INTERNAL FORCES Today’s Objective th Aug To€¦ · ¾The shear force at the left side of a concentrate force does . not equal. to the one at the right side of the force. The difference

Statics, Fourteenth EditionR.C. Hibbeler

Copyright ©2016 by Pearson Education, Inc.All rights reserved.

In-Class Activities:•Feed back on emails•Applications• Types of Internal Forces• Steps for Determining

Internal Forces• Reading Quiz•Concept Quiz•Attention Quiz•Homework•Group Problem Solving

Today’s Objective: 10th AugTo:1. Use the method of sections for

determining internal forces in 2-D load cases.

INTERNAL FORCES



Why are the beams tapered?

APPLICATIONS

How can we know when to taper and where to taper?

We need to know the forces at a given section, generally known as internal forces



Why might have this been done?

The shop crane is used to move heavy machine tools around the shop.The picture shows that an additional frame around the joint is added.

APPLICATIONS (continued)



The loads on the left and right sides of the section at B are equal in magnitude but opposite in direction. This is because when the two sides are reconnected, the net loads are zero at the section.

In two-dimensional cases, typical internalloads are normal or axial forces (N, acting perpendicular to the section), shear forces (V, acting along the surface), and the bending moment (M).

INTERNAL FORCES (continued)



1. In a multiforce member, the member is generally subjected

to an internal _________.

A) Normal force B) Shear force

C) Bending moment D) All of the above.

2. In mechanics, the force component V acting

tangent to, or along the face of, the section is

called the _________ .

A) Axial force B) Shear force

C) Normal force D) Bending moment

READING QUIZ



1. A column is loaded with a vertical 100 N force. At which sections are the internal loads the same?

A) P, Q, and R B) P and QC) Q and R D) None of the above.

•P

QR

100 N

2. A column is loaded with a horizontal 100 N force. At which section are the internal loads largest?

A) P B) QC) R D) S

P

QR

100 N

S

CONCEPT QUIZ



2. A column is loaded with a horizontal 100 N force. At which section are the internal loads the lowest?

A) P B) QC) R D) S

P

QR

100N

S

1. Determine the magnitude of the internal loads(normal, shear, and bending moment) at point C.

A) (100 N, 80 N, 80 N m) B) (100 N, 80 N, 40 N m)C) (80 N, 100 N, 40 N m)D) (80 N, 100 N, 0 N m )

•C

0.5m1 m

80 N

100 N

ATTENTION QUIZ



GROUP PROBLEM SOLVING I

Given: The loading on the beam.

Find: The internal forces at point C.

Plan: Follow the procedure!!

Solution:1. Plan on taking the imaginary cut at C. It will be easier to

work with the left section (point A to the cut at C) since the geometry is simpler.



GROUP PROBLEM SOLVING I (continued)

2. First, we need to determine Ax and Ay using a FBD of the entire frame.

Applying the E-of-E to this FBD, we get

→ + Σ Fx = Ax + 400 = 0 ; Ax = – 400 N

+ Σ MB = – Ay(5) – 400 (1.2) = 0 ; Ay = – 96 N

400 N

By

Ax

Ay

Free Body Diagram



GROUP PROBLEM SOLVING I (continued)

3. Now draw a FBD of the left section. Assume directions for VC, NC and MC as shown.

4. Applying the E-of-E to this FBD, we get

→ + Σ Fx = NC – 400 = 0; NC = 400 N

↑ + Σ Fy = – VC – 96 = 0; VC = – 96N

+ Σ MC = 96 (1.5) + MC = 0 ; MC = -144 N m

96 N VC

MCNC

1.5 m

A400 N

C



1. Take an imaginary cut at the place where you need to determine the internal forces. Then, decide which resulting section or piece will be easier to analyze.

2. If necessary, determine any support reactions or joint forces you need by drawing a FBD of the entire structure and solving for the unknown reactions.

3. Draw a FBD of the piece of the structure you’ve decided to analyze. Remember to show the N, V, and M loads at the “cut” surface.

