Question

A 250 mm wide × 600 mm deep rectangular concrete beam is prestressed by means of 4 high-tensile tendons, each of 14 mm diameter. The centre of the tendons is 200 mm from the soffit of the beam. The effective stress in each tendon is 700 N/mm².
The maximum bending moment (in kN-m), that can be applied to the section without causing tension at the soffit of the beam due to prestressing only, is ______ (rounded off to one decimal place).
Use \(\pi=3.14\)

Hint:

Stress at the soffit from prestress alone is \(P/A+Pe/Z\); the tendon here sits exactly at the kern point (\(e=D/6\)), so this simplifies to \(2P/A\).

Answer 1

Correct Answer: 86.2

Step 1: Find the prestressing force.
Each tendon has a diameter of 14 mm, so the cross-sectional area of one tendon is
\[ a_p=\frac{\pi}{4}d^2=\frac{3.14}{4}\times14^2=\frac{3.14}{4}\times196=153.86\text{ mm}^2 \] There are 4 identical tendons, so the total tendon area is
\[ A_p=4\times153.86=615.44\text{ mm}^2 \] The effective (after losses) stress in each tendon is 700 N/mm\(^2\), so the total prestressing force is
\[ P=A_p\times f_{pe}=615.44\times700=430808\text{ N}=430.808\text{ kN} \]
Step 2: Find the section properties of the beam.
The beam is 250 mm wide and 600 mm deep, so its cross-sectional area is
\[ A=b\times D=250\times600=150000\text{ mm}^2 \] For a rectangular section the centroid lies at mid-depth, 300 mm from either face, and the section modulus (same value for top and bottom fibre) is
\[ Z=\frac{bD^2}{6}=\frac{250\times600^2}{6}=\frac{90000000}{6}=15000000\text{ mm}^3 \]
Step 3: Find the eccentricity of the tendons.
The tendons are centred 200 mm from the soffit, while the centroid of the section is at 300 mm from the soffit (mid-depth).
So the tendons sit below the centroid by
\[ e=300-200=100\text{ mm} \]
Step 4: Write the stress at the soffit due to prestress alone.
An eccentric prestressing force produces two effects on any fibre: a uniform direct compressive stress \(P/A\), plus a bending stress \(Pe/Z\) from the eccentric moment \(P\times e\).
Because the tendon is below the centroid, the eccentric moment adds MORE compression at the soffit (the near fibre) and REDUCES compression at the top (the far fibre).
So the stress at the soffit due to prestress alone is
\[ f_{soffit}=\frac{P}{A}+\frac{Pe}{Z} \] \[ \frac{P}{A}=\frac{430808}{150000}=2.872\text{ N/mm}^2,\qquad \frac{Pe}{Z}=\frac{430808\times100}{15000000}=2.872\text{ N/mm}^2 \] \[ f_{soffit}=2.872+2.872=5.744\text{ N/mm}^2 \text{ (compressive)} \]
Step 5: Find the extra moment that just cancels this compression at the soffit.
Now imagine applying an external sagging bending moment \(M\) on this simply supported beam (from self-weight or service load). A sagging moment always puts the soffit fibre into TENSION and the top fibre into extra compression, which is opposite to what the prestress does at the soffit.
The stress this moment adds at the soffit is \(M/Z\) (tensile).
The soffit will just reach zero stress (the largest moment we can apply "without causing tension") when this tensile stress exactly cancels the compressive stress already there from prestress:
\[ \frac{M}{Z}=f_{soffit} \] \[ M=Z\times f_{soffit}=15000000\times5.744=86160000\text{ N-mm} \]
Step 6: Convert to kN-m.
\[ M=\frac{86160000}{10^6}=86.16\text{ kN-m}\approx86.2\text{ kN-m (rounded to one decimal place)} \]
Final Answer:
The maximum moment that can be applied without causing tension at the soffit is about 86.2 kN-m. \[ \boxed{M\approx86.2\text{ kN-m}} \]