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1 Problem

A simply supported prestressed concrete beam with a span of 15 m is designed to carry a superimposed dead load of 2.0 kN/m and a service live load of 5.0 kN/m. The beam is made of normal-weight concrete with a density of 24 kN/m³. The concrete strength at transfer and service are \(f'_{ci}\) = 28 MPa and \(f'_c\) = 35 MPa. The beam is post-tensioned with a draped tendon. The eccentricity of the tendon is 250 mm below the concrete centroid at midspan, and it linearly drapes to an eccentricity of 0 mm at the supports (anchored exactly at the neutral axis). The initial prestress force immediately after transfer is \(P_i\) = 850 kN. Time-dependent losses total 18% of the initial prestress force. The gross concrete properties are as follows.

Answer the following problems.

  1. Calculate the top and bottom stresses at midspan at immediate transfer (in MPa).
  2. Calculate the top and bottom stresses at supports at immediate transfer (in MPa).
  3. Calculate the top and bottom stresses at midspan at service level (in MPa).
  4. Calculate the top and bottom stresses at supports at service level (in MPa).
  5. From your answers in #1 to #4, check whether each stress is within allowable stress values per NSCP 2015 provisions (in MPa).

You may want to try to solve the problems first before looking at the solution below.

2 Solution

Handwritten solution page 1 Handwritten solution page 2

Preliminary Calculations

Self-Weight and Initial Moment:
First, we determine the self-weight of the beam (\(\omega_{SW}\)) and the corresponding moment at transfer (\(M_o\)) at midspan:

\[\omega_{SW} = \gamma_c A_c = (24\text{ kN/m}^3)(120 \times 10^3 \times 10^{-6}\text{ m}^2) = 2.88\text{ kN/m}\] \[M_o = \frac{\omega_{SW} L^2}{8} = \frac{(2.88\text{ kN/m})(15\text{ m})^2}{8} = 81\text{ kN}\cdot\text{m}\]

Tendon Drape Angle:
Because the tendon is draped, we calculate the angle \(\theta\) of the tendon from the supports to the midspan:

\[\theta = \tan^{-1} \left( \frac{250\text{ mm}}{7.5 \times 10^3\text{ mm}} \right) = 1.9092^\circ\]

(Note: The horizontal component of the prestressing force, \(P \cos\theta\), will be used for axial stress calculations.)

Stage 1: Stresses at Initial Transfer

A. Midspan Stresses at Initial Transfer

At transfer, the active loads are the initial prestressing force (\(P_i\)) and the beam's self-weight (\(M_o\)).

Top Fiber Stress (\(f_t\)):

\[f_t = -\frac{P_i}{A_c} \left[ 1 - \frac{e c_t}{r^2} \right] - \frac{M_o}{S_t}\] \[f_t = -\frac{850 \cos(1.9092^\circ) \times 10^3}{120 \times 10^3} \left[ 1 - \frac{(250)(400)}{41.67 \times 10^3} \right] - \frac{81 \times 10^6}{12.5 \times 10^6}\] \[f_t = 3.4298\text{ MPa}\]

Check against allowable initial tension:

\[3.4298\text{ MPa} > 0.25\sqrt{f'_{ci}} = 0.25\sqrt{28} = 1.3229\text{ MPa} \quad \rightarrow \quad \textbf{NG (Not Good)}\]

Bottom Fiber Stress (\(f_b\)):

\[f_b = -\frac{P_i}{A_c} \left[ 1 + \frac{e c_b}{r^2} \right] + \frac{M_o}{S_b}\] \[f_b = -\frac{850 \cos(1.9092^\circ) \times 10^3}{120 \times 10^3} \left[ 1 + \frac{(250)(400)}{41.67 \times 10^3} \right] + \frac{81 \times 10^6}{12.5 \times 10^6}\] \[f_b = -17.5886\text{ MPa}\]

Check against allowable initial compression:

\[-17.5886\text{ MPa} < -0.6f'_{ci} = -0.6(28) = -16.8\text{ MPa} \quad \rightarrow \quad \textbf{NG (Not Good)}\]

(Note: A more negative value indicates a compressive stress magnitude exceeding the allowable limit.)

B. Stresses at Supports at Immediate Transfer

At the supports, eccentricity is zero (\(e = 0\)) and the moment from self-weight is zero (\(M_o = 0\)). Only the uniform axial compression from the prestressing force remains.

\[f_t = -\frac{P_i}{A_c} \left[ 1 - \frac{e c_t}{r^2} \right] - \frac{M_o}{S_t} = -\frac{P_i}{A_c}\] \[f_t = -\frac{850 \cos(1.9092^\circ) \times 10^3}{120 \times 10^3} = -7.0794\text{ MPa}\] \[f_b = -\frac{P_i}{A_c} = -7.0794\text{ MPa}\]

Check against allowable initial compression at supports:

\[-7.0794\text{ MPa} > -0.7f'_{ci} = -0.7(28) = -19.6\text{ MPa} \quad \rightarrow \quad \textbf{OK}\]

Stage 2: Stresses at Service

Before calculating service stresses, determine the effective prestress force (\(P_e\)) after \(18\%\) losses, and the total service moment (\(M_t\)).

\[P_e = (1 - \%_{\text{loss}}) P_i = (1 - 0.18)(850\text{ kN}) = 697\text{ kN}\] \[M_t = \frac{(\omega_{SW} + \omega_{SDL} + \omega_{LL}) L^2}{8} = \frac{(2.88 + 2.0 + 5.0)(15)^2}{8} = 277.875\text{ kN}\cdot\text{m}\]

C. Midspan Stresses at Service

Top Fiber Stress (\(f_t\)):

\[f_t = -\frac{P_e}{A_c} \left[ 1 - \frac{e c_t}{r^2} \right] - \frac{M_t}{S_t}\] \[f_t = -\frac{697 \cos(1.9092^\circ) \times 10^3}{120 \times 10^3} \left[ 1 - \frac{(250)(400)}{41.67 \times 10^3} \right] - \frac{277.875 \times 10^6}{12.5 \times 10^6}\] \[f_t = -14.1040\text{ MPa}\]

Check against allowable service compression:

\[-14.1040\text{ MPa} > -0.6f'_c = -0.6(35) = -21\text{ MPa} \quad \rightarrow \quad \textbf{OK}\]

Bottom Fiber Stress (\(f_b\)):

\[f_b = -\frac{P_e}{A_c} \left[ 1 + \frac{e c_b}{r^2} \right] + \frac{M_t}{S_b}\] \[f_b = -\frac{697 \cos(1.9092^\circ) \times 10^3}{120 \times 10^3} \left[ 1 + \frac{(250)(400)}{41.67 \times 10^3} \right] + \frac{277.875 \times 10^6}{12.5 \times 10^6}\] \[f_b = 2.4937\text{ MPa}\]

Check against allowable service tension:

\[2.4937\text{ MPa} < 0.5\sqrt{f'_c} = 0.5\sqrt{35} = 2.9580\text{ MPa} \quad \rightarrow \quad \textbf{OK}\]

D. Stresses at Supports at Service

Similar to the initial transfer stage at the supports, eccentricity and moment are zero (\(e = 0\), \(M_t = 0\)).

\[f_t = -\frac{P_e}{A_c} = -\frac{697 \cos(1.9092^\circ) \times 10^3}{120 \times 10^3} = -5.8051\text{ MPa}\] \[f_b = -\frac{P_e}{A_c} = -5.8051\text{ MPa}\]

Check against allowable service compression:

\[-5.8051\text{ MPa} > -0.6f'_c = -21\text{ MPa} \quad \rightarrow \quad \textbf{OK}\]