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CIVL4903 – Floor Systems – Tutorial
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Week 2 Tutorial Reference Sheet – Rules of Thumb and Prestressing – Additional Notes
1. Span to Depth (L/D) Ratio Example
L/D ratios are very useful during the conceptual design stages. They express the approximate
total depth of a typical ‘type’ of section as a proportion of the span, and provide preliminary
guidance on section depths that will work efficiently in strength design, while also giving
acceptable deflections in most normal circumstances.
The total depth ‘D’ is normally assumed from the top of the slab to the soffit of the beam or
slab. Some typical L/D Ratio guidelines for reinforced and prestressed members are given
below.
Structure Type – Office Floors
Typical Applied Load = 3 + 1 = 4kPa
L/D Guidelines
RC P/T
Simply Supported band beams L/16 L/24
Continuous band beams Interior Spans L/20 L/30
Continuous band beams End Spans L/18 L/27
One Way Slabs Interior Spans
(L measured between beam edges)
L/26 L/38
1. Deriving Span to Depth Ratios – why do they work
1a. For constant stress with width and load / m constant
σ = M / Z
M = wL2 / 8 = k1 L
2
Z = bD2 / 6 = k2 D
2
where k1 and k2 are constants relating to the load and the width
Hence for constant stress σ = M / Z = k1 / k2 × (L / D)2 = k3(L / D)2
The stress remains the same for a constant L / D ratio regardless of the span.
For constant rate of deflection (for example, allowable deflection = span/250, etc.) with
beam width & load / m constant
δ = 5wL4 / 384 EI
I = bD3 / 12
And δ = (5w / 384 E {b / 12}) (L4 / D3 ) = k4 L4/D3
Hence δ / L = k4 (L /D)3
The rate of deflection remains the same for constant L / D regardless of the span.
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It is seen that the ‘basic stress parameter’ and the ‘deflection rate’ remain constant with
the same L/D regardless of span. In the end material strength or absolute deflection will
become the limiting factor.
1b. If the total service design load on simply supported RC band beam (L/D = 16) is
increased by 50% (i.e. w2 / w1 = 1.5), what approximate L/D ratio would you adopt to
achieve the same rate of deflection
Now D2 / D1 = (w2 / w1)
0.33 = 3√1.5 = 1.145
Hence D2 = 1.145 D1 and D1 = 0.873 D2
And L / D1 (Question 1a) = L / 0.873 D2 = 16
Hence L / D2 = 0.873 × 16 = 14
To retain a constant rate of deflection under increased load we need to increase the
depth of the member (as we would expect), which means that a lower L / D ratio is
appropriate.
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2. Load Balancing
Load balancing is an important concept in prestressed concrete design. Its main benefit is
helping to reduce the deflection of members, since a part of the load is carried on the draped
prestressing cables like a ‘suspension system’. The deflection of the concrete member is then
calculated using the ‘net’ load under service conditions.
While it is not applicable in the ultimate strength design, the load balancing concept provides
some useful rules of thumb for initial estimates of the amount of prestressing required
Consider the following load balancing example with a parabolically draped prestressing cable
in a simply supported beam.
T = total pre-stressing force in the strands under service conditions. (kN)
e = the eccentricity of the draped cable profile between supports. (m)
wp = the uplift provided by the curved profile of the strands. (kN/m)
C = the pre-compression applied to the concrete by the anchoring of the stressed
strands, which for equilibrium is equal to T
To calculate the load balancing uplift wp we can consider the free body equilibrium of half of
the beam:
In this exercise there are no externally applied loads. Hence to achieve internal equilibrium in
a statically determinate structure the centroid of the C force in the concrete will follow the
line of the T force in the strands.
Can you explain why this is so using simple concepts? One approach is to consider a
narrow cylindrical compression strut with a series of bends in it, and a tension cable running
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through the centre - a bicycle brake cable is a good example - and where the tension cable
applies a resistant compression to the outer tube. At any bends in the cable the changes in
direction of the C and T forces cause equal and opposite lateral loads which are therefore
balanced and in equilibrium, causing no bending to the outer tube.
