1. Chapter 7 Internal Forces ( Beams & Frames LECTURE 31 1 71 50N 50N
Dr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
LECTURE 31 1
Internal Forces
( Beams & Frames 71
Chapter 7
50N 4 3m 3 A B C 5m 5m
50N V M M N N V
RCx = 30N
RAy = 20N
RCy = 20N

2. Chapter 7 INTERNAL FORCES: e.g. LECTURE 31 2 72
Dr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
LECTURE 31 2
Chapter 7 72
INTERNAL FORCES:
These internal Forces depend on:-
1: Geometry of member.
2: Loading of member.
3: Location within member.
1- Axial (N) or (A)
2- Shear (V)
3- Moment (M)
( N )
( V )
( M )
40N 2m
e.g. A B 2m A
N=0 M V
20N 6m

3. EXAMPLE 1 Chapter 7 (1): Find external Reactions. (4): Find Internal
Dr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
LECTURE 31 3 73 5m A B 2m
20N/m
EXAMPLE 1
Chapter 7
(1): Find external
Reactions.
(4): Find
Internal
Forces.
20N/m
(2): Make Section
at required
location.
50N
50N A 2m
20N/m
50N M N V
(3): Select smaller
part of sectioned
member.

4. Chapter 7 Shear & B.M. Diagrams: x x 1 - Draw F. B. D. for Beam.
Dr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
Chapter 7
Shear & B.M. Diagrams:
1 - Draw F. B. D. for Beam.
2 - Solve for External Reaction.
3 - Determine Number of Segments.
4 - Decide on Global Coordinate System.
5 - Make x of each Segment.
6 - Solve for Internal Forces as Function of ( x ):
N = f ( x )
V = g ( x )
M = h ( x )
7 - Draw Internal Forces Vs. (x).
8 - Determine Max. & Min. Values of Functions
9 - Good Luck ! V M N V R x M N x R

5. Chapter 7 Using F · B · D : + Take Section @ x: x A 60N
Dr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
Chapter 7
20N/m
Using F · B · D : A B
B = 0 +
Ray = 60N () 6m
Fy = 0 +
RBY = 60N () F · B · D
120N
20N/m A B
RBx M
Fx = 0  RBx = 0 N
RBy
RAy A
Take x: N
Fx = 0  N = 0
Fy = 
x - V = 0 V 60 x + VN
60N _ 90
-60
 V = x +
Mx x == 0
MN·m
 M = x2

6. Chapter 7 EXAMPLE: + x + A Dr. Mustafa Y. Al-Mandil Department of
Civil Engineering
Chapter 7
EXAMPLE:
20N/m
Using F·B·D. A B 6m
RBy +
60N
20N/m
F·B·D. M
RBx 2m x N MB A NN V x
+60
Parabola + VN + MN _
Hyperbola
-120N·m

7. Chapter 7 Example - 3 50N A 20N _ _ Dr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
Chapter 7 4 3
50N
Example - 3 A C
Using FBD for ABC:
Fx = 0  Rcx = 30N (
c = 0 RAy = 20N (
Fy = 0  Rcy = 20N (
SEGMENT AB ( 0 x < 3 ) B
40N
30N B C A
Rcx A V N M x
20N 3m 3m
RAy
Rcy
Fx = 0  N = 0
Fy = 0 V = 20N
Mx = 0  M = 20x
SEGMENT BC ( 3 < x 6 ) x _ NN A x V N M
20N B
40N
30N
-30
-30
Fx = 0  N = N (c)
Fy = 0 V = - 20N
Mx = 0  M = 20x - 40 (x - 3)
= x - 40 x
= x
+20
+20 + _
-20
-20
+60 +
MN·m

8. Chapter 7 Example _ Dr. Mustafa Y. Al-Mandil Department of
Civil Engineering
Chapter 7
5N/m
20N
Example
80N·m B C A
MA = 0  RBY = 35N
Fy = 0 RAy = 15N
30N
20N
80N·m
SEGMENT AB ( 0 <x < 6 ) A x V N M
15N B
35N 4m
15N 6m
V = x
M = 15x -
+15
+20
+20 + + _
V(N)
SEGMENT BC ( 6 <x < 10 )
-15 x
5N/m V N M
22.5N·m
+80N·m
15N
35N
x - 6 6m + +
V = 20N
M = 20x - 120
MN·m