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Unformatted text preview: www.SunCam.com Copyright 2010 John P. Miller Page 15 of 49 Fundamentals of Post‐Tensioned Concrete Design for Buildings – Part One A SunCam online continuing education course 8 We can see that the effective posttensioning force, Fe, is a function of the span length
and the available drape, for a given balanced load. By observing the above equation,
we can see that the minimum effective posttensioning force will be found in spans with
small spans and large available drapes. However, if the smallest possible effective
posttensioning force is selected (the left end span would only require 3.42 kips/foot)
and used in all the spans, there would not be enough available drape to balance 60 psf
in all the other spans.
Referring to the diagram below, the effective posttensioning force has been calculated
for each span using the maximum available drape. Allowable upper limit of tendon profile Ideal Tendon Profile 10'‐0" 12'‐0" 9'‐0" Fe = 3.42 k/ft Fe = 3.70 k/ft Fe = 2.08 k/ft 13'‐0"
Fe = 4.35 k/ft Allowable lower limit of tendon profile 15'‐0" Fe = 7.71 k/ft FiveSpan Slab Example
This would be the optimal use of prestressing since all spans would make use of the
maximum amount of drape, and therefore the minimum amount of prestressing, to
balance a given load. However, it is neither practical nor desirable to use a different
amount of prestress in each span.
If we were to use the effective posttensioning force of 7.71 kips per foot in all spans
and adjusted the drapes to maintain the balanced load of 60 psf, we would have drapes
of 1.68", 0.95", 1.97", and 2.625" for the first four spans left to right. Only the right end
span maximizes the efficiency of the draped tendon to balance the load. Another way
of stating this is that the full available drape in every span would not be utilized and it
should be apparent that this would not result in the most efficient use of posttensioning.
www.SunCam.com Copyright 2010 John P. Miller Page 16 of 49 Fundamentals of Post‐Tensioned Concrete Design for Buildings – Part One A SunCam online continuing education course One technique that is commonly employed in continuous posttensioned structures is to
place additional tendons in the end spans to satisfy the controlling demand and use only
enough tendons in the interior spans to satisfy the lower demand there. This is a more
efficient use of posttensioning. Even though this may seem to be a minor difference, if
this is applied over a large multistory structure, the total savings can be significant.
So, to make this example as efficient as possible, we will only use the 7.71 kips per foot
in the right span and use 4.35 kips per foot in the other four spans. Thus, using the
equation 8 We find the drapes as shown in the figure below.
Fe = 4.35 k/ft Typical 10'‐0" Drape = 2.06" Tendon Profile 12'‐0" 9'‐0" Drape = 2.98" Drape = 1.67" 13'‐0"
Drape = 3.50" Fe = 7.71 k/ft this span only 15'‐0" Drape = 2.625" FiveSpan Slab Example
Balanced Load Moment Diagrams
As we learned earlier, once the balanced load is selected, the effective prestressing
force can be calculated for a given drape and span length. The member may then be
structurally analyzed with this equivalent set of tendon loads applied. The results of this
analysis may be combined with other load cases such as live load and superimposed
dead load.
Let's now consider the two span beam shown below. The spans are unequal and each
span requires a different effective posttensioning force. Let's determine the balanced
load moments. www.SunCam.com Copyright 2010 John P. Miller Page 17 of 49 Fundamentals of Post‐Tensioned Concrete Design for Buildings – Part One A SunCam online continuing education course 7"
Fe = 300 k Fe = 200 k 26" 30'‐0" 25'‐0" 30‐0" By inspection of the above diagram, we see that the drape for the left span is: 26 Left Span Drape 29.5 And the balanced load is computed as: 8 200
30 29.5
12 4.37 / Next, we can find the concentrated balanced load in the right span due to the harped
tendon by adding the vertical components on both sides of the harped point as follows:
B = 53.3 k 300 k 300 k 33" 26"
30' 25' 300 sin tan 2.75/30 300 sin tan 27.4 www.SunCam.com 25.9 300 sin 2.17/25 53.3 Copyright 2010 John P. Miller Page 18 of 49 Fundamentals of Post‐Tensioned Concrete Design for Buildings – Part One A SunCam online continuing education course Now we can construct the balanced loading diagram below. The left span has a
uniformly distributed upward load of 4.37 kips/ft and the right span has a concentrated
upward load of 53.3 kips.
4.37 k/ft 53.3 k Balanced Load Diagram Example
Using conventional elastic structural analysis, we find the moment diagram for the
above balanced loads to be the following: +515 ft‐k
‐267 ft‐k ‐488 ft‐k Balanced Moment Diagram
The above example illustrates the simplicity and straightforwardness of the bala...
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 Spring '14

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