This article shows a practical worked example on how to check the stress in a prestressed concrete beam using the combined stress approach. Verifying the stress will enable a designer to conclude whether or not the stress in a beam is within the acceptable stress limit.
Worked Example
The beam on which the stress verification will be carried out is a simply supported pre-tensioned I beam with tendons of constant eccentricity and constant prestressing force over the entire 20m span. The prestressing force and eccentricity are 1000KN and 300mm respectively. The short-term and long-term loss are taken as 10% and 25% of the jacking force respectively. The section properties of the beam are shown in the diagram below. The live load on the beam is 80KN/m2. The beam strengths during force transfer and after 28days are 25N/mm2 and 40N/mm2 respectively.
From the beam section above, some key dimensions of the beam are:
Breadth of top flange (bft) = 3000
Thickness of top flange (hft) = 200
Breadth of web (bw) = 300
Height of beam (htotal) = 2500
Thickness of bottom flange (hfb) = 300
Breadth of bottom flange (bfb) = 1000
span = 20m
It will not be ideal for obvious reasons to verify each and every point on the beam. Hence, two critical section which are the mid span, and the end span of the beam shall be verified for stress both at the top and bottom fiber. This verification shall be done for both transfer stage and service stage.
The steps towards calculating these is as follows:
-
Calculate the geometric properties of the beam
Area of the section
The total area of the section is calculated by adding up the area of the top flange, bottom flange, and the web.
Area of top flange (Atf) + Area of bottom flange (Abf) + Area of the web (Aweb)
(3000 – 300) X 200 + (1000 – 300) X 300 + 300 X 2500
540000 + 210000 + 750000
= 1500000mm²
Calculate the position of the centroid from the beam soffit.
Here we take the moment of area of the top flange (Atf), area of bottom flange (Abf), and area of the web about the soffit and equate it to the moment area of the overall section.
1500000 x ybot = 540000 x (2500 – 300/2) +210000 x (300/2) + 750000 x 2500/2
$y_{\text {bot }}=\frac{540000 \times(2500-300 / 2)+210000 \times(300 / 2)+750000 \times 2500 / 2}{1500000}$
ybot = 1510
Since the centroid is position at 1510mm from the soffit, hence the distance of the centroid to the top of the beam equals:
ytop = 2500 – 1510 = 990mm
Calculate the second moment area of the beam.
The second moment area of the section is calculated using the parallel axis theorem equation: I + A (x – y)²
((3000 – 300) X 200³)/12 + 540000((2500 – 300/2) – 1510)²) + ((1000 – 300) X 300³ + 210000(300/2 – 1510)²) + ((300 X 2500³)/12 + 750000(2500/2 – 1510)²)
$4.4 \times 10^{11}+4.3 \times 10^{11}+3.9 \times 10^{11}=1.26 \times 10^{12}$
Section Modulus to top and bottom flange
The section modulus to top is got by dividing the second moment area of the section by the distance of the centroid from the top of the beam.
$Z_t=1 / y_{\text {top }}=1.26 \times 10^{12} / 990=1273585859$
$Z_b=1 / y_{\text {bot }}=1.26 \times 10^{12} / 1510=835000000$
-
Estimation of Loads and Moment
Self-weight of beam
= Area x unit weight of concrete
$=1.5 \times 10^6 \times 10^{-6} \times 25=37.5 \mathrm{KNm}$
Moment due to self-weight
= (37.5 x 20²)/8 = 1875KNm
Moment due to dead and live load
= ((37.5+80) x 20² )/8 = (117.5 x 20²)/8 = 5875KNm
Prestress force at transfer
At transfer 10% of the jacking prestress force is lost.
Prestress at transfer (Pt) = 90/100 x 1000KN = 900KN
Load factor at transfer
ϒsup= 1.1
Prestress force at service
At service 25% of the jacking prestress force is lost.
