ft 6 ft = 12 k
+^ =^0
Vq = 1 k
Mq = 42 k-ft
Fq = - 4 k
=^0
M q V q 2 k/ft A
4 k
13 k
6 ft q q F q M q V q
A
4 k
13 k
6 ft q q F q 3 ft
Find Internal Forces at Point r Cut beam at point r 9 ft 2 ft 3 ft 4 ft 5 k E 23 k-ft 2 k/ft D B C
A
3 4
4 k
13 k
8 k
1 ft q (^) s t r r
FBDs of Segments AB r and r CDE
M r V r 2 k/ft A
4 k
13 k
9 ft 1 ft 3 ft 4 ft 5 k E 23 k-ft D C 3 4
8 k
t r r Find Internal Forces at Point r F r F r M r V r r r q 1 ft s
Internal Forces at Point r 2 k/ft A
4 k
13 k
9 ft r r q 1 ft 1 ft 3 ft 4 ft
E
23 k-ft D C 3 4
8 k
t r r s
Confirm that both segments are in
equilibrium
31 k-ft 5 k 5 k 31 k-ft 4 k 4 k
Find Internal Forces at Point s Cut beam at point s 9 ft 2 ft 3 ft 4 ft 5 k E 23 k-ft 2 k/ft D B C
A
3 4
4 k
13 k
8 k
1 ft q s t s r