In a welded plate girder, the web plate connected to flange along both the longitudinal edges. The minimum thickness of web plate of Fe 410 grade steel is required by considering serviceability criteria for un-stiffened plate girder, if the depth of the web plate is 2800 mm.
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- The T-beam in the floor system has a slab thickness of 90 mm and supported by beams 6.5 m span. The beam is casts monolithically with the slab. The spacing of beams is 2 m on centers and width the web is 400 mm. The effective depth of the beam is 600 mm. It is reinforced at tension side with 8-28mm Φ bars. fc’ = 28 MPa, fy = 415 MPa 1. Which of the following most nearly give the effective width of the flange? a. 1565 mm b. 1305 mm c. 1275 mm d. 1625 mm 2. Which of the following most nearly gives the depth of compression block? a. 62.25 mm b. 52.86 mm c. 60.26 mm d. 58.29 mm 3. Which of the following gives the nominal moment capacity of the beam? a. 1461.25 kN/m b. 1156.36 kN/m c. 1163.15 kN/m d. 1172.58 kN/mA structural steel member is in the form of tee. bf= 300mm , tf= 100mm, bw= 200mm, d=300, fy= 248 Mpa. Determine the following: a. location of plastic neutral axis from the top of the beam b. plastic section modulus c. plastic moment capacityUSE NSCP 2010 A simply supported beam is reinforced with 4 – 28 mm ø at the bottomand 2 – 28 mm ø at the top of the beam. Steel covering to centroid ofreinforcement is 70 mm at the top and bottom of the beam. The beamhas a total depth of 400 mm and a widthof 300 mm. fc’ = 30 MPa, fy = 415 MPa. Balanced steel ratio ρb = 0.031.Compute the ultimate moment capacity of the beam in kN-m. Usereduction factor of 0.90
- A girder supports concentrated dead and live loads. Determine the following required strengths for the girder based on the controlling load combinations.Use statics or Table 3-22a from the Steel Manual. The maximum bending moment, MD (kip-ft), due to dead load. The maximum bending moment, ML (kip-ft), due to live load. ASD required flexural strength, Ma (kip-ft) LRFD required flexural strength, Mu (kip-ft)Strength Of Materials Answer should be 45 kN/m; 0.26 cm :) A compound girder consists of a 45 cm by 18 cm steel joist, of weight 1000 N/m, with a steel plate 25 cm by 3 cm welded to each flange. If the ends are simply-supported and the effective span is 10 m, what is the maximum uniformly distributed load which can be supported by the girder? What weld thicknesses are required to support this load? Allowable longitudinal stress in plates = 110 MN/m2 Allowable shearing stress in welds = 60 MN/m2 Allowable shearing stress in web of girder = 75 MN/m2An 80-foot long plate girder (see below) is fabricated from a ½-inch x 78-inch web and two 3inch x 22-inch flanges. Continuous lateral support is provided. The steel is A992. The loading consists of a uniform service dead load of 1.0 kip/ft (including the self-weight), a uniform service live load of 2.0 kips/ft, and a concentrated service live load of 500 kips at midspan. Stiffeners are placed at each end and at 4 feet, 16 feet, and 28 feet from each end. Once stiffener is placed at midspan. Determine whether the flexural strength is adequate using LRFD.
- compute Compute the Euler's Buckling load andGoverning Nominal Compressive strength based on AISC E3-2 or E3-3 (kips)The section of simply supported beam is HE450B, under distributed uniform dead load PG=20 kN/m and live load PQ=60 kN/m. Beam flange is laterally supported against compression from support and midpoint of the span. a. Determine safety of the section for bending (compare maximum beam moment with bending moment capacity of the section) b. Check section safety against shear Material is S235.A simply supported beam is reinforced with 4 – 28 mm ø at the bottomand 2 – 28 mm ø at the top of the beam. Steel covering to centroid ofreinforcement is 70 mm at the top and bottom of the beam. The beamhas a total depth of 400 mm and a widthof 300 mm. fc’ = 30 MPa, fy = 415 MPa. Balanced steel ratio ρb = 0.031. Compute the ultimate moment capacity of the beam in kN-m. Usereduction factor of 0.90
- A Wide Flange Built Up beam is supported at both ends by a fixed support and has a length of 6.8m column to column. The flanges (450mmx20mm) are continuously welded to the web (500mmx18mm). The weight of the Built Up beam has a uniform load of 294N/m through-out its length. The brace is installed at both ends of the beam and has a space of 6.8m. The steel used is ASTM36. Determine the allowable bending stress of this beam.An overhang beam is loaded as shown below. The beam cross-section was butlt by attaching two (2) channels to a 9mm thick plate using 16mm rivets. The property of the channel is given below: Depth, D = 225 mm Flange Thickness, t= 9mm Flange Width, B, -112.5 mm Web Thickness, t. -9 mm The allowable flexural stress on the beam is 180 MPa. Rivets has a capacity of t= 100 MPa on shear, for bearing, o,-200 MPa on single shear, a, =260MPa on double shear. 1. determine the location of centroid, ȳ, from top of the beam in mm 2. determine the moment of inertia, I, of the section in mm4 3. determine the maximum allowable moment, Mall, inkn-m, base on the beam's cross section.A Beam is built up from the following plates. 450mm x 20mm as flanges and 500mm x 20mm as web. All plates are A36 steel with Fy=248 MPa and the flanges are continuously connected to web by means of fillet welds. The beam has a simple span of 4m. Laterally supported only at supports with total uniform load of 500 kN/m. Is the beam safe for bending?