A double-angle shape is shown in the figure. The steel isA36, and the holes are for 2-inch-diameter bolts.Assume that A₂ = 0.75An.Ae12Determine the design tensile strength for LRFD.Determine the allowable strength for ASD.CIVIL ENGINEERING- STEEL DESIGNALSection2L5 x 3 x 516 LLBB

Double angle section is shown in the figure above. Grade 60 steel is used. Diameter of bolts= 1/2…

A double-angle shape is shown in the figure. The steel is A36, and the holes are for 2-inch-diameter bolts. Assume that A₂ = 0.75An. 1 (2) Determine the design tensile strength for LRFD. Determine the allowable strength for ASD. CIVIL ENGINEERING - STEEL DESIGN Section 2L5 x 3 x 16 LLBB

A double-angle shape is shown in the figure. The steel is A36, and the holes are for 2-inch-diameter bolts. Assume that A₂ = 0.75An. 1 (2) Determine the design tensile strength for LRFD. Determine the allowable strength for ASD. CIVIL ENGINEERING - STEEL DESIGN Section 2L5 x 3 x 16 LLBB

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

determine: (a) the elastic section modulus and the yield moment. (b) the plastic section modulus and the plastic moment Mp. Bending is about the x-axis, and the steel is A572 Grade 50. Fy = 345 MPa Subject: Civil Engineering Steel Design
2. A steel plate is 360 mm wide and 20 mm thick with four bolts hole into the place as shown in the figure. Compute the following: a. Critical net area required by the NSCP specs. b. Max. critical net area required by the NSP specs. Note that, Max. net area is 85% of the critical net area Capacity of the joint if the allowable tensile stress is 0.75Fy. Use A36 steel Fy=248 MPa 25 mm= DIAMETER OF BOLTS 45 45mm fmmy C. 90mm 90mm 90mm Comm P
A tension plate shown below is used to support suspended load "T". Gusset Plate- F, = 248 MPa Fu = 400 MPa %3D 200 mm a) Determine the allowable tensile capacity of the plate if L= 240 mm. (Assume weld strength is satisfactory).
Problem2. The compression member is shown in figure. Find the following: a. The Euler stress Fe. b. The buckling stress Fcr c. The design strength d. The allowable strength e. Does the member satisfactorily meet the design requirements? Why? HSS 8x 8x4 ASTM AS00, Grade B steel (Fy = 46 ksi) 15'
A plate with width of 420 mm and thickness of 13 mm is to be connected to a plate of the same width and thickness by 30 mm diameter bolts, as shown in the Figure 1. The holes are 3mm larger than the bolt diameter. The plate is A36 steel with yield strength Fy = 248MPa. Assume allowable tensile stress on net area is 0.60Fy. It is required to determine the value of b such that the net width along bolts 1-2-3-4 is equl to the net width along bolts 1-2-4. W = 420 mm t = 13 mm a = 64 mm c = 104 mm d = 195 mm Bolt Diameter = 30 mm Holes Diameter = 30 mm + 3 = 33 mm ?? = 248 MPa Allowable Tensile Stress on ?? = 0.60Fy a. Calculate the vaue of b in millimeters. b. Calculate the value of the net area for tension in plates in square millimeters. c. Calculate the value of P so that the allowable tensile stress on net area will not be exceeded.
5. A bronze bar is fastened between a steel bar and an aluminum bar as shown in Figure 5. Axial loads are applied at the positions indicated. Find the largest value of P that will not exceed an overall deformation of 3.0 mm, or the following stresses: 140 MPa in the steel, 120 MPa in the bronze, and 80 MPa in the aluminum. Assume that the assembly is suitably braced to prevent buckling. Use Est = 200 GPa, Eal = 70 GPa, and Ebr = 83 GPa. Steel A = 480 mm² 1.0 m Bronze A = 650 mm² 3P 4P 2.0 m FIGURE 5 Aluminum A = 320 mm² 1.5 m 2P
An A36 steel plate with width of 400 mm and thickness of 12 mm is to beconnected to a plate of the same width and thickness by 34 mm diametersbolts, as shown in the figure (next slide). The holes are 2 mm larger thanthe bolt diameter. The yield strength of the steel plate is Fy = 248 MPa.Assume allowable tensile stress on net area is 0.60Fy. It is required todetermine the value of b such that the net width along bolts 1-2-3-4 isequal to the net width along bolts 1-2-4.a. Calculate the value of b in mm.b. Calculate the value of the net area for tension in plates in mm².c. Calculate the value of P so that the allowable tensile stress on net areawill not be exceeded.(Use at least 2 decimal points on your answers.)
Problem 4 Given two welded plates n A36 plates t= 3/ • fillet size w=1/4" Electrode Fexx= 70 ksi • plate width = 9" ● ● ↑ t=¾/2" = 2) = Pu= smaller of (1) A(2) w=1/4" Base Metal Strength, DRn is given by: ÞRn = 0.75(0.6Fu) (tBM)(Lw) t=3/8" Determine max Pu that can be applied to this connection ? hint: Dobtain Pue based on weld strength (transverse) BM strength = where tbm = thickness of the thinner connected plate Lw= length of the weld Fillet weld this side only i.e. lw= 9 in کہ ۔ Ри opy
What is the manufacturing tensional yield strength (lbf) of grade S135 drill pipe? a. This is a trick question because drill pipe size is not provided b. The answer can’t be calculated because the length of the pipe is not given c. The yield strength is 135000 psi and is a property of the steel d. The yield strength is 135000 psi and depends on the cross-sectional area
A PL 38 X 6 tension member is welded to a gusset plate as shown. The steel is A36 (Fy = 36ksi, Fu = 58ksi). a. The design strength, Pu based on gross area b. The design strength, Pu based on effective area PL % x 6 3/8" 6" Cross Sectional area of PL3/8x6 a) Blank 1 b) Blank 2 Blank 1 Add your answer Blank 2 Add your answer