Title TOPIC 3a_Analysis & Design of Beams.pdf ✅

Created with PTC Mathcad Express. See www.mathcad.com for more information. SUMMER 3231 QUIZ 3 - TORSION A. Load demands at critical section near the supports: Mu≔-264.40 kN ⋅m Vu≔209.10 kN Tu≔23.50 kN ⋅m B. Material and section properties: f'c≔34.5 MPa fy≔414 MPa db≔20 mm tb≔12 mm cc≔40 mm h≔760 mm b≔450 mm d≔h-cc-tb-d ⋅ b 0.5 Given section is for torsion. C. Area of steel required for flexure. A≔0.85⋅ f' ⋅ ⋅ ⋅ c b ((-0.5)) 0.90 B≔0.85⋅ f' ⋅ ⋅ ⋅ c b d 0.90 C≔Mu a1≔―――――――= -B+ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ 2 B - 2 4⋅A⋅C 2⋅A 32.658 mm a2≔―――――――= -B- ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ 2 B - 2 4⋅A⋅C 2⋅A 1363.342 mm a≔a1=32.658 mm β1≔0.85-―――⋅ = 0.05 7 MPa ⎛⎝f'c-28 MPa⎞ ⎠ 0.804 c≔―= a β1 40.641 mm εs≔0.003⋅――= d-c c 0.049 As≔―――――= 0.85⋅ f' ⋅ ⋅ c a b fy 1040.984 mm2 D. Will torsion be ignored? Acp≔b⋅h+580 mm ⋅180 mm =446400 mm2 pcp≔2⋅760 mm +2⋅450 mm +2⋅580 mm pcp=3580 mm Tuth≔0.75⋅0.083⋅1⋅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅⋅ 2 f' ⋅ c 1 MPa ―― Acp 2 pcp Tuth=20.352 kN ⋅m Tu>Tuth Torsion will be considered. Tucr≔0.75⋅0.33⋅1⋅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅⋅ 2 f' ⋅ c 1 MPa ―― Acp 2 pcp Tucr=80.919 kN ⋅m Tucr>Tu>Tuth Torsion need not be reduced. E. Is the section sufficient for the combined shear and torsion stresses?
Created with PTC Mathcad Express. See www.mathcad.com for more information. E. Is the section sufficient for the combined shear and torsion stresses? b'≔b-2⋅ cc-tb=358 mm h'≔h-2⋅ cc-tb=668 mm Aoh≔b'⋅h'=239144 mm2 ph≔2⋅b'+2⋅h'=2052 mm = 2 ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ + ⎛ ⎜ ⎝―― Vu b⋅d ⎞ ⎟ ⎠ 2 ⎛ ⎜ ⎜⎝ ――― T ⋅ u ph 1.7⋅Aoh 2 ⎞ ⎟ ⎟⎠ 2 0.83 MPa Vc≔0.17⋅1⋅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅⋅ ⋅ = 2 f' ⋅ c 1 MPa b d 313.636 kN 0.75⋅ = ⎛ ⎜ ⎝――+ Vc b⋅d 0.66⋅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ 2 f' ⋅ c 1 MPa ⎞ ⎟ ⎠ 3.656 MPa < 2 ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ + ⎛ ⎜ ⎝―― Vu b⋅d ⎞ ⎟ ⎠ 2 ⎛ ⎜ ⎜⎝ ――― T ⋅ u ph 1.7⋅Aoh 2 ⎞ ⎟ ⎟⎠ 2 0.75⋅ ⎛ ⎜ ⎝――+ Vc b⋅d 0.66⋅ ̅ ̅ 2 f'c ⎞ ⎟ ⎠ Section is sufficient. F. Required transverse reinforcement for shear: Vu=209.1 kN 0.75⋅Vc=235.227 kN 0.50⋅0.75⋅Vc=117.614 kN 0.75⋅Vc>Vu Avs≔0 ―mm― 2 mm ―= Av s 0 Vu>0.50⋅0.75⋅Vc Minimum shear reinforcement is required. G. Required transverse reinforcement for torsion: Ats≔――――――――――― Tu 0.75⋅2⋅0.85⋅A ⋅ ⋅ oh fy cot ((45 deg)) Ats=0.186 ―mm― 2 mm H. Required longitudinal reinforcement for torsion: Al≔――――――――――― T ⋅ u ph 0.75⋅2⋅0.85⋅A ⋅ ⋅ oh fy cot ((45 deg)) Al=382.01 mm2 I. Minimum transverse reinforcement Avmin1≔0.062⋅b⋅―――――= ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ 2 f' ⋅ c 1 MPa fy 0.396 ―mm― 2 mm Avmin2≔――――⋅ = 0.35 MPa fy b 0.38 ―mm― 2 mm Avmin≔Avmin1=0.396 ―mm― 2 mm J. Practical spacing of 12-mm closed stirrups in multiples of 25mm:
Created with PTC Mathcad Express. See www.mathcad.com for more information. J. Practical spacing of 12-mm closed stirrups in multiples of 25mm: Atb≔2⋅―⋅ = π 4 tb 2 226.195 mm2 s1≔――――= Atb Avs+2⋅Ats 607.512 mm s2≔――= Atb Avmin 571.438 mm s3≔―= ph 8 256.5 mm s4≔300 mm s≔250 mm K. Minimum longitudinal reinforcement for torsion: Almin1≔0.42⋅Acp ⋅―――――- = ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ 2 f' ⋅ c 1 MPa fy A ⋅ ⋅ ts ph ― fy fy 2277.996 mm2 Almin2≔0.42⋅Acp ⋅―――――- = ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ 2 f' ⋅ c 1 MPa fy ―――――⋅ ⋅ 0.175 MPa ⋅b fy ph ― fy fy 2269.68 mm2 Almin≔Almin2=2269.68 mm2 L. Determine the total number of 20-mm tension bars if 4 of the same bars are placed on the sides (2 each side). Ab≔―⋅ = π 4 db 2 314.159 mm2 A'l≔――――= Almin-4⋅Ab 2 506.522 mm2 A's≔As+A'l=1547.505 mm2 Asmin1≔0.25⋅―――――⋅ ⋅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ 2 f' ⋅ c 1 MPa fy b d Asmin1=1114.082 mm2 Asmin2≔―――⋅ ⋅ = 1.4 MPa fy b d 1062.174 mm2 n≔――= A's Ab 4.926 n'≔5