𝑑=𝑙 √(𝑊𝑢𝐵/("α" ɸ "f c " ′ 𝑏𝜔(1−0.59𝜔)))

((𝑊𝑢 𝐵) 𝑙𝑛 )/² 𝛼 = "f′c" 𝑏𝑑^2 "bd ω (1-0.59ω)ɸ ² "

(𝑀𝑢 )/ ="f'c bd ɸ ² ω (1-0.59ω)"

ℎ/1.1=〖 l √〗 (𝑊𝑢𝐵/(16∗0.9∗210∗𝐵/20 ∗0.14∗(1−0.59∗0.14)))

h = (𝑙 )/(4.01/√𝑊𝑢)𝑇= √𝐴𝑧+((𝑡1−𝑡2))/2

ť

²

ln(√(█(4@0.13)))

𝑇=√𝐴𝑧 + (t1-t2)

bD ((1.25∗245.00 𝑡))/((0.25) (0.28 𝑡/(𝑐𝑚^2 )))=4375.00 𝑐𝑚²

AZAP 𝑃/𝜎𝑛=(245 𝑡)/(30.3 𝑡/𝑚 )²

h= k z= 4 12.65ln ln

(█(2+𝑑𝑓@𝑏𝑜)) √(𝑓^′ 𝑐 𝑏𝑜𝑑)

ℎ/1.1 = l √(█(𝑊𝑢 𝐵@16∗0.9∗210∗𝐵/20∗0.14∗(1−0.59∗0.14)))

h = (𝑙 )/(4.01/√𝑊𝑢)h = ( )/((4/√0.1) ) = ( 𝑙 𝑙)/12.65=ℎ

h = (𝑙 )/√(4/0.13) = (𝑙 )/11.09=ℎh = (𝑙 )/((4/√0.11) ) = (𝑙 )/12.06=ℎh = ( )/((4/√0.1) ) = ( 𝑙 𝑙)/12.65=ℎ

Vc = 0.27(2+4/𝛽𝑐) √("f’" 𝑐) 𝑏𝑜 𝑑

Vc = 0.27(2+"α d" /𝑏𝑜) √("f’" 𝑐) 𝑏𝑜 𝑑Vc = 1.06√("f’" 𝑐) 𝑏𝑜 𝑑

ɸVc = 0.75∗0.53∗√280∗100∗9 𝑐𝑚=5986.30 𝑘𝑔=5.99 𝑡

𝑇= √2.85 + ((0,8−0,55))/2² = 2,987 m

𝑆= √2.85 − ((0,8−0,55))/2² = 2,725 m

𝑑=𝑙 √(𝑊𝑢𝐵/("α" ɸ "f c " ′ 𝑏𝜔(1−0.59𝜔)))

((𝑊𝑢 𝐵) 𝑙𝑛 )/² 𝛼 = "f′c" 𝑏𝑑^2 "bd ω (1-0.59ω)ɸ ² "

(𝑀𝑢 )/ ="f'c bd ɸ ² ω (1-0.59ω)"

ω = (𝜌 𝑓𝑦 )/(𝑓'𝑐 )

ℎ/1.1=〖 l √〗 (𝑊𝑢𝐵/(16∗0.9∗210∗𝐵/20 ∗0.14∗(1−0.59∗0.14)))

h = (𝑙 )/(4.01/√𝑊𝑢)𝑇= √𝐴𝑧+((𝑡1−𝑡2))/2 𝑆= √𝐴𝑧 −((𝑡1−𝑡2))/2

ln(√(█(4@0.13)))

bD ((1.25∗245.00 𝑡))/((0.25) (0.28 𝑡/(𝑐𝑚^2 )))=4375.00 𝑐𝑚²

AZAP 𝑃/𝜎𝑛=(245 𝑡)/(30.3 𝑡/𝑚 )²

ℎ/1.1 = l √(█(𝑊𝑢 𝐵@16∗0.9∗210∗𝐵/20∗0.14∗(1−0.59∗0.14)))√("ω" 𝑢)

h = (𝑙 )/((█(4@√0.11)) )=(𝑙 )/12.65=ℎ

h = (𝑙 )/(4.01/√𝑊𝑢)h = ( )/((4/√0.1) ) = ( 𝑙 𝑙)/12.65=ℎ

h = (𝑙 )/√(4/0.13) = (𝑙 )/11.09=ℎh = (𝑙 )/((4/√0.11) ) = (𝑙 )/12.06=ℎh = ( )/((4/√0.1) ) = ( 𝑙 𝑙)/12.65=ℎ

Vc = 0.27(2+4/𝛽𝑐) √("f’" 𝑐) 𝑏𝑜 𝑑

Vc = 0.27(2+"α d" /𝑏𝑜) √("f’" 𝑐) 𝑏𝑜 𝑑𝑑= √(𝑀𝑢/(40.5∗𝑏))

𝑑= √((2,38∗10 )/(40,5∗100)=7,67 𝑐𝑚)⁵

ɸVc = 0.75∗0.53∗√280∗100∗9 𝑐𝑚=5986.30 𝑘𝑔=5.99 𝑡

𝑇= √2.85 + ((0,8−0,55))/2² = 2,987 m

𝑆= √2.85 − ((0,8−0,55))/2² = 2,725 m

((𝑊𝑢 𝐵) 𝑙𝑛 )/² 𝛼 = "f′c" 𝑏𝑑^2 "bd ω (1-0.59ω)ɸ ² "

ℎ/1.1=〖 l √〗 (𝑊𝑢𝐵/(16∗0.9∗210∗𝐵/20 ∗0.14∗(1−0.59∗0.14)))

𝑆= √𝐴𝑧 −((𝑡1−𝑡2))/2

h = (𝑙 )/((█(4@√0.11)) )=(𝑙 )/12.65=ℎ

𝑑= √(𝑀𝑢/(40.5∗𝑏))

𝑑= √((2,38∗10 )/(40,5∗100)=7,67 𝑐𝑚)⁵

h = (𝑙 )/((█(4@√0.11)) )=(𝑙 )/12.65=ℎ

𝑑= √((2,38∗10 )/(40,5∗100)=7,67 𝑐𝑚)⁵