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
Statics, Fourteenth EditionR.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.All rights reserved.
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
Statics, Fourteenth EditionR.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.All rights reserved.
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)
Statics, Fourteenth EditionR.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.All rights reserved.
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)
Statics, Fourteenth EditionR.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.All rights reserved.
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
Statics, Fourteenth EditionR.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.All rights reserved.
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
Statics, Fourteenth EditionR.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.All rights reserved.
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
Statics, Fourteenth EditionR.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.All rights reserved.
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.
Statics, Fourteenth EditionR.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.All rights reserved.
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
Statics, Fourteenth EditionR.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.All rights reserved.
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
Statics, Fourteenth EditionR.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.All rights reserved.
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
Statics, Fourteenth EditionR.C. Hibbeler
Copyright ©2016 by Pearson Education, Inc.All rights reserved.
CIVE1143 – Structural Analysis
Member Internal Forces
by Dr. Xiaodong Huangoffice: 10.13.05C
Email: [email protected]
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
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
Consider the beam shown and determine the axial force, shear force and bending moment
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
Consider the beam shown and determine the axial force, shear force and bending moment
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
Consider the beam shown and determine the axial force, shear force and bending moment
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
FBD (the left part):
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
FBD (the left part):
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
FBD (the right part):
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
FBD (the right part):
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
12kN/m Determine the axial force, shear force and bending moment at C.
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
12kN/m Determine the axial force, shear force and bending moment at C.
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
FBD (the right part):
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
12kN/m Determine the axial force, shear force and bending moment at C.
1.5m
CNCV
CM
1.5m
B1.5m
FBD (the right part):
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
12kN/m Determine the axial force, shear force and bending moment at C.
1.5m
CNCV
CM
1.5m
B1.5m
FBD (the right part):
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.