Prestressed Concrete Examples

Concrete tension is positive and compression is negative. Tendon eccentricity ee is positive below the centroid, and external moment MM is positive for sagging. Each example identifies whether it uses transfer force PiP_i or effective service force PeP_e.

Prestress stress superposition

Concept and model scope

Tension is positive, compression is negative, e>0e>0 is below the centroid, and M>0M>0 is sagging.

The section is a fixed 300 mm × 600 mm rectangle. The active prestress is PiP_i at transfer and the loss-reduced force at service. Fiber stress is assembled from direct compression, eccentric prestress, and external bending.

This is a linear-elastic stress-superposition model; it does not perform transfer/service allowable-stress code checks, loss prediction, cracking redistribution, or ultimate resistance design.

Controls

Active effective force

960 kN

Pe = Pi(1 - supplied lumped long-term loss).

Top fiber

-8.33 MPa

compression

Bottom fiber

-2.33 MPa

compression

Both extreme fibers are compressed for this case. This one load state does not prove the member will remain uncracked at every construction, service, or overload condition.
Rectangular section and stress blocks for direct prestress, eccentric prestress, external moment, and total stressSection-P/A-5.3-5.3P·e/S+5.3-5.3M/S-8.3+8.3Total-8.3-2.3++=Compression plots left; tension plots right. Values are MPa.
Model scope: fixed rectangular 300 mm × 600 mm gross, uncracked elastic section; uniform tendon force at the section; user-supplied lumped long-term loss from PiP_i to PeP_e. The model omits friction variation along the member, anchorage zones, cracked-section properties, secondary moments, deflection, shear, and ultimate strength.

Example 1: Complete Transfer-Stage Fiber Stress Check

A 300 mm×600 mm300\text{ mm}\times600\text{ mm} rectangular beam has Pi=1200 kNP_i=1200\text{ kN} at e=+100 mme=+100\text{ mm}. At transfer, self-weight produces Mi=30 kN-mM_i=30\text{ kN-m}. Calculate both extreme-fiber stresses on the gross, uncracked section.

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Example 2: Effective Service Stress After Losses

For the beam in Example 1, take Pe=0.80Pi=960 kNP_e=0.80P_i=960\text{ kN} and a total sagging service moment Mser=150 kN-mM_{\mathrm{ser}}=150\text{ kN-m}. Calculate both fiber stresses.

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Example 3: A Tensile Service Stress Is a Classification Check

Use the same section, Pe=960 kNP_e=960\text{ kN}, and e=+100 mme=+100\text{ mm}, but increase the service moment to 250 kN-m250\text{ kN-m}. Determine the bottom stress and interpret it.

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Example 4: Eccentricity Sign Needed for Zero Bottom Stress Under Prestress Alone

A 300 mm×600 mm300\text{ mm}\times600\text{ mm} beam carries only a prestress force P=1500 kNP=1500\text{ kN}. Find the eccentricity that makes the bottom-fiber stress zero.

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Example 5: System-Specific Post-Tensioning Loss Sequence

A post-tensioned tendon is jacked to Pj=2000 kNP_j=2000\text{ kN}. At the section of interest, μα+kx=0.080\mu\alpha+kx=0.080. Anchorage seating is represented by a further 40 kN40\text{ kN} force reduction there. Subsequent creep, shrinkage, and relaxation reduce the transfer force by a combined 15%15\%. Find PiP_i and PeP_e under these explicitly supplied simplifications.

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Example 6: Elastic Shortening of a Pre-tensioned Tendon

At transfer, the concrete stress at the tendon level is fcgp=8.0 MPaf_{cgp}=8.0\text{ MPa}. Use Ec=30,000 MPaE_c=30{,}000\text{ MPa} and Ep=200,000 MPaE_p=200{,}000\text{ MPa} to estimate the compatible tendon stress loss.

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Example 7: Load Balancing with Overbalance

A simply supported 12 m12\text{ m} beam has a symmetric parabolic tendon with zero end eccentricity, midspan drape h=0.250 mh=0.250\text{ m}, and constant effective force Pe=1800 kNP_e=1800\text{ kN}. The uniform dead load is 10 kN/m10\text{ kN/m}.

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Example 8: Secondary Moment and the Meaning of Concordance

For a symmetric two-span beam with L=10 mL=10\text{ m} per span, prestressing produces a downward secondary reaction of 50 kN50\text{ kN} at the interior support. The balancing end reactions are 25 kN25\text{ kN} upward at each exterior support. Find the secondary moment at the interior support.

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