Introduction to RC Design
Dr Mizan Ahmed
Acknowledgment to Dr Trevor Htut
A global university with campuses in Asia and Western Australia
Perth | Dubai | Malaysia | Singapore
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Essential & Recommended Texts
AS 3600-2018 Concrete
Structures (Incorporating
Amendment Nos. 1 & 2)
Reinforced Concrete Design in
accordance with AS 3600-2009
https://www.ccaa.com.au/imis_prod/documents/IND
USTRY_GUIDE_T38_Reinforced_Concrete_Design_in_
Accordance_with_AS3600.pdf
Warner, R. F., Rangan, B. V.,
Hall, A. S. and Faulkes, K. A.
(1998). “Concrete
Structures”. Longman.
ISBN: 0 582 80247 4
Foster, S. J., Kilpatrick, A. E.
and Warner, R. F. (2021).
“Reinforced Concrete
Basics”, 3rd Edition,
Pearson. ISBN:
9780655703662
Design Philosophy
What are purposes of structures?
1. To provide shelter and protection (e.g., buildings, stadiums).
2. To facilitate an activity (e.g., observation tower, roller coaster).
3. To store materials (e.g., tanks, silos).
4. To provide access (e.g., bridges, tunnels).
5. To support (part of) equipment (e.g., telecommunication towers, offshore oil
platforms).
The structure must remain Stable, Safe and Serviceable under all actions and/or
combinations of actions that can be reasonably expected during its design life.
What are design requirements?
❑ Structural requirements
✓ Safety “strength” (no collapse or failure)
✓ Serviceability “stiffness” (limit vibrations, deflections, etc.)
❑ Other design requirements
✓ Economy (both initial cost and maintenance cost)
✓ Harmony
within the structure (e.g., mechanical, architectural &
structural systems)
within the physical surrounding and community
Limit State Design (LSD)
❖ Limit State: A design situation where the structure is at its limit of satisfactory
performance.
❖ LSD method requires satisfaction of two principal limit states: the ultimate limit
state (ULS) and the serviceability limit state (SLS).
❖ ULS is concerned with the safety of people and the structure. Examples of ULS
include fracture, collapse, loss of equilibrium and loss of stability.
❖ SLS is concerned with the functioning of the structure under normal use, the
comfort of people, and the appearance of the construction works. Examples of SLS
include excessive and/or unrecoverable deflection, cracking, vibration, or material
yielding.
❖ Each limit state has different Load combinations, Behaviour models, and
Performance limits.
Serviceability Limit State (SLS)
❖ Load combinations:
Consider loads (and combination of loads) that could conceivably occur during
the design life of the structure.
❖ Behaviour models and performance limits (when subjected to load
combinations):
Design to avoid excessive and/or unrecoverable deflection, cracking, vibration, or
material yielding.
❖ Consequence/penalty of non-compliance:
Inconvenience, offensive appearance.
Adopted from STEN3004 Lecture notes (Courtesy of Kerri Bland)
Ultimate Limit State (ULS)
❖ Load combinations:
Consider loads (and combination of loads) that would occur just before failure of
the structure (higher than ever expected to occur).
❖ Behaviour models and performance limits (when subjected to load
combinations):
Design to avoid material failure (breakage) or structural failure (collapse).
❖ Consequence/penalty of non-compliance:
Risk to life.
Adopted from STEN3004 Lecture notes (Courtesy of Kerri Bland)
Limit State Design (LSD)
Σ(ɣ . Nominal Load Effects) ≤ φ . Nominal Capacity
Design Action Effect (Load) ≤ Design Capacity
R* ≤ φ R
❖ Different types of loads have different probabilities of occurrence and different
degrees of variability.
❖ Load factors (ɣ) and load combinations are given in:
AS/NZS 1170.0 Structural design actions (Part 0: General Principles)
❖ Capacity reduction factor () is to consider variabilities due to:
✓ Variation material strength & defects
✓ Variation in section properties
✓ Modelling of connection behaviour.
✓ Effects on strength due to fabrication defects & erection process
Capacity reduction factors are given in AS3600:2018 Concrete Structures
Stability of Structures
Introduction to Stability of Structures
❖ A structure must not only transfer gravity forces to the ground; it must
also stand up against many lateral forces.
❖ Ability to satisfactorily resist horizontal or other disturbing forces which
Overturning
could cause overturning, uplift or cause the building to sway.
❖ To design for stability, the engineer must have a clear understanding of
how the horizontal or disturbing forces are transferred to the building
foundations.
Sliding
IStructE (2014)
Loads that may imposed lateral actions on the structure
IStructE (2014)
Stability Systems
IStructE (2014)
ICE (2012)
Vertical Bracing
Single Brace (Tension & Compression)
Cross Bracing (Tension only)
‘K’ Brace (Tension & Compression)
ICE (2012)
Recap on Loading
Load Allocation (Clause 6.10.3.4)
❖ For the design of supporting beams, slab loads are assigned based on 45°
pattern if all edges are continuous.
