Question

A rectangular beam having b=16 in and d=26 in spans 28 ft face to face of simple supports. It is reinforced for flexure with 6#11 bars that continue uninterrupted to the ends of the span. It is to carry service dead load 2.0 kips/ft (including self weight) and service live load 3.6 kips/ft, both uniformly distributed along the span. Design the shear reinforcement using vertical U stirrups. Economize the spacing of stirrups in appropriate number of bands. Material strengths are fy=60,000 psi and fc′=4,000 psi.

Where, Data is

A: b=16 in. B: d=26 in. C: 28 ft. D: 6#11 bars
E: 2.0 kips/ft (including self weight) F: 3.6 kips/ft

A rectangular beam having (A) and (B) spans (C) face to face of simple supports. It is reinforced for flexure with (D) that continue uninterrupted to the ends of the span. It is to carry service dead load (E) and service live load (F), both uniformly distributed along the span. Design the shear reinforcement using vertical U stirrups. Economize the spacing of stirrups in appropriate number of bands. Material strengths are fy=60,000 psi and f'=4,000 psi.

Answer 1

Que) Design laad Wu 1• 2 Wo_+ 1:6 WE 1:2 (2) + 116_13.6) bu 8.16 kipstift To find design shear force Vu >> Wy = 8.16 kipstift LI A B 28ft H 11424 kips 114.24 kips HG.26 + SFD Ky a 114.24 Shear force at force at 9-distance from suppost A 114.24 -8.16 x distance Design shear force will be shear at 'd from supporti We have d = 26 in d = 2 in = 2.16) FZ. Vu = 114.24 8.16 x 2.167 - 96.56 kips. shear be taken by .: This force w Csscanned with CamScanner concrete and stirrups.

Shear force resisted by..concrete ove 2 Sfc b.d P for ;-Vez 0.75*2 J4000 x 16x26 Ø Ves 39.465 kips. få 4 ksi o= 0.75 As Design Shear force Vu is force Vu is greater than I ØVc 2 96.56 > 1 x 59.465 2 th Strups must be provided Snear resisted by Stirrups . Vu Us > -Ve a 39.465 AN 96.56 0.75 Vs = 89.28 kips. CScanned with CamScanner

& To find Spacing of stirrups Use #4, 2 legged U shaped stirrups Ava 2x0.2= 0.4 in 2 Spacing Ar. fy.d Vs 0.4 x 60 x 26 89.28 Sa 6.989 in to Check for spacing from ACI code Smax min d 24" 4"] min / 21 24" 2 Smax 13.5" Also, 6 = min Av. fy 0.6260.000 0.4 X 60000 50. b 0.75 Jfcb 0.751 *J4000, *16 50 x 15 Smax : min 31.62 , 303 30" CSscanned with CamScanner

S = min {6.98",13.5",30"}

S = 6.98"

Therefore,f resisting shear force provide stirrups of #4 @6.5"