By Jianhu (Chris) Shen

Email: jianhu.shen@rmit.edu.auPhone: 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: jianhu.shen@rmit.edu.au

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