Prestress force at service = 0.75 x 1000 = 750KN
Load factor at service
ϒinf= 0.9
Stress Calculation at transfer
Mid Span top fiber
$\left(\frac{-P}{A}+\frac{P_\epsilon}{Z_t}\right) \Upsilon_{\text {sup }}-\frac{M_{\min }}{Z_t} \leq f_{t t}$
$
\left( \right. \frac{-\text{900}x\,\,10^3}{\text{15}x\,\,10^5}\,\,+\,\,\frac{\text{900}x\,\,10^3\,\,x\,\,300}{1273585859}\left. \right)1.1 \,\,-\,\,\frac{\text{1875}x\,\,10^6}{1273585859}
$
= (-0.6 + 0.212)1.1 – 1.47
= – 1.89MPa
Mid Span bottom fibre
$\left(\frac{-P}{A}+\frac{-P_\epsilon}{Z_b}\right) \Upsilon_{\text {sup }}+\frac{M_{\min }}{Z_b} \geq f_{t c}$
$
\left( \right. \frac{-\text{900}x\,\,10^3}{\text{15}x\,\,10^5}\,\,-\,\,\frac{\text{900}x\,\,10^3\,\,x\,\,300}{835000000}\left. \right)1.1 \,\,+\,\,\frac{\text{1875}x\,\,10^6}{835000000}
$
= (-0.6 – 0.3)1.1 + 2.2
= 1.2MPa
Support top fiber
$\left(\frac{-P}{A}+\frac{P_\epsilon}{Z_t}\right) \Upsilon_{\text {sup }}-\frac{M_{\min }}{Z_t} \leq f_{t t}$
$
\left( \right. \frac{-\text{900}x\,\,10^3}{\text{15}x\,\,10^5}\,\,+\,\,\frac{\text{900}x\,\,10^3\,\,x\,\,300}{1273585859}\left. \right)1.1 \,\,-\,\,\frac{\text{0}x\,\,10^6}{1273585859}
$
= (-0.6 + 0.2)1.1 + 0
= -0.4MPa
Support bottom fiber
$\left(\frac{-P}{A}+\frac{-P_\epsilon}{Z_b}\right) \Upsilon_{\text {sup }}+\frac{M_{\min }}{Z_b} \geq f_{t c}$
$
\left( \right. \frac{-\text{900}x\,\,10^3}{\text{15}x\,\,10^5}\,\,-\,\,\frac{\text{900}x\,\,10^3\,\,x\,\,300}{835000000}\left. \right)1.1 \,\,+\,\,\frac{\text{0}x\,\,10^6}{835000000}
$
= (-0.6 – 0.3)1.1 + 0
= -1.0MPa
Stress Calculation at Service
Mid Span top fiber
$\left(\frac{-P}{A}+\frac{P_\epsilon}{Z_t}\right) \Upsilon_{\text {inf }}-\frac{M_{\max }}{Z_t} \geq f_{t c}$
$
\left( \right. \frac{-\text{750}x\,\,10^3}{\text{15}x\,\,10^5}\,\,+\,\,\frac{\text{750}x\,\,10^3\,\,x\,\,300}{1273585859}\left. \right)0.9 \,\,-\,\,\frac{\text{5875}x\,\,10^6}{1273585859}
$
= (-0.5 + 0.18)0.9 – 4.6
= – 4.9MPa
Mid Span bottom fiber
$\left(\frac{-P}{A}+\frac{-P_\epsilon}{Z_b}\right) \Upsilon_{\text {sup }}+\frac{M_{\min }}{Z_b} \leq f_{t t}$
$
\left( \right. \frac{-\text{750}x\,\,10^3}{\text{15}x\,\,10^5}\,\,-\,\,\frac{\text{750}x\,\,10^3\,\,x\,\,300}{835000000}\left. \right)0.9 \,\,+\,\,\frac{\text{5875}x\,\,10^6}{835000000}
$
= (-0.5 – 0.27)0.9 + 7.0
= 6.34MPa
Support top fiber
$\left(\frac{-P}{A}+\frac{P_\epsilon}{Z_t}\right) \Upsilon_{\text {sup }}-\frac{M_{\min }}{Z_t} \geq f_{t c}$
$
\left( \right. \frac{-\text{750}x\,\,10^3}{\text{15}x\,\,10^5}\,\,+\,\,\frac{\text{750}x\,\,10^3\,\,x\,\,300}{1273585859}\left. \right)0.9 \,\,-\,\,\frac{\text{0}x\,\,10^6}{1273585859}
$
= (-0.5 + 0.18)0.9 + 0
= -0.29MPa
Support bottom fiber
$\left(\frac{-P}{A}+\frac{-P_\epsilon}{Z_b}\right) \Upsilon_{\text {sup }}+\frac{M_{\min }}{Z_b} \leq f_{t t}$
$
\left( \right. \frac{-\text{750}x\,\,10^3}{\text{15}x\,\,10^5}\,\,-\,\,\frac{\text{750}x\,\,10^3\,\,x\,\,300}{835000000}\left. \right)0.9 \,\,+\,\,\frac{\text{0}x\,\,10^6}{835000000}
$
= (-0.5 – 0.27)0.9 + 0
= – 0.69MPa
It should be noted that above a sign convention is adopted in the combined stress equation such that compressive stress is negative while tensile stress is positive. The moment due to external load and that due to eccentricity of prestress force are always opposite.
Discussion
Since the beam strength at transfer and in service are given as 28N/mm2 and 40N/mm2 respectively, the permissible stress at transfer and service are estimated below
Transfer
Compressive stress = 0.6fck = 0.6 x 28 = -15N/mm2
Tensile stress = 0.3 fck^3/2 = 0.3 x 28^3/2 = 2.6N/mm2
Service
Compressive stress = 0.6fck = 0.6 x 40 = -24N/mm2
Tensile stress = 0.3 fck^3/2 = 0.3 x 40^3/2 = 3.5N/mm2
It should be noted that the permissible stresses are assigned negative for compressive stress while the tensile stresses are assigned positive. The same applies to the stress in the beam: negative (-) is compressive, positive is (+) tensile.
Comparing the stresses against the permissible stresses.
At transfer stage the stresses at support (-0.4MPa and -1.0MPa), and at mid-span (-1.89MPa and 1.2MPa) falls within the limit of permissible stress. The section remains uncracked during transfer.
However, at service the stresses at the support are -0.29MPa and -0.69MPa, while that at the mid-span are -4.9MPa and 6.34MPa. The stress at the bottom fiber at service violates the limiting permissible tensile stress (i.e.: 6.3MPa > 3.5MPa), hence the section will crack at the bottom in service. The section can either be redesigned to prevent cracking altogether or bonded reinforcement can be introduced to limit the crack width, the favourable approach depends on the design requirement.
