From the Truss member shown, design all the tension members for ASD and LRFD, if the connection for the angle bars are shown. Use A36 STEEL. + 2 m 2m 100 mm 150 mm B SERVICE DEAD LOAD=300 KN SERVICE LIVE LOAD = 200 KN 2 m D 2 m AUMIV 2.5 m
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- For the flat plate shown in the figure below (edge beam are not used), design the hatchedframe (middle and column strips) in the N-S direction.Given:- fc = 28MPa , f y = 420MPa- Super imposed dead load = 4 kN/m^2- Live Load = 3 kN/m^2Does the column shown below have enough available strength to support the given working loads?Use LRFD. Take E=20,232 ksi ,ry=2.3 in, K=1,Fy=59.0 ksi, Ag=32.7 inAn unsymmetrical flexural member consists ofa 3 x 30 in. top flange, a 3 x 20 in. bottom flange, and a 60 * 2 in web A. Determine the distance from the top ofthe shape to the horizontal plastic neutral axis. B. If A572 Grade 50 steel is used, what is the plastic moment Mp for the horizontal plastic neutral axis?
- 1. An unsymmetrical flexural member consists of a 70 × 600 top flange, a 70 x 400 bottom flange, and a 12 × 180 web. a, Determine the Section Modulus. b. Determine the distance from the bottom of the shape to the horizontal plastic neutral axis. c. If A572 Grade 50 steel is used, what is the plastic moment MPy for the horizontal plastic neutral axis?2. A square aluminum bar is to support a load of 65kN on a length of 4.50 m. Assume pinned ends, Determine the length of each side E-70 GPa.Determine the Ultimate Moment Capacity of the given section with the following properties f'c=28 MPa fy=345 MPa Steel cover =d'=60mm
- Check the adequacy of the column shown in red colour with the 200UC59.5 section. The material is 300 Plus. Dead and live loads for all floors are the same as shown in Figure 1. All the support conditions are shown in the Figure. For bending about the minor axis, y, there is an intermediate secondary member that prevents its buckling and the column is simply supported. In case of buckling about the major axis, x, there is no intermediate lateral support and the column is pinned at the top and fixed at the bottom.1. An unsymmetrical flexural member consists of a 70 × 600 topflange, a 70 x 400 bottom flange, and a 12 × 180 web.a, Determine the Section Modulus.b. Determine the distance from the top of the shape to the horizontalplastic neutral axis.c. If A572 Grade 50 steel is used, what is the plastic moment MPy forthe horizontal plastic neutral axis?Calculate the service load capacity in tension for an L6x4x1/2 of A36 Grade 36 steel connected with 5/8-inch dia bolts in standard holes as shown in figure 2. The angle tension member is connected to a gusset plate, typical of truss joints. The gusset plate is the plate at the intersection of members to which they are connected. Use American Institute of Steel Construction Load and Resistance Factor Design, and assume the live load to the dead load ratio is 4.0.
- Q) A simply supported beam with rectangular section shown in the figure below carries a uniformly distributed service dead load of (50 kN/m) (including beam self rated service live load of (80 kN) at mid span, find: 1) Distance fo which the shear reinforcement will be applied. weight) and a conce 2) Spacing required for stirrups ®10mm at critical section. fo=22 Nimm® fy=420 N/mm? 80N +1. An unsymmetrical flexural member consists of a 70 × 600 top flange, a 70 x 400 bottom flange, and a 12 × 180 web. a, Determine the Section Modulus. b. Determine the distance from the bottom of the shape to the horizontal plastic neutral axis. ( Take note: Bottom Distance i need, not the top distance) c. If A572 Grade 50 steel is used, what is the plastic moment MPy for the horizontal plastic neutral axis?A tension member is required to carry an axial load of 800In. It consists of a pair of un equal angle placed back to back Symmetrically about 10mm thick gusset and longer legs connected by 20mm Diameter black bolt. Select a required unequal Angle From standard steel profile and Determine the tensile strength