Youare allowed to take books , hand-notes only. You have the rights to sélect ei

Question

Youare allowed to take books , hand-notes only. You have the rights to sélect either Eiglish units or St units. Sl units are in the brackets. All the design depend on ACt code. In this exam, you should use f-4ksi, Is-60ksi, fA30ksi fp-200MPa.), 9. -25MPa, f-20MPa igMPa), Ee3600ksi(3 x10MPa), Ei-29000ksi(2 density x of Questions 1 (totally 50%) Calculate the shear force for the reinforced concrete simple-supported beam and draw the behere envelope, Consider dead load as self-weight for oly and JI the live load. 2 kips/ft (3 kN/m) Section size is in m-n: (10%) · rm Im

Explanation / Answer

a)cross section of beam = 12"x25"

unit weight of beam = 150 pcf

self weight of beam in kip/ft = 150*(12/12)*(25/12)/1000 = 0.3125 kip/ft

live load on beam = 2 kips/ft

span length of beam = 20 ft

maximum shear force in bem will be at supports

shear force due to self weight =(0.3125*20/2)=3.125 kips

shear force in beam due to live load = (2*20/2)=20 kips

factored maximum shear force of beam per load combination 1.2D+1.6L = (1.2*3.125)+(1.6*20)=35.75 kips

Therefore, the beam should be designed for a shear force of 35.75 kips

b) Given stirrup spacing = 8 inches

grade of concrete = 4000 psi

width of beam = 12"

depth of beam = 25"

let us assume an effective depth of 23"

shear strength of concrete without shear reinforcement = 0.75*2*sqrt(4000)*12*23/1000=26.18 kips

shear force for which shear reinforcement should be designed = 35.75-26.18=9.57 kips

strength of stirrups = 30 ksi

spacing of stirrups =8"

Let area of shear reinforcement required be Av

9.57 =0.75*Av*30*23/8

Av = 0.148 in2

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