σ = σ + σ τ xy σ c2 σ 1 ω = τ = τ Disk of orthogonal reinforcement Consider a square disk of a unit length under shear and normal stresses = =Φ
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1 .. Disk o orthogonal reinorement Consider a square disk o a unit length under shear and normal stresses Conrete : Steel :,, s, s Total stresses are deomposed into onrete and steel stresses = + = + s = s prinipal onrete stresses are denoted as, reinorement ratio 3 r As = t C : r A s ω = bd 0.85 Tensile strength As Y = = t ' t ε 30
2 θ As Y = = t t ompressie strengths ( = + = + Y ( = + = + Y 3 pure shear steel : s = Y, s = 0, = dutile onrete : uniaial stress state = = os θ = ( + osθ = sin θ = os θ = ( + osθ = sin θ (, θ ( θ = sin θ = sinθ osθ t = As Y Σ F = 0 : os θ t = A s Y t = As Y Σ F = 0 : sin θ t = A From Eqs. 4 & 5 s Y As Y = t os θ = os θ - 6 As Y = t sin θ = sin θ - 7 Sine 6 = 7 3
3 = os θ sin θ = tan θ = μ - 8 = = ( + μ os θ = sinθosθ - 9 Substituting 6 into 9 ields As Y = sinθ osθ = tanθ t os θ = = = μ A s A s Unit length I 〡 II Shear kes Unit length First ield ondition (No rushing ailure o onrete < ( μ sinθosθ = + < + < + < Ma reinorement Under pure shear dierent approah! Aim is to ind the shear strength 3
4 Subjet to < s = s = s = i θ osθ Stress ree = tanθ = osθsinθ θ i θ sinθ = osθsinθ = tanθ = = to ind under ombined stresses 33
5 (i.e. or/and B point = + Con 0 0 = = At B Point s s Mohr s irle = = = = 0 As θ A s Equialent stress = ( = A s = os θ + sin θ = t As As, seond term : t t (First term : = =prinipal stress = 0 34
6 = = 0 = 0 = Keeping the steel stress And aring 0 = ( = ( = Y Sine = 3 + = ( + = ( ( ( + = 0-6 R = 0 = 0 Substituting Eq 5 into ields Eq 3 4 ( + = ( ( ( ( ( ( + (FG Line Eq 35
7 A F B E M K L N G + = ( D + = ( + H C - BFG Region at Point B s = s = Y = = 0 = 0 = + Con = ( φ φ in BFG region : = = s s Y < < 0 Conial ( ( + = 0 r 0 r FG Line ( + = 36
8 - In DEH Region At D Point 0 = = s s Y Keep : = 0 Equialent = + ( = s + = 0 & = s Y Y < s+ < Y ( + < < - = + - Sine ( ( + = = ( + ( + = + 4 ( + + ( = 0-4 We know that + = + = ( Substituting Eq 5 into ields ( + ( ( ( 37
9 At Line LG Line NC Line KH s = s = Y s s = Y = Y s = s = Y = 0, = : onl hange in steel stresses result in no ariation o along KH is onstant in KHNC In Region LMKN At point L = = ( = ( ( + = 0 = + = = ( ma A φ φ φ B F E M L + = ( G D K H N C + = 38
10 III ( + φ II ( φ I Substituting =0, into the equations representing ield surae ields ( = 3 For the third region ( ( + = ( + : EH Line = ( + - (* ( ( ( = ( = ( + = 0 = = 0w (** Substituting (* into (** ields ( [ ] = ( + = φ = I we disregard the speial ondition or 0 For the seond region (onstant = + 39
11 ( + ( ( + = 0 In remaining region ( + ( ( = 0 For more approimation + = + = ( ( ( + : ( ( Setting = 0 = ( + ( ( = + + = 0 Shear strength o Think Proess Consider isotropi Delete steel in -dir, add tensile ore 3 Add steel in -dir, substrat The ield ondition in the orthotropi disk 40 + φ
12 i i ' i i ' isotropi orthotropi = 0 + Same struture i i ' = - = + ' ( μ = - Where μ = Remind Eq s (75 and (76 ( ( + = 0-3 Substituting & into 3 ields ( ( ( ' ' ( ( μ 0 ( ( μ 0 μ + = 0 ' ' + = + = 4
13 isotropi orthotropi (, (, t t II I A D + t = + + t A = η t η = Where (, + + = η + η t For Ⅰ region : η + η t t t t A (, t For Ⅱ region : η + η t 4
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