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- A 100 mm x 200 mm wooden beam with a clear span of 7.2 m is embedded firmly ea one end and is simply supported on the other end as shown in the figure. If the allowable stresses are respectively 13.8 MPa and 1.11 MPa for bending and shear, ODetermine the safe maximum uniform load the beam can carry. Yeight of wood is 9.9 kN/m3 Find the reaction at B. OFind the max, moment of the beam.Reinforce Concrete: A rectangular isolated beam has a clear span of 8.0m and carries a service live load of 60KN/m and 15KN/m including its own weight. The beam has a width of 350mm and an effective depth of 700mm. It is reinforced with 6 – 28mm rebars in two rows. Total depth of the beam is 800mm with F’c = 27.6MPa, Fy = 414MPa, Fyt = 280MPa. Use sand lightweight concrete having a unit weight of 2400Kg/m3. a. Using Detailed calculation method for shear strength for concrete, determine the shear force to be resisted by the shear reinforcement at the critical section from the face of support. b. Determine the spacing of 10mm shear reinforcement required by the beam due to its load. c. Determine the point where no shear reinforcement are needed measured from the center of the beam.A single-span which carries a uniform load of 26 kN/m applied on the narrow side of the beam has an unsupported span of 4 m. It is made up of yakal 200 mm x 350 mm. It is also subjected to a compressive load of 400 kN at its centroid. Use 80% Stress Grade wood. The beam is fixed at both ends.
- A cantilever timber beam of 4m carries a total load of 10 KN/m (including its weight ). It has a width of 200mm and a depth of 260mm for its nominal dimensions and its dressed dimensions have 10-mm deductions for the width and height dimensions. The wooden section is made of 80% stress grade Narra ( such that Fb = 15 MPa, Fv = 1.50 MPa , &allowable = L/180, and E = 7310 MPa ). Check if the beam is adequate in (i.) bending (ii.) shearing (iii.) deflection O (i.) inadequate (ii.) adequate (iii.) inadequate O (i.) adequate (ii.) adequate (iii.) adequate O (i.) adequate (ii.) inadequate (jii.) adequate O (i.) adequate (ii.) adequate (iii.) inadequateAn Am. long beam is on hinged at its left end and on fixed support at its right end. Its flexural capacity is 610 kN.m. whilte its web shear capacity is 642 kN. Beam flexural rigidity is Y x 1012 N.mm². #1 What is the allowable uniformly distributed load on the beam so that the flexural capacity is not exceeded? #2 Based on web shear capacity, what is the allowable uniformly distributed load (kN/m)? #3 When the support at the left end is removed, a cantilever beam results. How much should the beam be cambered at the free end to offset the end deflection if the load W = 210 kN/m? Use: A = 8 Y=29Strength 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/m2
- A cantilever beam, 60 mm wide by 100 mm high and 2 m long, curri UDL of 3000 N/m over entire span. Compute the magnitude of the maximum flexural stress.An 2m. long beam is on hinged at its left end and on fixed support at its right end. Its flexural capacity is 610 kN.m. whilte its web shear capacity is 642 kN. Beam flexural rigidity is 52x 1012 N.mm². #1 What is the allowable uniformly distributed load on the beam so that the flexural capacity is not exceeded? #2 Based on web shear capacity, what is the allowable uniformly distributed load (kN/m)? #3 When the support at the left end is removed, a cantilever beam results. How much should the beam be cambered at the free end to offset the end deflection if the load W210 kN/m?Q5: Answer Only One A precast T beam is to be used as a bridge over a small roadway. Concrete dimensions are bw= 400 mm, hf 125 mm, and h= 750 mm. The effective depth d= 600 mm. Concrete and steel strengths are f 'c= 38 MPa and fy = 420 MPa, respectively. Determine the design moment capacity of the girder. The beam is used on a 5 m simple span. A) b = 1300 mm B) b = 1200 mm %3D %3D %3D %3D %3D
- About Uncracked and Craked Section 1. A doubly reinforced beam has dimension of 300mmw x 600mmh. It is reinforced with 2-25mm diam compression bars and 4 25mm diam tension bars. Concrete cover from centroid of steel bar is 60mm. On its section has a bending moment of 300KN-m. Using working stress design method what are the stresses acting on the concrete, compressive and tensile steel bars? used n: 8A composite beam has a simple span of 6 m. This consists of a Wide Flange steel section and a 12 mm concrete slab whose width is 1.5 m. Assume shoring was provided. | Wide Flange A = 9480 mm² D= 400 mm Unit Wt. = 23KN/m 1,= 270x10°mm* w= 0.74 KN/m | F,-138 MPa Concrete f~-20.7 MPa n=10 a) Determine the moment of inertia (transformed steel section) b) Compute the moment capacity of the composite beam. c) Compute the allowable live load moment of the composite beam.Light-grade steel channel was used as a purlin of a truss. The top chord of the truss is inclined I V: 4 H and distance between trusses is equal to 6 m. The purlin has a weight of 79 N/m and spaced at 1.2 m. on centers. The dead load including the roof materials is 720 Pa, live load of 1000 Pa and wind load of 1.2 1.2 1440 Pa. Coefficient of Purlins pressure at leeward and windward are 0.6 and 0.2 respectively. Assume all loads passes through the centroid of the section. Truss Properties of C 200 x 76 mm Sx = 6.19 x 104 mm Sy = 1.38 x 104 mm W = 79 N/m 12 12 1.2 I Allowable bending stress Fbx= Fby = 207 MPa Truss %3D 6m O Calculate the bending stress, fox, for dead load and live load combination (D+ L). Calculate the bending stress, foy, for dead load and live load combination (D + L). O Calculate the maximum ratio of actual to the allowable bending stress for load combination 0.75 (D + L + W) at the windward side. fbx = 151.14 MPa fby = 169.6 MPa Interaction = 1.25