This article presents a worked example on the construction of magnel diagram for the design of a prestressed beam section. The beam that shall be designed is the same as that on which stress check was carried out here. The cross-section of the beam is provided in the figure below for easy reference.
Geometric Properties of the Beam
The geometry properties are also re-provided below, you can check the previous post here to study how the geometric properties are calculated.
Area of section (A) = = 1500000 mm2
Distance of centroid from soffit (yb) = 1510mm
Distance of centroid to top fiber (yt) = 990mm
Second moment area (I) =1.26 x 1012 mm4
Section modulus to top fiber (Zt) = 1273585859 mm6
Section modulus to bottom fiber (Zb) = 835000000mm6
Estimation of Internal forces and Load factors
The internal forces also remain the same, these are also reproduced below for easy follow up.
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
Load factor at service
ϒinf= 0.9
Estimation of Permissible Stresses
The permissible stresses in the beam are to be estimated as follows
Permissible stress at transfer
Compressive stress = 0.6fck = 0.6 x 25 = -15N/mm2
Tensile stress = 0.3 fck^3/2 = 0.3 x 25^ (3/2) = 2.6N/mm2
Permissible stress at 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
η = 1- %loss at transfer/ 1 – % loss at service = 0.75/0.25 = 0.83
Evaluation of Magnel Equations and Construction of Magnel diagram
The magnel Equation for at both transfer and service are evaluated and plotted as shown in the following paragraphs
Magnel equation for mid-span
At transfer
Top
$
\frac{10^8}{P}\,\,\geqslant \,\,\frac{\left( -\frac{1}{A}\,\,+\,\,\frac{e}{Z_t} \right)}{\frac{\eta \,\,M_{\min}}{\gamma _{sup\,\,}Z_t}+\frac{\eta \,\,f_{tt}}{\gamma _{sup}}}x 10^8
$
Substituting the values of parameters in the equation
$
\frac{10^8}{P}\,\,\geqslant \,\,\frac{\left( -\frac{1}{1500000}\,\,+\,\,\frac{e}{1273585859} \right)}{\frac{\0.93 x\,\,1875 x 10^6}{1.1 x {\,\,}1273585859}+\frac{\0.93\,\,2.6{}}{1.1{}}}x 10^8
$
Substituting different values of eccentricity into the equation gives corresponding values of 108/P as shown in the table below
-300 – 29.4
-200 -24.4
-100 -21.8
0 -16.7
200 -11.5
400 -6.4
600 -1.3
800 3.9
1000 9.0
1200 11.6
1400 14.1
The values in the table are plotted as shown below.
$
\frac{10^8}{P}\,\,\geqslant \,\,\frac{\left( -\frac{1}{A}\,\,-\,\,\frac{e}{Z_b} \right)}{\frac{\eta \,\,M_{\min}}{\gamma _{sup\,\,}Z_b}+\frac{\eta \,\,f_{tc}}{\gamma _{sup}}}x 10^8
$
Substituting the values of parameters in the equation
$
\frac{10^8}{P}\,\,\geqslant \,\,\frac{\left( -\frac{1}{1500000}\,\,-\,\,\frac{e}{ 835000000} \right)}{\frac{\0.93 \,\,1875 x 10^6}{1.1 x {\,\,}835000000}+\frac{\0.93\,\,-15{}}{1.1{}}}x 10^8
$
Substituting different values of eccentricity into the equation gives corresponding values of 108 /P as shown in the table below
-300 2.4
-200 3.3
-100 4.2
0 5.1
200 7.0
400 8.8
600 10.6
800 12.4
1000 14.3
1200 16.1
1300 17.0
1400 17.9
The values in the table are plotted as shown below.
At Service
$
\frac{10^8}{P}\,\,\leqslant \,\,\frac{\left( -\frac{1}{A}\,\,+\,\,\frac{e}{Z_t} \right)}{\frac{\,\,M_{\max}}{\gamma _{inf\,\,}Z_t}+\frac{\eta \,\,f_{sc}}{\gamma _{inf}}}x 10^8
$
Substituting the values of parameters in the equation
$
\frac{10^8}{P}\,\,\leqslant \,\,\frac{\left( -\frac{1}{1500000}\,\,+\,\,\frac{e}{1273585859} \right)}{\frac{\,\,5875 x 10^6}{0.9 x {\,\,}1273585859}+\frac{\0.93 \,\,-24{}}{0.9{}}}x 10^8
$
Substituting different values of eccentricity into the equation gives corresponding values of 108 /P as shown in the table below
-300 4.2
-200 3.8
-100 3.5
0 3.1
200 2.3
400 1.6
600 0.9
800 0.2
1000 -0.6
1200 -1.3
1300 -1.6
1400 -2.0
The values in the table are plotted as shown below.
