Giáo trình Strength of Materials - Chapter 2: Internal forces

CHAPTER 2: INTERNAL FORCES 
2.1 Free Body Diagram 
2.2 Internal Forces & Method of Section 
2.3 Sign Conventions 
2.4 Procedure 
2.5 Diagrams of internal forces 
2.6 Relationships between loads, shear force and bending 
moment 
2.1 FREE BODY DIAGRAM 
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. 
2.2 INTERNAL FORCES & METHOD OF SECTIONS 
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. 
CROSS SECTION METHOD 
2.2 INTERNAL FORCES & METHOD OF SECTIONS 
2.2 INTERNAL FORCES & METHOD OF SECTIONS 
CROSS SECTION METHOD 
Point O is often chosen as the 
centroid of the sectioned area 
2.2 INTERNAL FORCES & METHOD OF SECTIONS 
CROSS SECTION METHOD 
2.2 INTERNAL FORCES & METHOD OF SECTIONS 
CROSS SECTION METHOD 
Four types of internal forces can be defined: 
Normal force, N. This force acts 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 
2.2 INTERNAL FORCES & METHOD OF SECTIONS 
CROSS SECTION METHOD 
• If the body is subjected to a co-planar force system then only normal 
force N, 01 shear force V, and 01 bending moment M component exist 
on the section. 
2.2 INTERNAL FORCES & METHOD OF SECTIONS 
CROSS SECTION METHOD 
M
x
 > 0 
N
z
 > 0 
Q
y
 > 0 
A 
O 
P
3
P
2
P
1
z 
B 
P
4
P
5
P
6
z 
N
z
 > 0 
M
x
 > 0 
Q
y
 > 0 
O 
P1 
P2 
P3 
P4 
P5 
P6 
Consider a bar at “balance” state (ie. free body diagram) 
An imaginary cross section 



6
1
0
i
iyF



6
1
0
i
izF
 


6
1
0
i
ix Fm
y 
y 
Fi : P1, P2, P6 
Fi : P1, P2, P3, Nz, Qy, Mx 
Fi : P4, P5, P6, Nz, Qy, Mx 
2.3 SIGN CONVENTION 
CROSS SECTION METHOD 
2.3 SIGN CONVENTION 
CROSS SECTION METHOD 
2.3 SIGN CONVENTION 
CROSS SECTION METHOD 
2.3 SIGN CONVENTION 
CROSS SECTION METHOD 
Shear force: clockwise 
Bending moment: compresses the upper part of the bar or 
elongates the lower part 
Normal force: elongates 
2.3 SIGN CONVENTION 
CROSS SECTION METHOD 
• If the internal shear force causes a 
clockwise rotation of the beam segment. 
Then it is considered to positive. 
• If the internal moment causes compression 
in the top fibers then it is considered to be 
positive 
2.3 SIGN CONVENTION 
CROSS SECTION METHOD 
Stress Under General Loadings 
• A member subjected to a general 
combination of loads is cut into 
two segments by a plane passing 
through Q 
• For equilibrium, an equal and 
opposite internal force and stress 
distribution must be exerted on 
the other segment of the member. 
A
V
A
V
A
F
x
z
A
xz
x
y
A
xy
x
A
x











limlim
lim
00
0


• The distribution of internal stress 
components may be defined as, 
2.4 PROCEDURE 
CROSS SECTION METHOD 
Determination of support reactions by studying the 
equilibrium of the whole structure 
 (Xác định phản lực liên kết bằng cách xét cân 
 bằng tòan hệ) 
Imagine a section passing through the body 
(Tưởng tượng mặt cắt qua vật thể) 
Equilibrium of one divided part 
 (Xét cân bằng một phần bị chia) 
2.4 PROCEDURE 
CROSS SECTION METHOD 
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) 
2.4 PROCEDURE 
CROSS SECTION METHOD 
Determine the reactions 
using the equilibrium 
conditions of the overall 
structure 
Cut the beam at the cross section 
at which shear force and bending 
moment are to be determined. 
Draw a free-body diagram 
Set up equilibrium equations of the F.B.D. to 
determine shear force and bending moment at the 
cross section 
2.4 PROCEDURE 
CROSS SECTION METHOD 
Example: Determine internal forces on the cross section at C 
(Xác định nội lực tại tiết diện C) 
x CF 0 N 0    
y C
C C
F 0 V 58,8N
M 0 M 5,69N.m
     
     


2.5 DIAGRAMS OF INTERNAL FORCES 
CROSS SECTION METHOD 
Beams are long straight bars having constant cross section 
area and support loads that are applied perpendicular to its 
longitudinal axis 
2.5 DIAGRAMS OF INTERNAL FORCES 
In order to properly design a beam, the maximum values for V and M in the 
beam have to be found. This could be done through the shear force and 
bending moment. 
At each location z, values of V(z) and M(z) are obtained by using the 
procedure of determining internal forces on the cross section at z. 
V and M vary throughout the length of the beam. This means that V = V(z) 
and M = M(z). 
Graphs are plotted as values of V or M versus distance z along the axis of 
the beam. 
Graphs are called shear force and 
bending moment diagrams. 
2.5 DIAGRAMS OF INTERNAL FORCES 
EXAMPLE 1: Cantilevered beam and concentrate load 
2.5 DIAGRAMS OF INTERNAL FORCES 
EXAMPLE 2: Simply supported beam and concentrate load 
 Remind: 
Given: beam AB, length L, concentrate load P 
at distance L1 (from A) 
Problem: plot V and M 
diagram? 
EXAMPLE 3: Cantilevered beam and uniformed distribution load 
EXAMPLE 4: Simply supported beam and uniformed distribution load 
2.5 DIAGRAMS OF INTERNAL FORCES 
REMARKS 
If we let the cross section to move from left end to right end of the beam and 
always consider the left-hand side segment then: 
1. Whenever we see a external concentrate force or concentrate moment, 
there will be a sudden change of the shear diagram or moment diagram. 
Value of the change in the diagram is equal to that of force or moment. 
Direction of the change in the diagram follows that of the change of the 
force or moment. 
2. Whenever we see a change of external force or moment (including 
reaction force), it is necessary to add one more time of considering the 
internal force formulation i.e. the internal force diagrams will have one more 
segment. 
2.6 RELATIONSHIPS BETWEEN LOADS, SHEAR 
FORCE DIAGRAM AND BENDING MOMENT 
DIAGRAM 
2.6 RELATIONSHIPS BETWEEN LOADS, SHEAR 
FORCE DIAGRAM AND BENDING MOMENT 
DIAGRAM 
The concentrated loads 
cause abrupt changes in 
the shear force wherever 
they are located. 
As the differentials are small, the 
bending moment does not change 
as we pass through the point of 
application of a concentrated load.