beams analysis

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REF; TECHIE ASSOCIATES Bs.6399 PROJECT No: PROPOSED COMMERCIAL DEVELOPMENT ON LR.No. THIKA prt 1 CLIENT: DR ELEMENT: BEAM 6 ANALYSIS Breadth,b= 200 mm Fire resistance= 1.5hrs Table Depth,h= 450 mm Cover= 20 mm Span= 7.9 m Slab thickness= 150 mm Wall height= 2100 mm 200 mm Imposed load,LL= 1.5 (BS 6399, 5.1.1 Table 1) Partitions= 1.0 (BS 6399, 5.1.4) Finishes= 1.0 gk(KN/m) qk(KN/m) Span point slab s/w=24 x 0.5 x 7.9x 150/1000 14.2 1 0.0 walling=20 x 2100/1000 x 200/1000 8.4 2 0.0 Partitions=1 x 0.5x7.9 3.95 3 0.0 Finishes=1 x 0.5 x7.9 3.95 4 0.0 Beam s/w=24 x 200/1000 x 450/1000 2.2 5 0.0 Imposed loads=1.5 x 0.5 x 7.9 5.93 6 0.0 32.7 5.93 Design load,n=1.4gk + 1.6qk= = 55.23 KN/m Left support= C F.E.M's= Right support P grids E F span 1 2 3 4 5 6 Length,m 1.317 3.574 2.811 0 0 0 s.f 0.76 0.28 0.36 d.f 0.73 0.27 0.44 0.56 F.E.M's 0 15.92 -58.79 58.79 ### 36.4 1st d 0 31.33 11.5 -9.87 -13 -36 1st c/o 15.7 0 -4.9 5.772 ### -6.3 2nd d -16 3.6 1.3 5.465 6.9 6.28 2nd c/o 1.8 -7.83 2.7323 0.665 3.14 3.47 3rd d -1.8 3.7 1.3731 -1.67 -2.1 -3.5 3rd c/o 1.86 -0.9 -0.8 0.687 -1.7 -1.1 4th d -1.9 1.271 0.4682 0.462 0.6 1.06 support 0 47.12 -47 60.3 ### 0 moments span -11.58 34.48 24.41 moments X 0 0.659 1.317 3.104 4.9 6.3 7.7 M 0 -11.6 -47 34 ### 24 0 WL/2 36.4 36 99 99 78 77.6 M/S -36 35.8 -3.7 -3.69 21.5 -21 V 0.59 -72 95 -95 99.1 -56 Density of concrete=24KN/m 3 Wall density=20KN/m 3 Wall thickness Wt= KN/m 2 KN/m 2 KN/m 2 nL 2 /12 + Pa 2 b/L 2

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Page 1: Beams Analysis

REF; TECHIE ASSOCIATESBs.6399 PROJECT No: PROPOSED COMMERCIAL DEVELOPMENT ON LR.No. THIKA

prt 1 CLIENT: DRELEMENT: BEAM 6 OUTPUTANALYSISBreadth,b= 200 mm Fire resistance= 1.5hrs Table 3.4Depth,h= 450 mm Cover= 20 mm

Span= 7.9 m

Slab thickness= 150 mm

Wall height= 2100 mm 200 mm

Imposed load,LL= 1.5 (BS 6399, 5.1.1 Table 1)

Partitions= 1.0 (BS 6399, 5.1.4)

Finishes= 1.0gk(KN/m) qk(KN/m) Span point load X

slab s/w=24 x 0.5 x 7.9x 150/1000 14.2 1 0.0 KN 0walling=20 x 2100/1000 x 200/1000 8.4 2 0.0 KN 0Partitions=1 x 0.5x7.9 3.95 3 0.0 KN 0Finishes=1 x 0.5 x7.9 3.95 4 0.0 KN 0Beam s/w=24 x 200/1000 x 450/1000 2.2 5 0.0 KN 0Imposed loads=1.5 x 0.5 x 7.9 5.93 6 0.0 KN 0

