2 internal force theory 2015

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

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

CROSS SECTION METHOD 2.2 INTERNAL FORCES & METHOD OF SECTIONS

2.2 INTERNAL FORCES & METHOD OF SECTIONS

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

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

Consider a bar at “balance” state (ie free body diagram)

An imaginary cross section

2.3 SIGN CONVENTION

CROSS SECTION METHOD

2.3 SIGN CONVENTION

2.3 SIGN CONVENTION

CROSS SECTION METHOD

2.3 SIGN CONVENTION

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

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

lim

0 0

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

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

Example: Determine internal forces on the cross section at C

(Xác định nội lực tại tiết diện C)

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

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