CIVL311 CIVL911 2018 Week2 Analysis and Design of Beams for Serviceability 4 slides PDF

Preview

Summary

Download CIVL311 CIVL911 2018 Week2 Analysis and Design of Beams for Serviceability 4 slides PDF

Description

CIVL 311: STRUCTURAL DESIGN 1 CIVL 911: STRUCTURAL DESIGN BASED ON AUSTRALIAN STANDARDS AUTUMN 2018 (WEEK 2): ANALYSIS AND DESIGN OF BEAMS FOR SERVICEABILITY

A/Prof. Neaz Sheikh Room 4.127 Email: [email protected] Consultation time: Tuesday 12:30 pm – 14:30 pm Friday 11:30 am – 13:30 pm (Online)

Important Notes These lecture notes do not substitute recommended textbooks. These lecture notes are not peer-reviewed. They may contain errors. The lecturer is not responsible for the learning of students who rely exclusively on the lecture notes It is highly recommended that students review the lecture materials using the recommended textbooks. Lecture notes are based on Foster et al. (2010) 2

Week 2 Essential Reading Foster et al. (2010) Chapter 3: Beams

Loo and Chowdhury (2013) Part 1- Chapter 4: Deflection of beams and crack control Is it the right time for you to enrol CIVL311/CIVL911? Very Important: Have you completed on-line QUIZ?

AS 3600 Section 8: Design of Beams for Strength and Serviceability

Have you completed the on-line QUIZ?

Tutorial Feedback: Week 1

Agenda for today Recap on week 1 Topics for Today:

Warner et al (1998)

1. Flexural behaviour of beams under load

Self Weight = [500x400 + 150x (4000-400)] x25x10-6= 18.5 kN/m Additional weight= 1 kN/m2 = 1x4= 4 kN/m Total dead load= 18.5+ 4= 22.5 kN/m

2. Properties of cross section (a) Modular ratio method (b) Equilibrium and compatibility method 3. Deflection of beams under load 4. Crack Control

wb

Live Load= 3 kN/m2 = 3x4= 12 kN/m

wc

4000 mm L2= 10 m

L3= 3.5 m

5

Tutorial Feedback: Recap Week 1

4000 mm

Case 1: 1.35 FG

4000 mm

wb =1.35FG= 27.68 kN/m

wc =1.35FG= 27.68 kN/m

4000 mm

MB= 325.18 kN-m MC= -345.94 kN-m

4000 mm

Self Weight = [500x400 + 150x (4000-400)] x25x10-6= 18.5 kN/m

4000 mm

Additional weight= 0.5 kN/m2 = 0.5x4= 2 kN/m

4000 mm

L2= 12 m

L3= 5 m

Case 2: 1.2FG+1.5FQ wb =1.2FG+1.5FQ=42.6 kN/m

wc =1.2FG= 24.6 kN/m

Total dead load= 18.5+ 2= 20.5 kN/m wb

wc

Live Load= 3 kN/m2 = 3x4= 12 kN/m

MB= 613.05 kN-m

L2= 12 m

L3= 5 m wc = 1.2FG+1.5FQ=42.6 kN/m

4000 mm L2= 12 m

MC= -532.5 kN-m

L3= 5 m L2= 12 m

L3= 5 m

Tutorial Feedback: Recap Week 1

Motivation for Week 2 Lecture

Case 3: Alternative Load Combination: not as per AS300

 How does a beam behave under load? wb =1.2FG+1.5FQ=42.6 kN/m

 Is it better to use more reinforcement in beams?

wc =1.0FG= 20.5 kN/m

 Is the cross-section of RC beam a composite section? If yes, how can we apply elastic beam theory to analyse the cross-section

MB= 638.68 kN-m L2= 12 m

 Why the value of neutral axis depth obtained from modular ratio method different from equilibrium and compatibility method? Is the difference significant?

L3= 5 m

 What is total deflection, short-term deflection, and incremental deflection?

wc = 1.2F G+1.5FQ=42.6 kN/m

How can I control cracking in the beam? Can I calculate crack width?

MC= -532.5 kN-m L2= 12 m

L3= 5 m

Common Types of Beam Sections

Flexural Behaviour of Beams Under Load

11

12 Foster et al. (2010)

Flexural Behaviour of Beam Under Load

Flexural Behaviour of Beam Under Load

Pre-cracking Behaviour Beam Beam

z

Free body diagram showing internal moment as a compression and tension force couple Bending moment diagram

=E

C=T M=Tz = Cz

Cross-section

Elastic beam theory =My/I

Strains

Stresses

 Strain in the concrete and steel at the same level are almost the same  Stress in the steel is greater than the stress in the adjacent concrete

Free body diagram showing internal moment and shear force MacGregor and Wight (2007)

14

13

Foster et al. (2010)

Post-cracking: service load Behaviour cr =f’ct.f – cs

Post-cracking: service load Behaviour Moment-curvature relationship of a beam segment

C=T

z= lever arm Tension stiffening between primary and secondary cracks

M=Tz = Cz 15 Foster et al. (2010)

16 Foster et al. (2010)

Overload behaviour Overload behaviour

dn

C

C

z T

stres s

T=Astfsy

Reinforcement

C=T

T=Astfsy

M=Tz = Cz Strain e

Changes in compression stress block with increasing moment Overload Conditionthe section can still accept a small increment of moment and Mu>My

At all stages of loading continuity of deformation: average strain is linearly distributed over the depth of the beam Force Equilibrium:

17

C=T

Foster et al. (2010)

18 Foster et al. (2010)

Beam Failure

Under-reinforced and over-reinforced beams dn

C

Under-reinforced beam: The tensile reinforcement yields at a moment My which is less than the ultimate moment Mu. The failure is described as primary tension failure (MyMu) .

Balanced failure: Tensile steel yields at the same time that the capacity of the concrete compressive region reached (My=Mu) . Both balanced failure and primary compression failure are brittle and hence undesirable. 19 Foster et al. (2010)

20

Analysis and Design for Serviceability For Reinforced concrete, the prime serviceability consideration relate to deflections and cracking of concrete.

Elastic Section Analysis

Deflection control and crack control require a study of the stress and deformation that occur in cross-section and local region.

=My/I

Hence, we need to carry out cross-sectional analysis

21

22

PROPERTIES OF BEAM CROSS-SECTION: Composite Beams

Elastic beam theory =My/I

Beams constructed of two or more different materials are called composite beams Flexure formula cannot be applied directly to determine normal stress in a composite beam Look at the formulae for the second moment of area of a rectangular cross-section at the centroidal axis and at the base!

Thus a method is required to “transform” a beam’s crosssection into one made of a single material, then the flexure formula can be applied 23

24 Foster et al. (2010)

Composite Beam: modular ratio method • Height remains the same, but stiffer portion (material 1) of beam widened to carry equivalent load to that n carried by material 1.

Transformed Section : RC Beam

E1 E2

• Dimensionless factor n, is called the modular ratio. It indicates that crosssection, with a width b on original beam, be increased to a width of b2 = nb in region where material 1 is being transformed into material 2. • Once “transformed”, the normal-stress distribution over the transformed xsection will be linear as shown. Limitation: only applicable to linearelastic section analysis

Ast

M...