Lecture_14 - 3.2 Shear Strength of Beams with Shear...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3.2 Shear Strength of Beams with Shear Reinforcement • The complementary shear stress gives rise to diagonal tensile and diagonal compressive stresses. • Taking a simplified view, concrete is weak in tension, and shear failure is then caused by a failure in diagonal tension with cracks running at about 45°. • Shear reinforcement should be provided by bars that cross the cracks. Theoretically, either vertical links (stirrups) or inclined bars will serve this purpose. 3.2.1 Types of shear reinforcement in beams (1) Links or vertical stirrups (2) Bend-up bars (inclined bars) (3) Combination system (links + bend-up bars) 115 The shear force V is resisted by shear in concrete and shear in steel: V = Vc + Vs (3.5) where the shear resistance of the concrete Vc = vcbv d (3.6) 3.2.2 Shear resistance with links (stirrups) − The 45 o truss model • The design and analysis for shear are based on the concept of an analogous truss. In the truss shown, d tension d ~ 45º Links T b (stirrups) compression (compression strut) V Fig. 3.2-1 The 45° truss analogy o The links are spaced at a distance equal to d, so that the diagonal concrete compression members are at an angle of about 45°, which generally agrees with the experimental observations of the cracking of reinforced concrete beams close to the supports. o The web bars are considered to form the tension members of the imaginary truss, while the trusts in the concrete constitute the compression members. 116 o The shear resistance of the vertical links (stirrups) is Vs = 0.95 f yv Asv (3.7) where Asv is the cross-sectional area of the stirrups. o When the stirrup spacing is less than d, the number of stirrups crossing a 45°−crack is about d / s v , where sv is the stirrup spacing. Fig. 3.2-2 Stirrups crossing a 45° diagonal crack o The shear resistance of the stirrups is Vs = 0.95 f yv Asv ⋅ d ⎛ Asv ⎞ =⎜ ⎟ 0.95 f yv d sv ⎝ sv ⎠ (3.8) • The ultimate shear strength (shear resistance) of the cross-section with vertical stirrups is V = Vc + Vs ⎛A ⎞ = vcbv d + ⎜ sv ⎟ 0.95 f yv d ⎝ sv ⎠ concrete resistance (3.9) web reinforcement resistance 117 Q The design shear stress is v= V bv d The shear resistance of the cross-section presented by Eq. (3.9) can then be written as (v − vc )bv d = ( Asv / sv ) 0.95 f yv d (3.10) o Eq. (3.10) is expressed in the BS 8110’s form, Asv (v − vc )bv = sv 0.95 f yv (3.11) ⎛A ⎞ where ⎜ sv ⎟ can be considered as a relative web reinforcement ⎝ sv ⎠ ratio. o The actual stirrup ratio is defined by ρv = Asv sv bv (3.12) Total area = Asv bv sv /2 sv sv /2 • Stirrup sizes are usually from 8-mm to 16-mm diameter, and they are often of mild steel to minimise the radii of bends. o Minimum links should be provided as (Clause ⎛ Asv ⎞ 0.4bv ⎜ ⎟= ⎝ sv ⎠ min 0.95 f yv (3.13) 118 By comparing Eqs (3.11) and (3.13), it can be seen that (v – vc) = 0.4 N/mm2, indicating minimum links should at least provide a design shear resistance of 0.4 N/mm2 (i.e. 0.4 N/mm2 is the minimum design shear resistance). Reason: where the web reinforcement is used, it remains unstressed until diagonal cracking occurs. If the amount of web steel is too small, the sudden stress increase may cause the instant yielding of the web bars. o Maximum shear stress (Clause The design shear stress, v = V/(bv d), at any cross-section must in no case exceed 0.8√fcu or 5 N / mm 2 , whichever is the lesser, whatever shear reinforcement is provided (This limit has included an allowance for γm = 1.25), i.e. v= V ≤ 0.8 f cu or 5 N/mm 2 bv d (3.14) whichever is the lesser. This upper limit prevents crashing of the concrete in the direction of the maximum principal compressive stresses (in diagonal compression in general). When the design shear stress v = V is exceeded, the crossbv d section sizes should be increased. 119 o The maximum shear stress presented in Clause of the Hong Kong code (Buildings Department, Code of Practice for Structural Use of Concrete, 2004, HKSAR): v= V ≤ 0.8 f cu or 7 N/mm 2 bv d whichever is the lesser, whatever shear reinforcement is provided (this limit has included an allowance for γm = 1.25). o Maximum spacing of links (Clause In the direction of span, sv,max = 0.75d. Thus, sv ≤ 0.75d (3.15) This condition ensures that at least one stirrup intercepts one diagonal crack. • Form and area of shear reinforcement (BS 8110: Part 1, Table 3.7) 120 3.2.3 Enhanced shear strength of sections close to supports (Clause • Shear failure at sections of beams without shear reinforcement will normally occur on a plane inclined at an angle of about 30° to the horizontal. If the angle of failure plane is forced to be inclined more steeply than this (due to the load or because the section where the shear is to be checked, such as X-X in Fig. 3.1-9, is close to a support), the shear forced required to produce failure is increased, i.e. the shear capacity of these sections is increased. NOTE: The shear causing failure is that acting on section X-X Fig. 3.1-9 Shear failure near supports • This enhancement of shear strength may be taken into account in the design of sections within a distance of 2d from the face of a support by increasing the design concrete shear stress, from vc to (2d/av)⋅vc, provided that the shear stress v at the face of the support remains the lesser of 0.8√fcu or 5 N/mm2. (The enhanced shear resistance is partly because the concrete in diagonal compression resists shear. In this case, the shear span ratio is small.) 121 • Simplified approach (Clause o Take the critical section for design at a distance d form the face of the support using the code specified value vc . d d Maximum shear stress at face of support Critical section for simplified approach Fig. 3.1-10 Critical section for design o Calculate the shear reinforcement required at this critical section; then provide this reinforcement at all sections up to the face of the support. In other words, the shear stress at the critical section is used for design to replace the maximum shear stress at the face of the support. 122 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online