## Curtailment

a) End span, bottom reinforcement

Assuming end support to be simply supported, 50% of As should extend into the support. <9.3.1.2(1) >

Try H12 @ 300 (376 mm2 / m) at supports In theory, 50% curtailment of reinforcement may take place al <9.3.1.2(1) Note, 9.2.1.3 (2)>

from where the moment of resistance of the section with the remaining 50% would be adequate to resist the applied bending moment. In practice, it is usual to determine the curtailment distance as being al from where MEd = MEdmax / 2. Thus: for a single simply supported span supporting a UDL of n,

M^ax = 0.086n/2; Ra = 0.4n/ At distance, X, from end support, moment, MeĀ«@x = RaX - nX2 / 2

0.043n/2 = 0.4n/xl - nx2/2 / 2 0.043 = 0.4x - x2 / 2 0 = 0.043 - 0.4x + x2 / 2

** Detail MS2 of SMDSC[21], suggests 50% of T1 legs of U-bars should extend 0.3I (= say 1800 mm) from face of support by placing U-bars alternately reversed.

ft Detail MS2 of SMDSC[21], suggests 50% of T1 legs of U-bars should extend 0.3I (= say 1800 mm) from face of support by placing U-bars alternately reversed.

.'. at end support 50% moment occurs at 0.13 x span 0.13 x 5975 = 777 mm

Shift rule: for slabs al may be taken as d ( = 144 mm)

.'. curtail to 50% of required reinforcement at 777 - 144 = 633 mm from centreline of support

Say 500 mm from face of support A

.'. in end span at 1st internal support 50% moment occurs at 0.66 x span

Shift rule: for slabs al may be taken as d (= 144 mm) .'. curtail to 50% of required reinforcement at 3944 + 144 = 4088 mm from support A or 5975 - 4088 = 987 mm from centreline of support B Say 850 mm from face of support B

Figure 3.5

Curtailment of bottom reinforcement: actions, bending moments, forces in reinforcement and curtailment

Figure 3.5

Curtailment of bottom reinforcement: actions, bending moments, forces in reinforcement and curtailment b) 1st interior support, top reinforcement

Presuming 50% curtailment of reinforcement is required this may take place al from where the moment of resistance of the section with the remaining 50% would be adequate. However, it is usual to determine the curtailment distance as being al from where MEd = MEdmax / 2. Thus, for the 1st interior support supporting a UDL of n, MEd.maxT = 0.086n/2; RB = 0.6n/ At distance Y from end support, moment, [email protected] = MEd.maXT - RaY + nY2 / 2

0.086n/2 / 2 = 0.086n/2 - 0.6n/Y + nY2 / 2 Assuming Y = yl

0.043n/2 = 0.086n/2 - 0.6n/y/ + ny212 / 2 0 = 0.043 - 0.6y + y2 / 2

.'. at end support 50% moment occurs at 0.08 x span 0.08 x 5975 = 478 mm

Shift rule: for slabs, al may be taken as d 144 mm

.'. curtail to 50% of required reinforcement at478 + 144 = 622 mm from centreline of support 50% of reinforcement may be curtailed at, say, 600 mm from either face of support B

100% curtailment may take place al from where there is no hogging moment. Thus:

0 = 0.086 - 0.6y + y2 / 2 y = 0.166 (or 1.034), say 0.17

.'. at end support 50% moment occurs at 0.17 x span 0.17 x 5975 = 1016 mm Shift rule: for slabs al may be taken as d

.'. curtail to 100% of required reinforcement at 1016 + 144 = 1160 mm from centreline of support

100% of reinforcement may be curtailed at, say 1100 mm from either face of support B

c) Support B bottom steel at support

At the support 25% of span steel required <9.3.1.1(4), 9.2.1.5(1), 9.2.1.4(1)> 0.25 x 639 = 160 mm2

Ae,min as before = 216 mm2 / m <9.3.1.1, 9.2.1.1>

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