For Information Only
AS 3600-2009 Supp1
Load Allocation (Clause 6.10.3.4)
❖ If one edge is discontinuous, an adjustment of 10% extra on continuous
edges is made, i.e. Load factor of 1.1 on the continuous edges and a
reduction of 20% (factor of 0.8) on discontinuous edge.
For Information Only
AS 3600-2009 Supp1
Load Allocation (Clause 6.10.3.4)
❖ If two edge is discontinuous, an adjustment of 10% extra on continuous
edges is made, i.e. Load factor of 1.1 on the continuous edges and a
reduction of 25% (factor of 0.75) on discontinuous edges.
❖ For other cases the code recommends the elastic analysis.
Decrease load by
25%
For Information Only
AS 3600-2009 Supp1
Load Allocation (Clause 6.10.3.4)
The loading on support beams and hence moments in those beams can be found
from the triangular and trapezoidal patterns in Figure 6.10.3.4 of AS3600. This
gives equivalent uniformly distributed loads as
w*x =
w*  Lnx/3
w*y =
[w*  Lnx/6]  [3 - (Lnx/Lny)2]
where w* is the loading kPa on the slab panel, and w*x is the load in kN/m run of a
beam from that one panel in the shorter direction. The loading of the beam itself
and potentially the slab panel on the other side should also be added.
w*x
For Information Only
AS 3600-2009 Supp1
w*y
Loading on Structure
Example 1: Determine the design loading on the office floor slab and various continuous
beams shown below. Take superimposed permanent load as 1 kPa and density of
reinforced concrete as 25 kN/m3.
2
1
4
3
7000
8000
7000
A
3500
A
B
A
C
D
E
4200
B
500
150
400
C
4200
4200
4200
3500
Section A-A
3500
D
E
A
Office Floor Plan
3500
Loading on Structure
Load on Slab:
Slab self-weight = 0.15 m (thickness) × 25 kN/m3 (density) = 3.75 kPa (kN/m2)
Superimposed permanent load
 Permanent load on slab
= 1.0 kPa
= 4.75 kPa
From Table 3.1 of AS/NZS 1170.1, for general office use
Imposed load on slab
= 3.0 kPa
From Clause 4.2.2 of AS/NZS 1170.0, for strength limit states,
Design load, w* = Maximum of [1.35G or 1.2G + 1.5Q or 1.2G + 1.5lQ]
Note: Wind load, earthquake load and snow load are not applicable for this floor slab.
From Table 4.1 of AS/NZS 1170.0, for office floors, l = 0.4
Design load,
*
w
= Max. [1.35 × 4.75; 1.2 × 4.75 + 1.5 × 3; 1.2 × 4.75 + 1.5 × 0.4 × 3]
= Max. [ 6.41 kPa; 10.2 kPa; 7.5 kPa] = 10.2 kPa
Loading on Structure
B
A
C
D
E
Load on beams:
Permanent load from slab, G = 4.75 kPa
500
150
Imposed load from slab, Q = 3.0 kPa
400
3500
4200
Design load from slab, w* = 10.2 kPa
Beam self-weight = (0.5 m – 0.15 m) × 0.4 m × 25
Tributary width on Beam along Grid
Tributary width on Beam along Grid
Tributary width on Beam along Grid
3
kN/m =
3.5 kN/m
0.4 𝑚
3.5 𝑚
A&E=
+
= 1.95 m
2
2
4.2 𝑚
3.5 𝑚
B&D=
+
= 3.85 m
2
2
4.2 𝑚
4.2 𝑚
C=
+
= 4.2 m
2
2
4200
3500
Loading on Structure
2
1
Beam along Grid A & E:
Permanent load, G = 4.75 kN/m2 × 1.95 m + 3.5 kN/m
4
3
7000
8000
7000
A
A
3500
= 12.8 kN/m
B
4200
Imposed load, Q = 3.0 × 1.95 = 5.9 kN/m
UDL design load, w* = 1.2 × 12.8 + 1.5 × 5.9 = 24.2 kN/m
Beam along Grid B & D:
C
Imposed load, Q = 3.0 × 3.85 = 11.6 kN/m
D
3500
= 21.8 kN/m
4200
Permanent load, G = 4.75 kN/m2 × 3.85 m + 3.5 kN/m
E
UDL design load, w* = 1.2 × 21.8 + 1.5 × 11.6 = 43.6 kN/m
Beam along Grid C:
Permanent load, G = 4.75 kN/m2 × 4.2 m + 3.5 kN/m = 23.5 kN/m
Imposed load, Q = 3.0 × 4.2 = 12.6 kN/m
UDL design load, w* = 1.2 × 23.5 + 1.5 × 12.6 = 47.1 kN/m
A