What are the functions of grout inside tendon ducts?

What are the functions of grout inside tendon ducts?
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1 Answer

Answer :

Grout in prestressing works serves the following purposes:
(i) Protect the tendon against corrosion.
(ii) Improve the ultimate capacity of tendon.
(iii) Provide a bond between the structural member and the tendon.
(iv) In case of failure, the anchorage is not subject to all strain energy.
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is the pre-stressed force applied to the tendon of a rectangular pre-stressed beam whose area of cross section is and sectional modulus is . The maximum stress in the beam, subjected to a maximum bending moment , is (A) f = (P/A) + (Z/M) (B) f = (A/P) + (M/Z) (C) f = (P/A) + (M/Z) (D) f = (P/A) + (M/6Z)

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If the maximum dip of a parabolic tendon carrying tension P is h and the effective length of the pre-stressed beam is L, the upward uniform pressure will be (A) 8hp/l (B) 8hp/l² (C) 8hl/p (D) 8hl/p²

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Last Answer : Answer: Option B

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If the tendon is placed at an eccentricity e below the centroidal axis of the longitudinal axis of a rectangular beam (sectional modulus Z and stressed load P in tendon) the stress at the extreme top edge (A) Is increased by PZ/e (B) Is increased by Pe/Z (C) Is decreased by Pe/Z (D) Remains unchanged

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A pre-stressed rectangular beam which carries two concentrated loads W at L/3 from either end, is provided with a bent tendon with tension P such that central one-third portion of the tendon remains parallel to the longitudinal axis, the maximum dip h is (A) WL/P (B) WL/2P (C) WL/3P (D) WL/4P

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If is the sectional area of a pre-stressed rectangular beam provided with a tendon pre-stressed by a force through its centroidal longitudinal axis, the compressive stress in concrete, is (A) P/A (B) A/P (C) P/2A (D) 2A/P

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is the pre-stressed force applied to tendon of a rectangular pre-stressed beam whose area of cross section is and sectional modulus is . The minimum stress on the beam subjected to a maximum bending moment is (A) f = (P/A) - (Z/M) (B) f = (A/P) - (M/Z) (C) f = (P/A) - (M/Z) (D) f = (P/A) - (M/6Z)

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