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Environmental Engineering

by Harsha Philips
Type: NoteInstitute: Mg university Course: B.Tech Specialization: Civil EngineeringViews: 11Uploaded: 5 months agoAdd to Favourite

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Harsha Philips
Harsha Philips
Zp  2 (550 x50 x725  700 x 22 x 700 )  5.065 x10 7 2 1x5.065 x10 7 x 250  11511.3 kNm  8826 kNm 1.1x10 6 7.Check for shear capacity Md  Nominal shear  Vn  Vp  Av fyw 3 m o  1400 x 22 x 250 3 x1.1x10 3 safe.  4041  1227 kN 8. Check for end bearing According to the code specification , assuming the slope of dispersion as 1:2.5, the dispersion length = n2 = 2.5x50 = 125 mm. Minimum stiffener bearing length provided by support = b1=550/2 =275 mm Local shear capacity of web = Fw =(b1+n2 )tw(fy/ mo ) Fw =(275+125 ) 22(250/1.10)=(2000x103)N =2000kN>support reaction (V) = 1276 kN Hence bearing stiffeners are not required. Problem 2 The effective span of a through type plate girder two lane highway bridge is 30 m. The R.C.C slab is 250 mm thick including wearing coat. Foot paths of 1.50 m width are provided on both sides of 6.80 m wide carriage way. Cross girders are provided at 3.0 m c/c and the stringers are at 2.45 m c/c. The spacing between main girders is 9.80 m. Design the maximum section of plate girder to carry IRC class A loading. Step 1 Total width of slab = 9.80 m. Number of cross girders ( 3.0 m cc) = 11 nos No of Stringers in one panel (2.45 m c/c)= 5 nos. Step 2 Dead load B.M and S.F. Total weight of R.C.C slab = 30 x 9.80 x 0.25 x 24 = 1764 kN (taking unit weight of slab as 24 kN/m3 Weight of stringers (assumed at 1kN/m) = 5x30x1 =150kN Weight of cross girders(assumed at 3 kN/m) = 11x9.8x3=323.4 kN Self weight of plate girders by Fuller’s formula = 0.2L+1 =0.2x30+1 =7 kN/m 13
Total weight of P.G = 7x30 =210 kN Total D.L = 2447.4 kN D.L per girder = 2448/2=1224 kN Moment due to D.L = M1 = S.F due to D.L = WL 1224 x30   4589 kNm 8 8 W 1224   612 kN 2 2 Step 4.Live Load B.M and S.F Transverse location of loading For the design of maximum section of the plate girders, the loads are so placed that the reaction on one plate girder is the maximum. Min. clearance from kerb = 0.15+0.25 = 0.40 c/c distance of two wheels of one train = 1.80 m Minimum distance between adjacent edges of wheels of two trains = 1.20  0.4 g  0.40  (6.80  5.50)  0.92m 2 c/c distance between two wheels =0.92+0.50=1.42 m Taking moment about B P (7.9  6.1  4.68  2.88)  2.2 P 9.8 9 I ..F   0.207 13.5  L Reaction at A including impact  2.2 x1.207  2.655 P RA  P represents one wheel load of IRC class A loading Longitudinal location of loading 14
Af  Mm o 211.56 x10 6 x1.1   2069 fy d 250 x 450 To keep the flange in plastic condition b  8.4 tf bf  tw 2 bf 8 we get  8.4 2t f    1, outstand, b  Taking b  8.4 tf b f  16.8t f  8   A f  16.8t f  8 t f  2069 mm 2 16.8t f  8t f  2069  0 2 t f  10.86 Provide 12 mm thick plate 2069  172 12 Provide 225 x 12 mm flange bf  Outstand of flange  b  225 - 8  112.5 mm 2 b 112.5   9.375 semicompac t tf 12 Zp  2(225x12x206  225x8x225/2  1.517 x 10 6 mm 3 Step 6. Check for bending strength Md  b Zp fy 1x1.517 x10 6 x 250   344kNm  211.56 kNm safe. m o 1.1x10 3 Step 7.check for shear capacity Nominal shear strengthV n  Vp  Avfw 3m o  450 x8 x 250 3x1.1  472kN  313.21kN saf Step 8. Check for end bearing According to the code specification , assuming the slope of dispersion as 1:2.5, the dispersion length = n2 = 2.5x12 = 30 mm. Minimum stiffener bearing length provided by support = b1=225/2 =112.5mm f  Local shear capacity of web = Fw = (b1  n 2 ) t w  y   o  Fw =(30+112.5 ) 8(250/1.10)=(259x103)N Fw=259 kN<313 kN . Provide end bearing stiffener 7. Design of end stiffeners 21
(a) For buckling resistance 225 - 8  108.5 mm 2 Try a pair of 100 x8 mm flats. Outstand of flange  now 14 t f  14 x8  112 mm Core area of stiffener along with effective area of web ( 20 t w ), assuming web area is available only on one side. Area for buckling rsistance  (2 x 100  (112.5  20 x 8 )8  3780 mm 2 1 1 3 x8100  100  8  272.58 3  6.01x10 6 mm 4 12 12 0.7 x3000   52.67 From table 9c in IS 800 39.87 183  1682.67  179 N/mm 2 fcd  183  10 Ix  r I  A 6.01x10 6  39.87 mm 3780 Buckling resistance  179 x 3780  676620 N  676.62 kn  313 kN Hence the stiiffener is safe. (b) Check for bearing capacity of stiffener. (cl. 8.7.5.2) Fpsd  Fpsd A q f yq where A q  Area of stiffner in contact with flange  2 x 100 x 8  1600 mm 2 0.8 x1.1 1600 x 250   454.54 x10 3 N  454 kN  313 kN safe 0.8 x1.1 Problem 4 The effective span of a through type plate girder two lane highway bridge is 30 m. The R.C.C slab is 250 mm thick including wearing coat. Foot paths are provided on both sides of the carriage way Cross girders are provided at 3.0 m c/c and the stringers are at 2.45 m cc. The spacing between main girders is 9.80 m. Design the maximum sections for the stringers if the bridge is to carry IRC class A loading. Nov 2016 Problem 5 The effective span of a deck type plate girder two lane highway bridge is 30 m. The reinforced concrete slab is 250 mm. thick inclusive of wearing coat. Foot paths are provided on either sides of the carriage way . Design maximum section of plate girder , if the bridge is to carry IRC class A loading. Nov 2016 Problem 6 The effective span of a through type plate girder two lane highway bridge is 30 m. The R.C.C slab is 250 mm thick including wearing coat. Foot paths are provided on both sides of the carriage way Cross girders are provided at 3.0 m c/c and the stringers are at 2.45 m cc. The spacing between main girders 22

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