4. Apply the E-of-E to the FBD (drawn in step 3) and solve for the unknown internal loads.

STEPS FOR DETERMINING INTERNAL FORCES



CIVE1143 – Structural Analysis

Member Internal Forces

by Dr. Xiaodong Huangoffice: 10.13.05C

Email: [email protected]

Scott Knight

Text Box

READ IN CONJUNCTION WITH PREVIOUS SLIDES

Revision: Equilibrium equationsF

A B

A structure is in equilibrium when it maintains a balance of force and moment. In 2D cases, there are in general three equations of equilibrium for a given member:

Three equilibrium equations (simultaneous equations) can determine three reaction forces (unknowns).

Statically determinate structures: number of reactions (r) = equilibrium equations (3n)(n – the number of members)

Statically indeterminate structures: number of reactions (r) > equilibrium equations (3n)degree of indeterminacy = r – 3n

∑ = 0xF+ ∑ = 0yF+ ∑ = 0M+

Revision: Articulated structures

Reactions: 6Equations: 2x3=6

Statically determinate structure

Disassemble articulated structure into its basic simple components and then compare the total number of reaction forces (unknowns) with the number of available equilibrium equations. For example

F

F

Internal Forces

• So far we have concentrated on external forces applied to structures – the applied loads and support reactions.

• In order for the structure to transmit the external loads to the ground, internal forces must be developed within the individual members.That is the main reason that structures would be deformed with applied loads.

• The aim of the design process is to produce a structure that iscapable of carry all these internal forces, ie Axial Force, Shear Forceand Bending Moment.

• Internal forces can be determined by using the method of sections.

The Nature of Member Internal Forces

Axial force Nx

Transverse Shear force resolved into components Vy and Vz

Three moments about each of the axes:

Mx tends to cause the member to twist about its longitudinal axis.

My causes the member bending in horizontal (x-z) plane.

Mz causes bending in vertical (x-y) plane.

x

y

z

xNyV

zV

xM

yM

zM

o

The Nature of Member Internal ForcesIn two dimensions, the internal forces at each section are

Transverse Shear force V

Bending moment MN

V

M

Normal force N

N

V

M

The Nature of Member Internal ForcesTo determinate the internal forces at a point such as B, we need to imaginarily cut the beam through the point into two segments. Thus the internal loading will be exposed on the free-body diagram of the segment.

A B

1P 2P

xA

yAAM

xA

yAA B

AMN

V

M

N

V

MB

1P2P

OR

Sign ConventionTo understand the sign convention must remember that axial force, shear force and bending moment are internal forces, and must not be confused with force and applied moment which are external forces

The sign convention for external forces and applied moments (external forces)

Forces are positive in positive axis direction

Applied moment are positive in anti-clockwise direction

yF

xFM

Scott Knight

Text Box

BIT CONFUSING PART

Sign ConventionN N

N N

V V

V V

M M

M MPositive axial force

(tension) Positive shear force Positive bending moment

N

V

MN

V

M

MMTension

Compression

xA

P

A BC

yAyB

B

yB

CNCV

CM

P

AxA

yA

CN

CV

CM

Calculation of the Internal Forces

xA

P

A BC

yAyB

P

AxA

yA

CN

CV

CM

• Identify a section in the structure where you wish to determine the values of the internal forces.

• Consider the free body diagram from this section to the end of the beam obtained by “cutting” the body through this section. You may consider either the right or the left part of the beam obtained, whichever gives the easier calculation.

•Use the sign convention defined earlier to provide the positive direction of these forces. Treat the 3 internal forces at the section as unknowns.

• 3 unknown internal forces can be determined by 3 equilibrium equations.

∑ = 0xF+ ∑ = 0yF+ ∑ = 0M+

B

yB

CNCV

CM

OR

Example 16kN

A B C

Determine the axial force, shear force and bending moment at point B and C.

