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**Q.1 A 6 m long simply-supported beam is prestressed as shown in the figure.The beam carries a uniformly distributed load of 6 kN/m over its entire span. If the effective flexural rigidity EI = 2 x 10**

Span of PSC beam = 6 m

Total UDL = 6 kN/m

eccentricity = e = 50 mm

(a) Slope of beam due to P-force

(b) Slope of beam due to UDL

(c) Net slope of beam

Solution:

Note: Tendon profile shall follow the shape of bending moment diagram.

âˆ´

e = 150 mm

e =0.15 m

Solution:

Prestressing force,

Stress in concrete at the location of steel

Loss of stress

= m x f

Loss of prestressing force

(a) 4.166 and 20.833

(b) -4.166 and -20.833

(c) 4.166 and -20.833

(d) -4.166 and 20.833

Ans.

Given, net loss = 10%

Final prestressing force after loss = 750 kN

Hence, initial prestressing force

Initial stresses at top and bottom

Stress at top = 8.33 - 12.50 = - 4.17 (tensile)

Stress at bottom = 8.33 + 12.50 = 20.83 (Comp.)

Solution:

Since the tensile stress at bottom face of the beam is zero.

Since the prestressing force is located at 200 mm from the bottom face of the beam.

âˆ´ Eccentricity = 300 - 200 = 100 mm

The stress (in N/mm

(a) tensile 2.90

(b) compressive 2.90

(c) tensile 4.32

(d) compressive 4.32

Ans.

Total stress at Bottom =

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