Design for Shear for Prestressed Concrete Beam

Design for Shear for Prestressed Concrete Beam

Introduction The behaviour of prestressed beams at failure in shear is distinctly different from their behaviour in flexure. The beam will tend to fail abruptly without sufficient warning and the diagonal cracks that develop are considerably wider than the flexural cracks. Shear forces result in shear stress. Such a stress can result in principal tensile stresses at the critical section which can exceed the tensile strength of the concrete When the tensile strength of the concrete is exceeded cracks will formed.

Cracking Patterns & Failure Modes Cracking in prestressed concrete beams at ultimate load depends on the local magnitudes of moment and shear. In regions where the moment is large and shear is small, vertical flexural cracks appear after the normal tensile stress in the extreme concrete fibres exceeds the tensile strength of concrete. This type of cracks shown as type A in figure. Where both the moment and shear force are relatively large, flexural cracks which are vertical at the extreme fibres become inclined as they extend deeper into the beam owing to the presence of shear stresses in the beam web. These inclined cracks, which are often quite flat in a prestressed beam are called flexure-shear cracks and are designated crack type B.

Cracking Patterns & Failure Modes If adequate shear reinforcement is not provided, a flexure-shear crack may lead to a so-called shear-compression failure, in which the area of concrete in compression above the advancing inclined crack is so reduced as to be no longer adequate to carry compression force resulting from flexure. Another type of inclined crack sometimes occurs in the web of a prestressed beam in the regions where moment is small and shear is large, such as the cracks designated type C adjacent to discontinuous support and near the point of contraflexure in the figure. In such locatio, high principal tensile stress may cause inclined cracking in the mid-depth region of the beam before flexural cracking occurs in the extreme fibres. These cracks are known as web-shear cracks and occur most often in beams with relatively thin webs.

Cracking pattern

Effect of Prestressing in Shear The longitudinal compression introduced by prestress delays the formation of each of the crack types shown previously. The effect of prestress on the formation and direction of inclined cracks can be seen by examining the stresses acting on a small selement located at the centroidal axis of the uncracked beam as shown in figure of next page.

Effect of Prestressing in Shear Using a simple Mohr s circle construction, the principal stresses and their directions are well establish. When the principal tensile stress 1 reaches the tensile strength of concrete, cracking occurs and the crakcs form in the direction perpendicular to the direction of 1. When the prestress is zero, 1 is equal to shear stress and acts at 45 o to the beam axis. If diagonal cracking occurs, it will be perpendicular to the principal tensile stress.

Effect of Prestressing in Shear When the prestress is not zero, the normal compressive stress ( = P/A) reduces the principal tension 1. the angle between the principal stress direction and the beam axis increases and consequently if cracking occurs, the inclined crack is flatter. Prestress tehrefore improves the effectiveness of any transverse reinforcement (strirrups) that may be used to increase the shear strength of a beam. With prestress causing the inclined crack to be flatter, a larger number of the vertical stirrup legs are crossed by the crack and consequently a larger tensile force can be carried across the crack.

Shear Analysis Uncracked Vco Cracked Vcr

Uncracked Sections Small element A at the centroidal of a simply supported prestressed concrete member subjected to compressive stress, f cp and a shear stress, f s. and from Mohr circle, The allowable principal tensile stress is given in BS8110 as For a rectangular section

Uncracked Sections Where Vco = design ultimate shear resistance of a section uncracked in flexure f cp = design compressive stress at centroidal axis due to prestress = Pe/A f t = maximum design principle tensile stress, b v = breadth of the member or for T, I and L beams used width of the web If grouted duct is present in the web b v = b w 0.67d d (d d = diameter of duct)

Cracked Section Clause 4.3.8.5 BS8110 gives the following empirical equation for the ultimate shear resistance of a section cracked in flexure: Mo = moment which produces zero stress at extreme tension fibre f pt = level of prestress in concrete at the tensile face the value of V cr should be taken as less than 0.1b v d f cu

Cracked Section or using Table 3.8 where m = 1.25 Should not be taken as greater than 3 Should not be taken as less than 0.67 for members without shear reinforcement Should not be taken as less than 1 for members with shear reinforcement providing a design shear resistance of 0.4 N/mm2 For characteristic concrete strengths greater than 25N/mm2, the values in Table 3.8 may be multiplied by (f cu /25) 1/3. The value of f cu should not be taken as greater than 40.