$
\frac{10^8}{P}\,\,\geqslant \,\,\frac{\left( -\frac{1}{A}\,\,-\,\,\frac{e}{Z_b} \right)}{\frac{\,-M_{\max}}{\gamma _{inf\,\,}Z_b}+\frac{\eta \,\,f_{st}}{\gamma_{inf}}}x 10^8
$
Substituting the values of parameters in the equation
$
\frac{10^8}{P}\,\,\geqslant \,\,\frac{\left( -\frac{1}{1500000}\,\,-\,\,\frac{e}{ 835000000} \right)}{\frac{\,-5875 x 10^6}{0.9 x {\,\,}835000000}+\frac{\0.93\,\,3.5{}}{0.9{}}}x 10^8
$
Substituting different values of eccentricity into the equation gives corresponding values of 108 /P as shown in the table below
-300 7.8
-200 10.9
-100 13.9
0 17.0
200 23.9
400 21.2
600 35.3
800 41.5
1000 47.6
1200 53.7
1300 56.7
1400 59.8
The values in the table are plotted as shown below.
Estimation of limiting eccentricity
The cover to the tendon is taken as 100mm.
Let’s take 110mm as possible allowance for multilayered tendons: 100 + 110 = 210
Maximum eccentricity possible = 1510 – 200 = 1300mm
This is plotted as shown below:
Estimation of the jacking prestress force
When these plots are combined together then a magnel diagram is produced with feasible region coloured as shown below:
From the magnel diagram 108 /P is taken as 55
P = (108 /55)/1000 = 1818KN
This is the prestress at service after long-term loss. To get the stress during jacking operation, the loss has to be accounted for.
Since the loss at service is 0.25
Pjack = 1818/0.25 = 2424KN
Detailing of tendons
Ultimate strength of tendon (fpk) = 1860MPa
Yield strength of tendon (fpk0.1) = 0.85fpk = 1581MPa
Design strength of tendon (fpd) = Fpk0.1/1.15 = 1374MPa
Capacity of tendon = fpd x As
= 1374 x 112 x = 154KN
No of tendons required = 2424/154 = 15.7
We shall take this as 16 tendons.
The 16 tendons shall be placed in the section as shown below.
From the section above, the centroid of the tendons to the soffit of the beam is 175mm. Consequently, the eccentricity from the center of the beam is 1335mm. This is greater than 1300 but appreciably less than 1410 which is the factual eccentricity allowing for 100mm cover. 110mm was only added as a precaution in anticipation against multilayered tendon.
On the other hand, since there are 16 tendons in the beam, the actual jacking force is:
16 x 154 = 2464KN
P = 2464KN e = 1335mm
Stress Check
Let’s run a stress check to scrutinize whether the stresses in the beam violate the permissible stresses.
Stress Calculation at transfer
P = 0.9 x Pjack = 0.9 x 2464 = 2217.6
e = 1335mm
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{2217.6}x\,\,10^3}{\text{15}x\,\,10^5}\,\,+\,\,\frac{\text{2217.6}x\,\,10^3\,\,x\,\,1335}{1273585859}\left. \right)1.1 \,\,-\,\,\frac{\text{1875}x\,\,10^6}{1273585859}
$
= (-1.48 + 2.32)1.1 – 1.47
= – 0.54MPa
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{2217.6}x\,\,10^3}{\text{15}x\,\,10^5}\,\,-\,\,\frac{\text{2217.6}x\,\,10^3\,\,x\,\,1335}{835000000}\left. \right)1.1 \,\,+\,\,\frac{\text{1875}x\,\,10^6}{835000000}
$
= (-1.48 – 3.54)1.1 + 2.2
= -3.28MPa
Stress Calculation at Service
P = 0.75 x Pjack = 0.75 x 2464 = 1848KN
e = 1335mm
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{ 1848}x\,\,10^3}{\text{15}x\,\,10^5}\,\,+\,\,\frac{\text{ 1848}x\,\,10^3\,\,x\,\,1335}{1273585859}\left. \right)0.9 \,\,-\,\,\frac{\text{5875}x\,\,10^6}{1273585859}
$
= (-1.2 + 1.9)0.9 – 4.6
= – 3.97MPa
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{ 1848}x\,\,10^3}{\text{15}x\,\,10^5}\,\,-\,\,\frac{\text{ 1848}x\,\,10^3\,\,x\,\,1335}{835000000}\left. \right)0.9 \,\,+\,\,\frac{\text{5875}x\,\,10^6}{835000000}
$
= (-1.2 – 2.95)0.9 + 7.0
= 3.26MPa
Conclusion
Applying prestressing force of 2464KN at eccentricity of 1335mm, the stress in the beam is within constraint at both transfer stage and serviceability stage. This is contrary to what is obtained in the stress verification here where the bottom fiber of the beam at service stage fails stress check when prestressing force of 1000KN is applied at eccentricity of 300mm.
You should also click here to study a worked example on the design of the beam for bending