D

9kNm

3m 6m

Example 16kN

A B C

yA yD

Reaction Forces

D

9kNm

3m 6m

:0∑ =yF+

09669 =+×+×− yA

kN5=yA

:0∑ =DM+

06 =+− yy DA

kN1=yD

Example 16kN

A B C

ABN

BV

BM

Segment AB

D

9kNm

3m 6m5kN 1kN

5kN

B

:0∑ =xF+

:0∑ =yF+

:0∑ =BM+

0=BN

05 =− BVkN5=BV

kNm15=BM

FBD:

035 =+×− BM

N

V

MN

V

MSign Convention:

Example 16kN

A B C

ACN

CV

CM

Segment AC

D

9kNm

3m 6m5kN 1kN

5kN

C

:0∑ =xF+

:0∑ =yF+

:0∑ =BM+

0=CN

065 =−− CVkN1−=CV

kNm15=CM

FBD:

035 =+×− CM

6kN

N

V

MN

V

MSign Convention:

Example 16kN

A B C D

9kNm

3m 6m5kN 1kN

kN1−=CV kNm15=CMkN5=BV kNm15=BM

The shear force at the left side of a concentrate force does not equal to the one at the right side of the force. The difference between two shear force is equal to the concentrate force, e.g. 5kN-(-1kN)=6kN.

The bending moment at the left side of a concentrate force equals to the one at the right side of the force.

Example 2

20kN

A B C

5kN/m

4m 4m1 2 3a 4 53b

Consider the beam shown and determine the axial force, shear force and bending moment

10kN

Example 2

20kN

A B C

5kN/m

4m 4m1 2 3a 4 53b


10kN

Support Reactions:

20kN

A B C

5kN/m

10kNxA

yAyB

The resultant force of distributed load is equivalent to the area under the loading diagram, and act through the geometric center of this area.

20kN4mkN/m5 =×20kN

FBD:

Example 2

20kN

A B C

5kN/m

4m 4m1 2 3a 4 53b


10kN

Support Reactions:

20kN

A B C

5kN/m

10kNxA

yAyB

20kN

:0∑ =xF+

:0∑ =yF+

:0∑ =AM+

010 =−xA

02020 =−+− yy BA

kN10−=yA

kN50=yB08204220 =×−×+×− yB

kN10=xAFBD:

Example 2

20kN

A B C

5kN/m

4m 4m1 2 3a 4 53b


10kN

Support Reactions:

20kN

A B C

5kN/m

10kN

50kN

:0∑ =xF+

:0∑ =yF+

:0∑ =AM+

010 =−xA

02020 =−+− yy BA

kN10−=yA

kN50=yB08204220 =×−×+×− yB

kN10=xA

10kN

10kN

FBD:

Example 2

20kN

A B C

5kN/m

4m 4m1 2 3a 4 53b

10kN

Section 1

20kN

A B C

5kN/m

10kN

50kN

:0∑ =xF+

:0∑ =yF+

:0∑ =M+

010 1 =+ N

010 1 =−− V

kN101 −=V

01 =M

kN101 −=N

10kN

10kN

FBD:

FBD (the left part):

A

10kN

10kN

N

V

MN

V

MSign Convention:

1N

1V

1M

The bending moment is zero at the end pin (or roller) support.

Example 2

20kN

A B C

5kN/m

4m 4m1 2 3a 4 53b

10kN

Section 2

20kN

A B C

5kN/m

10kN

50kN10kN

10kN

FBD:

A

10kN

10kN

N

V

MN

V

MSign Convention:

2N

2V

2M

10kN

:0∑ =xF+

:0∑ =yF+

:0∑ =M+

010 2 =+ N

01010 2 =−−− VkN202 −=V

kNm302 −=M

kN102 −=N

0110210 2 =+×+× M

Note: Keep the distributed loading where it is until after the section is made.

5kN/m


Example 2

20kN

A B C

5kN/m

4m 4m1 2 3a 4 53b

10kN

Section 3a

20kN

A B C

5kN/m

10kN

50kN10kN

10kN

FBD:

A

10kN

10kN

N

V

MN

V

MSign Convention:

aN3

aV3

aM 3

20kN

:0∑ =xF+

:0∑ =yF+

:0∑ =M+

010 3 =+ aN

02010 3 =−−− VkN303 −=aV

kNm803 −=aM

kN103 −=aN

0220410 3 =+×+× M

Note: Keep the distributed loading where it is until after the section is made.

