by jianhu (chris) shen · shear force and bending moment diagrams key information about sfd and bmd...
TRANSCRIPT
By Jianhu (Chris) Shen
Email: [email protected]: 99250421Office: 10.12.25 (city) 251.2.67 (Bundoora)
Mode of delivery and consultation � Lectures – mainly concepts, methodology, knowledge
� Tutorials –Practise by solving problems
�Brief review on lecture notes to be used in tutorial
�Feedback and comments on previous Quiz
�Quiz on previous tutorial
�Demonstrate the solution process for tutorial problems
�Practise time and answer questions for tutorial problems
� My consultations Time Wednesday: 2:00 PM to 4:00 PM
Other time, please confirm available with email: [email protected]
An Overview for Structural Analysis
DD/F
Double integration method
Slope/Deflection
Equations
BMD/F; SFD/F;
Statics-FBD
Materials, Geometries, Sections, Applied Load
Section properties
Internal forcesN, Q, M
Equilibrium Eq
Constitutive Eq
Stress Transformation
Bending Stress
Stress
Safety of an Structure
Strength Rigidity Stability
Deflection
Brief review on lecture notes
� Key internal forces for beams (N, V, M)
� Variation of shear force and bending moment along beam
� Shear force and bending moment functions
� Shear force and bending moment diagrams
� Solution procedure
Simplified consideration for beam
Special cases:• Beam with buckling, • beam & column, • Frames• …………….
Beam only
V and M
Identifying the critical location for
beams
https://www.youtube.com/watch?v=SdpjUunqel4
From the tutorial problem last week, we knew the shear forces and bending moments are different at different location on a beam
How to locate the critical points to design the beam accordingly?
SFF, BMF and SFD, BMD
Functions for shear force and
bending moment-coordinates
Coordinate system:
Default Coordinate system:Y1
X1O1
Y1
X1O1
Y3X3
O3
Y2
X2O2
Y
XO
Functions for shear force and
bending moment-validate range
( ) mxxVV 40,1 ≤≤=
( ) mxmxVV 64,2 <≤=
( ) mxmxVV 86,3 ≤<= ( ) mxxMM 40,1 ≤≤=
( ) mxmxMM 64,2 ≤≤=
( ) mxmxMM 86,3 ≤≤=
Functions for shear force and
bending moment-validate range
( ) mxxVV 40, 111 ≤≤=
( ) mxxVV 20,22 <≤=
( ) mxxVV 20, 333 ≤<= ( ) mxxMM 40, 111 ≤≤=
( ) mxxMM 20, 222 ≤≤=
( ) mxxMM 20, 333 ≤≤=
Y1
X1O1
Y2
X2O2
Y3
X3O3
Tips: The functions will be different at different coordinate system. A good choice will make the functions simpler.
Shear force and bending moment
diagramsKey information about SFD and BMD
• Plot variation of bending moment on the compression side
• Indicate the shape of line strips-flat line, inclined line, curved line (parabolic, cubic).
• Provide all important values on your diagram, i.e., local or overall maximum values of SF and BM and their location,
Steps for tutorial problems
Feedback and comments on
previous Quiz
Overall PerformanceTotal correct answers (%) 54.63%
Total incorrect answers (%)45.37%
Average score (points) 4758.32 points
Leader Board
Quiz 3-Atriculated Structures
Rank Student No. Name PointsRight Answer
1 (4) S3653948 Chan Chul Zhen 8822 9
2 (8) s3609811 Shaikh Haider Ali 8100 8
3 (10)s3602554 Gill Amanjeet 7898 8
Most difficult quiz questions
43.4%
Most difficult quiz questions
43.4%
Most difficult quiz questions
5.7%
Most difficult quiz questions
44.2%
Most difficult quiz questions
32.1%
Now Quiz Time
1. Use the internet explore in your mobile, ipad, computer etc to explore the following webpagehttp://kahoot.it
2. Using the pin shown in the screen of your teacher to enter the quiz.
3. Please use your student number as your
nickname, otherwise we cannot allocate marks to you. After you enter your name, your name should be shown on the screen of your teacher’s computer.
4. Then just wait the teacher to start the quiz.
Tutorial Problem 1-: Simple problem
Question 1: Simple problem: Given the shear force and bending moment diagram shown below, develop equations to obtain specific values of shear and BM at 0.5m, 1.5m and 2.5 m from the left end and verify your answers: (tutors to demonstrate the equations in the class)
Solution-P1-1
1. Solve reaction forcesIt is easy to get the reaction forces at A and B because of the symmetric of the supports and loads.