Design Ultimate Shear Resistance, Vc According to Clause 4.3.8.3 BS8110 Vc = Vco at uncracked section (M < Mo) Vc = is the smaller of Vco and Vcr at cracked section (M Mo) For deflected tendon, the vertical component of the prstressing will help to resist the shear force. The total shear resistance then becomes : Vc + P e sin where is the angle of inclination of the prestressing tendon For parabolic profile, e(x) = (-4 /L 2 )x 2 + (4 /L)x (x) = (-8d/L2)x + (4 /L) in radian where = abs(e s e ms )

Parabolic Profile

Shear Design Steps 1. Draw the shear force and bending moment diagram 2. Check maximum allowable shear stress (Clause 4.3.8.2) v = V/b v d 0.8 f cu or 5N/mm2 3. Plot Mo on the BMD. Mo will varies along the span if the tendon profile is not straight. Determine the cracked (M Mo) and uncracked (M < Mo) 4. Calculate Vco and Vcr 5. Determine the ultimate shear resistance of the prestressed beam as follows : cracked region : Vc = V cr or V co + P e sin uncracked region : Vc = V co + P e sin

Shear Design Steps 6. If V 0.5Vc, shear reinforcement is not required (Clause 4.3.8.6) 7. If 0.5Vc < V Vc + 0.4bvd, nominal shear reinforcement is required. Refer Cl. 4.3.8.7, use 8. If V > Vc + 0.4b v d, shear reinforcement is required (Cl. 4.3.8.8). Use d t is the depth from the extreme compression fibre either to the longitudinal bars of to the centroid of the tendons, whichever is greater. 9. Spacing of shear reinforcement, Sv the lesser of 0.75d t or 4 x web thickness, but when V > 1.8Vc, the max spacing should be reduced to 0.5d t. 10. The lateral spacing of the individual legs of the links provided at a cross-section should not exceed d t.

Example 1 A prestressed concrete T-beam shown in figure is simply supported over a span of 28m, has been designed to carry in addition to its own weight, a characteristic dead load of 4kN/m and a characteristic imposed load of 10kN/m. The beam is pretensioned with 14 nos 15.7mm diameter 7-wire super strands (A ps = 150mm2) but due to debonding only 7 of the strands are active at a section 2m from support. The effective prestressing force, Pe for these 7 strands is 1044kN. Use the following data: f cu = 50 N/mm 2, A = 5.08 x 10 5 mm 2, I = 134x10 9 mm 4, y 2 = 912mm, f pu = 1770 N/mm 2, fyv = 250 N/mm 2, Design the section for shear.

Solution 1. Loads Selfweight, W sw = 24 kn/m 3 x 0.508 m 2 = 12.19 kn/m Ultimate load, W = 1.4(12.19 + 4) +1.6(10) = 38.67 kn/m 2. Shear and moment at 2m from support V (x) = W(0.5L-x) V(2) = 38.67 (0.5x28 2) = 464 kn M(x) = 0.5W(Lx x 2 ) M(2) = 0.5x38.67 (28x2 2 2 ) = 1005 knm

Solution Calculate Mo e = 814mm, d = 1500-98 = 1402mm < M = 1005 knm Section is cracked in flexure

Solution Calculation of Vco

Solution Calculation of Vcr Ok Use 400/d = 1 f cu = 50 N/mm 2 > 40 N/mm 2 ; use f cu = 40 N/mm2 520 kn And Vcr > 0.1b v d f cu = 174 kn ok

Solution Shear resistance provided by the concrete, Vc Vc = smaller between Vco (420 kn) and Vcr (520 kn) Vc = 420 kn Design of shear reinforcement V = 464 kn, 0.5Vc = 210kN, Vc +0.4b v d = 518 kn Where 0.5Vc < V < Vc +0.4b v d Only nominal links required Using 10mm diameter stirrup/links, Asv = 157mm2 Use R10-450mm c/c < 0.75dt = 0.75(1500-98) = 1051.5mm

Example 2 The beam shown below supports an ultimate load, including selfweight of 85kN/m over a span of 15m and has a final prestress force of 2000kN. Determine the shear reinforcement required. Use the following data: fcu = 40N/mm2, A = 2.9 x 10 5 mm 2, I = 3.54 x 10 10 mm4, A ps = 2010 mm 2, f pe /f pu = 0.6

Solution Draw BMD and SFD M(x) = 0.5w(Lx-x 2 ) & V(x) = w(0.5l x) Where w = 85 kn/m Check maximum allowable shear stress (Cl. 4.3.8.2) V = 637.5 kn < 0.8 fcu or 5 N/mm2 OK

Solution Plot Mo on BMD Mo vary along the length of beam since the tendon profile is parabolic which produce the different eccentricity along the length of beam. Eccentricities of tendon along parabolic tendon profile given by: Where = 425mm

Solution Distribution of M and Mo along beam span

Variation of V along beam span

Solution Nominal Shear Reinforcement Use R8, < 0.75dt =0.75(908) = 681mm Shear reinforcement Use R8 350 mm V-Vc = 107.12 kn < 0.75dt =0.75(772) = 579mm Use R8 150 mm