Question

A rectangular reinforced concrete beam of span 16 ft supports a concentrated load of 40 kips at mid-point of the beam and a uniformly-distributed load of 4 kips/ft over the entire span. Given: (a) The breadth of the beam is 18 inches. (b) Concrete compressive strength, f 'c, is 3500 psi. (c) Rebar is Grade 40 steel

i) Determine the maximum moment and maximum shear acting on the beam.

ii) Determine the minimum effective depth 'd' of the beam section (rounded to the nearest integer).

iii) Specify the gauge number of the longitudinal tensile reinforcement. Use six rebars.

iv) Determine if the development length of the tensile reinforcement is satisfactory.

v) Determine the maximum spacing of web (shear) reinforcement. Use No. 4 stirrups.

vi) Determine the required minimum extent of shear reinforcement into the beam.

vii) Sketch a longitudinal section and a transverse section of the beam, Show beam size, effective and actual depths, tensile reinforcement and location, and stirrup location and spacing.

Answer 1

Given data I= 16ft, Plauorips , w ukielft., b18" fc': 3500psi fy =40,000psi. Let assume the the type of load is not given given load is i whether load the ultimate load since it is dead load or live GOR 3 ) lucifer mo Ra = R$ = [(ux16)740 B А. 8' C -* 81 2. 16' RA - RB 52k maximum shear force occurs at supports. Vumax RA = RB 521 moment occurs at Bending midspan , since maximum loading the symmetrical Mimox WIL hall 8. * WOX16 ū ux162 8 288. - A6&-Ft. f U uole lclft poema mo som 52k. 206 + 2ok 526 SED 288 kc-ft BMD

6 As per Aci 318 iu the Table 9.3.1.1 deflection minimum depth beam to avoid required (16x12 Il 16. hmin 12in. [hmina b]. so, let us take 16 we Inow that d=15b [Thumb rule depth d=2 tin, dar 1.5x18 = 27 in. h=27+115 Cover) = 28.5 in cil) Rna mu r ber 288x103x12 lbrin Required 18 x 272 =263.37 PSI steel ratio ez 0.85fc' La 2Rn e- 0.85*3500 fy 0.8fc 40,000 드 2X263.37 0.00736 0-8x3500 emax 17 0.858ife! لا 87000 87000 otfy 11 4 0-8540-85*3500 40,000 87000 87000+10,00 F0.043 Pmin. 3F (ov) 200 2000 = 0.00uu can 0.005, emin ceclmax ; hence ok. bars are used, area of Asrq = Preqxbxd 0.0073 6+18x277 If 6 No's of Ab - 0.59616 so, use #7bar ; Abi Asprovided 6x0.6= 3.6oinn. 3.57696 inr each bara 3:57696 6 0.6oinr. a

iv) Required development length length (d=34, Ut ve Us Pra (btkh db lo for bottom bars 1.0 for uncoated bars Charts) Ue = uo ilfe db. location factor coating factor size factor. es 1.0 for #7 bars I a lo for normal weight 1 krra UDAN sn. UOX 2X0-20 13 x 6. 0 205 Cbr150 Cbikar do 11 1.50 +0.205 0.875 -1.quo to kako < 2.50 06 (d db 11 3 x 40,000 (0X1oxio цо TOX3500 = 26-138. 1.quo 4- 26. 138x db - 22.876. Ld a depth of beam, dength of the beam So it satisfies the development

Vse shear strength of stimps =Vn -vc So I provide V we tnow Shear reinforcement design that Vn= Vetvs Vu.tovna (uctus) 52 Vu क - 69.33 k. 0.75 69.336 where Ve= shear str - VC+VS. strength of concrete 2 Jfc bd.=253500x18x27 -57.5our =) 69,33-57504 spacing of Stirrups S Aufyd VS 11.8261 (0:2)x2x 40,000x27 11.826x103 db = 36.529 in. uff bd. For Vs--vn. Vc susfi bd ; 115.0081 so, Smax Smap - d/2 szuin. 27 / 13.5 in. #u - 2 2 legged stirrups at B.5 in

maximum entent is Shear where reinforcement vi)so, the Provided where Unive. that of beam up to where Vn - 57.50u length span stirrups are to be provided Vua vn =) Vua Vnxda 43.128k. Vn = 57.50u = vu Vu = 43.128. RA 43.128 (Uxu) = 52 - 43.128 =) n = 2.2 1844 support stirrups are requires required up to 2.22ft on sa from bothsides vii) #u@ 13.5"olc support 22.22' * NO t 2.221 Stirrupal #7@6 No's 16 Longitudinal section tr #u@ 13.5"ole 27 12 #7 @ UND'S 1.5 * 18" Cross section @ AA Reinforcement details