5kN/m


Example 2

20kN

A B C

5kN/m

4m 4m1 2 3a 4 53b

10kN

Section 3b

20kN

A B C

5kN/m

10kN

50kN10kN

10kN

FBD:

10kN

N

V

MN

V

MSign Convention:

:0∑ =xF+

:0∑ =yF+

:0∑ =M+

0103 =−− bN

0203 =−bVkN203 =bV

kNm803 −=bM

kN103 −=bN

04203 =×−− bM

FBD (the right part):

20kN

CbN3

bV3

bM 3

Example 2

20kN

A B C

5kN/m

4m 4m1 2 3a 4 53b

10kN

Section 4

20kN

A B C

5kN/m

10kN

50kN10kN

10kN

FBD:

10kN

N

V

MN

V

MSign Convention:

:0∑ =xF+

:0∑ =yF+

:0∑ =M+

0104 =−− N

0204 =−VkN204 =V

kNm404 −=M

kN104 −=N

02204 =×−−M


20kN

C4N

4V

4M

Example 2

20kN

A B C

5kN/m

4m 4m1 2 3a 4 53b

10kN

Section 5

20kN

A B C

5kN/m

10kN

50kN10kN

10kN

FBD:

10kN

N

V

MN

V

MSign Convention:

:0∑ =xF+

:0∑ =yF+

:0∑ =M+

0105 =−− N

0205 =−VkN205 =V

05 =M

kN105 −=N

05 =−M


20kN

C5N

5V

5M

Example 2

20kN

A B C

5kN/m

4m 4m1 2 3a 4 53b

10kN

kN205 =V 05 =M

kN101 −=V 01 =MkN202 −=V kNm302 −=MkN303 −=aV kNm803 −=aM

kN203 =bV kNm803 −=bMkN204 =V kNm404 −=M20kN

-10kN

SF:

-20kN

20kN

-30kN

0

-30kNm0

-80kNm-40kNm

BM:

Summary:

Example 3

A BC

12kN/m Determine the axial force, shear force and bending moment at C.

1.5m 1.5m

Example 3

A BC


1.5m

N

V

MN

V

MSign Convention:

CNCV

CM

1.5m

It is not necessary to find the support reactions at Asince segment BC can be used to determine internal forces. Draw the Free-Body Diagram (FBD) of segment BC.

BC1.5m

FBD (the right part):To determine the internal forces, the intensity of triangular distributed load at C should be found first.wC = ?

Example 3

A BC


1.5m

CNCV

CM

1.5m

The intensity of triangular distributed load at C can be determined using similar triangles from the geometry.

BC1.5m


wC = ?1.5m

12kN/m wC

3m

If two triangles are similar, then the corresponding sides are in the same ratio. Thus

35.1

12=Cw kN/m6=Cw

6kN/m

Example 3

A BC


1.5m

CNCV

CM

1.5m

B1.5m


6kN/m

The resultant force of distributed load is equivalent to the area under the loading diagram, and act through the geometric center of this area.

kN5.42

5.16=

×=area

0.5m

4.5kN

Example 3

A BC


1.5m

CNCV

CM

1.5m

B1.5m


0.5m

4.5kN

:0∑ =xF+

:0∑ =yF+

:0∑ =M+

05.4 =−CV

kN5.4=CV

kNm25.2−=CM

0=CN

05.05.4 =×−− CM

Equilibrium Equations:

SummaryDiffering from the internal force in truss with axial force only,

there are three type of internal forces in beam namely axial force N, shear force V and bending moment M.

The internal forces (axial force, shear force and bending moment) at a given section can be determined using the method of sections. Note: Follow the sign convention of the internal forces.

However, in order to be able to design members of structures it is essential that the values of N, V and M are known at any part of the member. In other words, we need to determine the shear force and bending moment diagrams which will be discussed in the next lecture.

Documents

INTERNAL FORCES Today’s Objective th Aug To€¦ · ¾The shear force at the left side of a concentrate force does . not equal. to the one at the right side of the force. The difference