Taking moments about the left side support:
(Ca = Cc = T ) × e = wp (L / 2) × (L / 4)
Hence T × e = wpL
2/8 and wp = 8 × T × e / L
2
This formula can be applied in the same way for (indeterminate) continuous beams using the
total drape in any span.
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3. Span / Depth Ratios for Prestressed Concrete Members
a. In the lecture it was mentioned that the moment of inertia of a cracked reinforced
concrete section is about 50% of the value for an uncracked section.
If the recommended L/D ratio for a simply supported RC band beam is 16, what is
an appropriate L/D ratio for a PT member to be used for the same span and
loading?
Assume that the PT beam remains uncracked under service conditions, and ignore any
load balancing effects at this stage.
Considering the deflection of a simple beam:
δ = 5wL4 / 384 EI
For a RC beam Ieff ≈ 0.5 Igross = 0.5 bD13 / 12
For a PT beam Ieff = Igross = bD2
3 / 12
For the same deflection
0.5 bD1
3 / 12 = bD2
3 / 12
hence (D2 / D1)
3 = 0.5
and D2 =
3√0.5 × D1 = 0.79 D1
Hence for the same span
D2 = 0.79 [D1 = L/16] ≈ L/20
and L / D2 for a PT member = 20
b. The total design service load on the above beam ws = 60kN/m, and the load balancing
uplift applied by the draped prestressing tendons wp = 30 kN/m.
What is the net load that should be used in calculating the deflection of the beam?
The net load used in calculating the deflection of the beam is the difference between the
downward design service load, and the load balancing uplift from the prestressing
tendons. Hence
w net = (ws - wp) = (60 – 30)kN/m = 30 kN/m
a. The following formula was given in Question 1b for calculating the depth of members
under different loading in order to achieve the same rate of deflection:
D2 / D1 = (w2 / w1)
0.33
Estimate further adjustments to the appropriate L/D ratio for the prestressed
member in Question 3a, when load balancing effects are also taken into account.
The effective rate of loading on the prestressed beam in 3b. is halved due to the load
balancing. For a constant rate of deflection, we know from 1b. that:
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D2 / D1 = (w2 / w1)
0.33 = 3√0.5 = 0.79
Hence D2 = 0.79 × (0.79 D1) [D1 = RC member from answer 3a. above]
And L / D2 = L / 0.64 D1 = 16 / 0.64 = 24
Summary:
In these simple exercises we have seen how the deflection performance of prestressed
members is improved in two ways compared with reinforced members:
• In exercise 3a the member depth was reduced to about 80% because it was uncracked due
to the pre-compression of the section.
• In exercise 3c we saw how load balancing relieves a significant proportion of the load on
the actual concrete section, leading to a further reduction of depth for the same deflection.
The overall effect is that the depth of prestressed members may be as low as 65% to 70% of
that of equivalent reinforced sections.
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c. Ultimate Strength Calculation
The ultimate bending strength of prestressed members follows the same principle as all
bending members – there is a force couple C = T, separated by a lever arm, la. Compare the
following with what you know about the bending of reinforced concrete sections, and
make sure you are familiar with the basic principles.
- Tpu = Apt x σpu
- Apt = total area of prestressing strands. For example in AS3600 table 3.3.1, the area of
each 12.7mm 7 wire ordinary strand is 98.6mm2.
- σpu = the maximum stress developed in the tendons. This is given in AS3600 8.1.7,
expressed as a proportion of the characteristic breaking stress of the strands fpb.
Note that k1 is given in the code, and k2 expresses the strength relationship between
the total tensile (prestress plus any reinforcement) and compressive strengths in the
section – i.e this reflects how heavily the section is reinforced.
- The rectangular compression stress block is derived in the same way as for reinforced
concrete.
- In light to moderately utilised concrete sections, the following ratios can be used to
establish approximate prestressing quantities in schematic design:
la ≈ 0.95 do and σpu ≈ 0.95 fpb.
In schematic design it is simplest to use the ‘characteristic breaking load’ from Table 3.3.1,
which is 184kN per strand for the 12.7mm strands that are used most commonly in Australian
building construction. Hence Tpu per strand = 0.95 x 184 = 175kN per strand.