Section
Total ultimate load on beam = 200 x 8.0 = 1600kN
Support reaction = 1600/2 800 kN
Shear, Vtl at face of support = 800 - 200 x 0.3/2 = 770 kN
Shear. Vu distance ct from face of support = 770 200 x 0.65 = 640 kN
1. Check the crushing strength VRd maxof the concrete diagonal strut at the face of the beams support.
From equation 5.6 with 0 = 22
0.124 x 350 x 650(1 - 30/250)30 = 745 kN (<Vnf = 770kN)
From equation 5.7 with 0 45
= 0.18 x 350 x 650( 1 30/250)30 = 1081 kN (>VEf=770kN)
2. Determine angle 0 From equation 5.8(a)
or alternatively from equation 5.8(b)
3. Determine shear resistance of the links
The cross-sectional area /lsw of a 12mm bar = 113 mm2. Thus for the two legs of the link and a spacing of 175 mm
(or alternatively the value could have been obtained from table A4 in the Appendix)
From equation 5.11 the shear resistance, VRJ s of the links is given by
Therefore shear resistance of links 781 kN.
Design shear, VH<i distance d from the face of the support = 640 kN (< 781 kN). Therefore, the beam can support, in shear, the ultimate load of 200 kN/m.
4. Additional longitudinal (ensile force in the tension steel
It is necessary to check that the bottom tension steel has a sufficient length of curtailment and anchorage to resist the additional horizontal tension Ah\t caused by the design shear. These additional tension forces are calculated from equation 5.12. Therefore
This force is added to the MEd/«" diagram, as described in section 7.9. to ensure there is sufficient curtailment of the tension reinforcement and iis anchorage bond ^_length at the supports, as described in section 5.2._j
To resist shearing forces, longitudinal tension bars may be bent up near to the suppon-as shown in figure 5.5. The bent-up bars and the concrete in compression are considers to act as an analogous lattice girder and the shear resistance of the bars is determined b> taking a section X-X through the girder.
(a) Single System x | Anchorage length
(a) Single System
(b) Multiple System
Figure 5.5
Bent up bars
(b) Multiple System
From the geometry of part (a) of figure 5.5. the spacing of the bent-up bars is:
and at the section X-X the shear resistance of a single bent-up bar (Vwd) must equal the shear force (VV.a)-
= Vfci = /j.wd<4Sw sin a = A,w sin a = 0.87/ykAsw sin a where is the cross-sectional area of the bent-up bar.
For a multiple system of bent-up bars, as in part (b) of figure 5.5. the shear resistance is increased proportionately to the spacing, .v. Hence:
0,, , . 0.9i/(cota + cotfl) VEd - 0.87/ykAsw sin a x-1
Vfct
This equation is analogous to equation (5.9) for the shear resistance of shear links. In a similar way it can be shown that, based on crushing of the concrete in the compressive struts, the analogous equation to (5.4) is given by:
and the additional tensile force to be provided by the provision of additional tension steel is given by a modified version of equation 5.12:
EC2 also requires that the maximum longitudinal spacing of bent-up bars is limited to 0.6i/(l +cola) and specifies that at least 50 per cent of the required shear reinforcement should be in the form of shear links.
5.1.4 Shear between the web and flange of a flanged section
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