1. CE 201 - STATICS LECTURE
31 1
71
Dr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
Internal Forces
( Beams & Frames
50N
50N
3m
5m
5m
A B C
3
4
M M
N N
V
V
RAy = 20N
RCy = 20N
RCx = 30N

2. CE 201 - STATICS LECTURE
31 2
72
Dr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
1- Axial (N) or (A)
2- Shear (V)
3- Moment (M)
( N )
( V )
( M )
These internal Forces depend on:-
1: Geometry of member.
2: Loading of member.
3: Location within member.
mN
N
y
x
40M0ΣM
20V0ΣF
0N0ΣF
⋅
=⇒=
=⇒=
=⇒=
2m
6m
A B
40N
2m
A
N=0
M
V
20N

3. CE 201 - STATICS LECTURE
31 3
73
Dr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
5m
A B
2m
20N/m
0 0
20N/m
50N
50N
A
2m
20N/m
50N
M
N
V
(1): Find external
Reactions.
(2): Make Section
at required
location.
(3): Select smaller
part of sectioned
member.
(4): Find
Internal
Forces.
mN
C
N
y
x
6040100M
02501220M0ΣM
104050V
0V220500ΣF
0N0ΣF
⋅
=−=∴
=×−××+⇒=
↓=−=∴
=−×−⇒=
=⇒=

4. CE 201 - STATICSDr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
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 Section @ 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 !
M
N
V
R
x
M
N
VR x

5. CE 201 - STATICSDr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
Using F · B · D :
+
Ray = 60N
(↑)
ΣFy = 0 +↑
RBY = 60N
(↑) F · B · D
ΣFx = 0 ⇒ RBx = 0
ΣΜB = 0
Take Section @ x:
ΣFx = 0 ⇒ N = 0
ΣFy = 0 + ↑
60 - 20 x - V = 0
∴ V = 60 - 20x
∴ M = 60 - 10x2
ΣMx - 60x == 0





2
x
M
N
V
60N
x
20N/m
A
6m
A B
20N/m
120N
A B
RAy
RBy
RBx
N
+
_
60
-60
VN
MN·m
90
+

6. CE 201 - STATICSDr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
EXAMPLE:
Using F·B·D.
mN
B
B
B
N
Byy
Bxx
120M
0260M
0ΣM
60R0ΣF
0R0ΣF
⋅
=∴
=×−
=
↑=⇒=
=⇒=
6
20x
mN3
x
2
y
x
36
20x
M
0
3
x
2
x
6
20x
M
0ΣM
N
6
10x
2
x
6
20x
V0ΣF
0N0ΣF
⋅





 −
=∴
=

















−−
=






=





⇒=
=⇒=
+
+
M
N
V
x
A
20N/m
6m
A
B
20N/m
60N
NN
+
_
+60
VN
MN
F·B·D.
2m
x
RBy
RBx
MB
Parabola
Hyperbola
-120N·m

7. CE 201 - STATICSDr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
Using FBD for ABC:
ΣFx = 0 ⇒ Rcx = 30N
(←)
ΣΜc = 0 ⇒ RAy = 20N
(↑)
ΣFy = 0 ⇒ Rcy = 20N
(↑)
⇒ SEGMENT AB ( 0 ≤ x < 3 )
ΣFx = 0 ⇒ N = 0
ΣFy = 0 ⇒ V = 20N
ΣMx = 0 ⇒ M = 20x
⇒ SEGMENT BC ( 3 < x ≤ 6 )
4
3
50N
A
B
C
BA C Rcx
40N
30N
3m
3m
RcyRAy
x
_ NN
-30-30
+20+20
-20 -20
_
+
+
+60
MN·m
A
V
N
Mx
20N
ΣFx = 0 ⇒ N = 0 - 30N
(c)
ΣFy = 0 ⇒ V = - 20N
ΣMx = 0 ⇒ M = 20x - 40 (x - 3)
= 120 + 20 x - 40 x
= 120 - 20 x
A
x V
N
M
20N
B
40N
30N

8. CE 201 - STATICSDr. Mustafa Y. Al-Mandil
Department of
Civil Engineering
ΣMA = 0 ⇒ RBY = 35N
ΣFy = 0 ⇒ RAy = 15N
SEGMENT AB ( 0 < x < 6 )
SEGMENT BC ( 6 < x < 10 )
V = 15 - 5x
M = 15x -
2
5 2
x
A
x V
N
M
15N
B
V
N
M
x
15N 35N
5N/m
x - 6
6m
V = 20N
M = 20x - 120
20N
A
B C
V(N)
-15
+20 +20
_
+
+
MN·m
5N/m
80N·m
30N
6m 4m
15N
80N·m
20N
35N
+
+
22.5N·m
+15
+80N·m