32.7 5.93 KN 0Design load,n=1.4gk + 1.6qk=

= 55.23 KN/m Left support= C

F.E.M's= Right support= Pgrids E F

span 1 2 3 4 5 6Length,m 1.317 3.574 2.811 0 0 0s.f 0.76 0.28 0.36 d.f 0.73 0.27 0.44 0.56 F.E.M's 0 15.92 -58.792 58.79 -36.4 36.4 1st d 0 31.33 11.5 -9.872 -12.6 -361st c/o 15.7 0 -4.9 5.772 -18.2 -6.282nd d -16 3.6 1.3 5.465 6.9 6.282nd c/o 1.8 -7.832 2.7323 0.665 3.138 3.473rd d -1.8 3.7 1.3731 -1.674 -2.13 -3.473rd c/o 1.86 -0.902 -0.8 0.687 -1.74 -1.064th d -1.86 1.271 0.4682 0.462 0.6 1.06support 0 47.12 -47 60.3 -60.3 0momentsspan -11.58 34.48 24.41moments

X 0 0.6585 1.317 3.104 4.9 6.3 7.7M 0 -11.6 -47 34 -60.3 24 0

WL/2 36.4 36 99 99 78 77.6M/S -36 35.8 -3.7 -3.687 21.45 -21V 0.59 -72 95 -95 99.08 -56

Density of concrete=24KN/m3

Wall density=20KN/m3

Wall thickness Wt=

KN/m2

KN/m2

KN/m2

nL2/12 + Pa2b/L2

Page 2: Beams Analysis

REF; TECHIE ASSOCIATESBs.8110 PROJECT No: PROPOSED COMMERCIAL DEVELOPMENT ON LR.No. THIKA

prt 1 CLIENT: DRELEMENT: BEAM 6design

25 425 b= 200 mm h= 450 mmTABLE Cover on reinforcement= 20 mm d=h-20-8-16/2

3.3 Assume main reinforcement= Y 16 d= 414 mmAssume shear reinforcement= Y 8 d'=c+8+16/2 d'= 36

3.4.4.4= 134 KNm (Moment capacity)

K=or 0.95d whichever is less

3.4.4.4 As= M As'= (M-Mu)0.95FyZ (0.95 x fy x (d-d'))

Table Vc=3.7 0.87fy

3.4.4.4 K'= 0.156 Z= If K>K' Z=0.5+√(0.25-K'/0.9)GRIDS E F 0.00 0.00 0.00 0.00 0

X 0 0.6585 1.317 3.104 4.9 6.3 7.7M 0 -11.6 -47.1 34 -60.3 24.4 0K 0 -0.01 0.055 0.04 0.07 0.03 0.0

1.00 1.01 0.93 0.95 0.91 0.97 1.00Z,mm 393.3 393.3 386.9 393.3 379 393 393

0 -73 301.6 217.1 394 154 0A's reqd. 0 0 0 0 0 0 0

DESIGN FOR SHEAR 100

Table V V=V/bd Vc Vc+0.4 Sv3.8 99 1.20 402 0.49 0.9 464.4

CHECK FOR DEFLECTIONbasic span/effective depth= 3574 mm = 8.63 < 26 OK

414 mm

fcu= N/mm2 fy= N/mm2

Mu=0.156fcubd2

M/fcubd2

Z=d{0.5+√(0.25 - K/0.9 ) }

0.79[100As/bd]1/3 x(400/d)1/4 Asv ≥ 0.4b vym Sv

If K≤K' 0.5+√(0.25-K/0.9)

{0.5+√(0.25 - K/0.9 ) }

As reqd.

For shear reinforcement assume 2 legs of Y8,Asv= mm2

AS

Page 3: Beams Analysis

REF; TECHIE ASSOCIATESPROJECT No:ELEMENT: RING BEAM

Beam sections fcu fy200 30 460250 25 425300 20 410400 15450500600700800

bar areas As No. SLABY12 113.0 2 150Y16 201 3 175Y20 314 4 200Y25 491 5Y32 803.84 6

78901

0.5hr P1hr C1.5hrs F2hrs3hrs4hrs