kNBA yy 20==
Solution-P1-2
Y1 X1
O1
2. Set coordinate system to solve functions of shear force and bending moment
kNV 20=
+↑=∑ 0yF
mx 10 <<
+−=∑ CAntiM X 0
xxM 20)( =
X
Apply cut at X and select AX as free body and plot FBD:
Solution-P1-3 Y1 X1
O1
01020 =−+−V
+↑=∑ 0yF
mx 21 <<
+−=∑ CAntiM X 0
( ) 011020)( =−+− xxxM
X
Apply cut at new location X and select AX as free body and plot FBD:
kNV 10=
xxM 1010)( +=
Solution-P1-4Continue this process, and we can get:
<<−<<−
<<<<<<
=
mxmkN
mxmkN
mxmkN
mxmkN
mxmkN
V
54,20
43,10
32,0
21,10
10,20
≤≤−≤≤−≤≤⋅
≤≤+≤≤
=
mxmx
mxmx
mxmmkN
mxmx
mxmx
xM
54,20100
43,1060
32,30
21,1010
21,20
)(
Solution-P1-5Of course you can work out those functions directly from
diagrams
mxkNV 10,20 <<=
mxmkNV 21,10 <<=
mxmV 32,0 <<=
Y1X1
O1
mxmkNV 43,10 <<−=
mxmkNV 54,20 <<−=
Solution-P1-6For bending moment functions
mxxxM 10,20)( ≤≤=
( )mxmx
xxM
21,1010
11020)(
≤≤+=−+=
mxmmkNxM 32,30)( ≤≤⋅=
Y1X1
O1
( )mxmx
xxM
43,1060
31030)(
<<−=−−=
( )mxmx
xxM
54,20100
42020)(
<<−=−−=
Solution-P1-7Checking
kNV
mx
20)5.0(
10
=≤≤
mkNM
mx
⋅=×=≤≤
105.020)5.0(
10
=
=
kNV
mx
10)5.1(
21
=≤≤ =
mkNM
mx
⋅=×+=≤≤
255.11010)5.1(
21
=
kNV
mx
0)5.2(
32
=≤≤ =
mkNM
mx
⋅=≤≤
30)5.2(
32
=
Question 2: Slightly more
complexDraw the shear and moment diagrams for the beam
Do it by hand on white board
Question 3: More complex
problems – beams and frames
Draw the shear and moment diagrams for the beam
Solution-P3-1
1. Solve reaction forces
kNBy 25.206=
+−=∑ CAntiM A 0
0150112084320 =−×−×+×− yB
320kN
kNBy 75.133=
+−=∑ CAntiM B 0
0150112084320 =−×−×−× yA
Solution-P3-2Y1
X1
O1
X
+↑=∑ 0yF
04075.133)( =−+− xxV
+−=∑ CAntiM X 0
075.1332
40)( =+− xx
xxM
22075.133)( xxxM −=
mx 80 ≤≤
xxV 4075.133)( −=mx 80 <<
Solution-P3-3Y1
X1
O1
X
+↑=∑ 0yF
020)( =−xV
+−=∑ CAntiM X 0
( ) 01501120)( =−−−− xxM
37020)( −= xxM
mx 118 ≤≤
kNxV 20)( =mx 118 <<
P4-More complex problems -
frames Draw the shear and moment diagrams for each of the three members of the frame
Assume the frame is pin connected at A, C and D and there is fixed joint at B.
Solution-P4-11. Solve reaction forces, start
analysis with member CD:Note the loads is this member is symmetric along y, so
Cy
Cx
Dy
Dx
90kN
kNCD xx 452
90 ===
For the whole structure:
+−=∑ CAntiM A 0
05.1505.3401902455 =×−×−×+×+×yD
Dy
45kN
Ay
Ax
90kN
kNDy 7=
kN
DC yy
7−=
−=
kNAy 83=kNAx 45=Then,
Solution-P4-2
For frame section ABC, we need to divide it into beams: Cut at B and use BA as free body:
83kN
45kN
NB
MB
VB
kNNB 83−=
kNVB 45−=
kNM B 180=
Solution-P4-3
7kN
45kN
7kN
45kN
83kN
45kN
83kN180kN m
45kN
83kN
180kN m
45kN 7kN
45kN
x
VM
x
VM
x
VM
Note: By default, we assume the outside surface of frame is positive direction for V(x) and M(x)
Solution-P4-4
83kN180kN m
45kN
83kN
45kN
x
VM
x(m)
M(kN•m)
180
45x(m)
V(kN)
AB
A
B
Solution
-P4-5 83kN
180kN m
45kN
7kN
45kNx
VM
x(m)
M(kN•m)
55.5
7
83
33
180
10.5
x(m)
V(kN)
CB
BC
Solution-P4-6
7kN
45kN
7kN
45kN
xVM
x(m)
V(kN)
CD
C
D
x(m)
M(kN•m)67.5
45
45
Practise problem 1Plot shear force and bending moment diagrams for each beam in the frame. Assume the support at A is a pin and D is a roller
Solution guide for Practise problem
Solution guide for Practise problem
Practise problem 2Draw the shear and moment diagrams for the following beam. Self-check your answers by using sections from opposite ends.
Solution guide for Practise problem