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Describes step-by-step method for anchoring stub for transmission tower
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STUB SUTETFull description
Describes step-by-step method for anchoring stub for transmission towerFull description
Describes step-by-step method for anchoring stub for transmission towerFull description
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Drawing clothes, draperies, and costumes Principles that govern the folds in drapery - The seven types of folds - Sculpture helps to reveal the structure of folds - Drawing folds in actio…Full description
C LIENT:
DESIGN IG N OF OF STUB C LEAT
KSEB
SPECIA EC IAL L TOWER TYPE "MLS" (0 -27 -27 )
o
o
400/ 400/ 220 KV MCMV MCMV T/ L
C HEC K AGA AG AINSTUPROOTING OF STUB DESIG N UPLIFT UPLIFT = 695427x1=695427 Stub Sec tion tio n =SL(H)250x2 L(H)250x250x 50x35 35 Stub d ep th belo b elow w G L = 1200 1200 mm (a) Ult. Load resisted by stub in slab due to Bond Us =[D x {Xx2+ Xx2+(X(X-Ts)x2}s)x2}- Np x {X+(X-T X+(X-Ts)}x k] x s wherein, wherein, X =fla fla nge width of stub stub D = Dep th of o f stub stub in s slab lab . s = Ult. perm permissible ble bond bond st stress ess between between stub and and concret concrete e= 12.24 Kg/cm2 [Refe r: Cla use 26.2.1.1; IS456 : 2000] Ts = Thic hic kness ness of stub tub sec sec tion. tion. Np = No of o f cleats c leats pa ir k =Fla Fla nge width of o f clea c leatt sec sec tion. tion. Us =2 x [ 150 x {25 x 2 + (25 - 3.5) x 2}- 8 x {25 + (25 - 3.5)}x 3.5)}x 12 ] x 12.24 12.24 232217 Kg = lt. Loa Load d resisted by c leat leat (b) Ult. Use Use outer ute r clea clea t = 16 nos no s. 12 120 0 x 120 120 x 10 -- 450 mm long long HT Use Use inne innerr cle clea a t = 16 nos. no s. 120 x 120 x 10 10 -- 450 450 mm long lon g HT Provide Provide 3 nos of 24 dia b olts of c lass lass 5.6 p er clea c leatt in DS ea ring str strength ength of Conc C oncrete rete (i) Ult. Bearing Uc = fbx(L0+ fbx(L0+Li) x Np Np x (k-C (k- C t) wherein, wherein, fb = Ult. Ult. Bear ea ring ing str stres ess s in c onc rete = 0.45 .45fck = 0.45 .45x20x10.19 .1972 = 91.78 .78 Kg/cm2 [Refe r: Cla use 34.4; IS456 IS456 : 2000] L0=Length eng th of outer oute r clea t Li =Length Leng th of inner inne r c leat lea t C t=Thickness of c leat lea t sec sec tion Uc = 91.78 x (45 + 45) x 8 x (12 - 1) 726898 Kgs -----= -------- (i) (ii) Ult. Shear strength of Bolts Ub = Np x Nb x Sp x Ab x fs wher wherei ein n, fs = Sheari earing stress of Bolt olt = 310 MPa MPa =310 x 10.1972 = SP = No. of Shea r Pla nes = 2 Nb = No o f b olts p er p a ir of c lea t = 3 Ab = Area o f Bolt = 4.52 c m2 Ub = 8 x 2 x 4.52 x 3161 x 3 685811 Kgs. =
3161 Kg/cm2 [Refe r: Cla use 5.2.1; IS12427 IS12427 : 1988] & Cla use use 7.9.2; 7.9.2; CBIP CBIP Ma nua l]
------ (ii)
ea ring strength of Bolts in Stu Stub b or Cle C lea at (iii) Ult. Bearing Ub = Np x Nb x f x tmin x fb wherei erein n, fb =Min Minm Bear earing stress of Bolt olt or HT steel eel= 620 Mpa Mpa = φ = ∆ιαµετερ οφ βολτσ = 24 mm tmin = M inm thic kn kness o f Stub or Pa ir of C le lea t = Ub =( 3 x 8 ) x 2.4 2.4 x 2 x 6322 728294 Kgs. ----- (iii) = ec tive strength of stub and and clea c leatt ( c ) Effect = Us + Leas Lea st o f the streng strength th of c a se [ (i), (ii), (iii) ] = 232217 232217 +685811 685811 = 918028 > 695427 SAF
6322 Kg/cm2 [Refer: Clau se 7.9.2; CBIP CBIP Ma nua l]