The member shown in Figure P6.5-7 has lateral support at points A, B, and C. Bend- ing is about the strong axis. The loads are service loads, and the uniform load includes the weight of the member. A992 steel is used. Is the member adequate? a. Use LRFD. b. Use ASD. 7*D, 18*L -1.5kAD, 3.54ML 70*D, 170°L B 16' W10 x 100
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- Ex: For a cantilever beam of length (L) subject 1 a concentrated load (P), shown in fgure, Find - 1 Max.tension and compression bending stress in beam section 2. Determine the proper spacing of the nails used 10 fix the flange boards to the web. The nals o be used can safely resist (F) of shearing force, given in 3 - “Table below. I»A cantilever beam AB of length L = 6.5 ft supportsa trapezoidal distributed load of peak intensity q,and minimum intensity q/2, that includes the weight ofthe beam (see figure). The beam is a steel W12 X14wide-flange shape (see Table F-1(a), Appendix F).Calculate the maximum permissible load q basedupon (a) an allowable bending stress σallow =18 ksiand (b) an allowable shear stress τallow = 7.5 ksi.Note: Obtain the moment of inertia and section modulusof the beam from Table F-1(a).The steel is supported by the steel tie rod in AB beam B. Steel connection tension rod is placed 2 meters to the left of B and C sliding bracket is placed and P is loaded between AC It is. By ignoring the weights of beams and connecting rods, they can be determine the largest P load it can carry. The diameter of the BD rod is 16 mm. E = 200GPa I = 150x(10^6)mm^4
- A steel bracket of solid circular crosssection is subjected to two loads, each of which isP = 4.5 kN at D (see figure). Let the dimension variablebe b = 240 mm.(a) Find the minimum permissible diameter dminof the bracket if the allowable normal stress is110 MPa.(b) Repeat part (a), including the weight ofthe bracket. The weight density of steel is77.0 kN/m3.An unsymmetrical flexural member consists of a ½ × 12 top flange, a ½ × 7 bottom flange, and a ⅜ × 16 web. a. Determine the distance y from the top of the shape to the horizontal plastic neutral axis. b. If A572 Grade 50 steel is used, what is the plastic moment M, for the horizontal plastic neutral axis? c. Compute the plastic section modulus Z with respect to the minor principal axis. Pls help!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?
- Read the question carefully and give me right solution according to the question. A W12x79 of A573 Grade 60 (Fy=415 MPa) steel is used as a compression member. It is 8 m long, pinned at the top fixed at the bottom, and has additional support in the weak direction at mid-height. Properties of the section are as follows: A = 14,500 mm^2 Ix = 258.6 x 10^6 mm^4 Iy = 84.375 x 10^6 mm^4 Calculate the effective slenderness ratio with respect to strong axis buckling using theoretical value of k.Q5. Joint A is roller and C is pin supportedPlot BMD & SFDAn 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?
- Select a rectangular (not square) HSS for use as a 15-foot-long compression member that must resist a service dead load of 35 kips and a service live load of 80 kips. The member will be pinned at each end, with additional support in the weak direction at midheight. Use A500 Grade C steel. a. Use LRFD. b. Use ASDA W12 x 79 of A572 Grade 60 steel is used as a compression member. It is 28 feet long, pinned at each end, and has additional support in the weak direction at a point 12 feet from the top. Can this member resist a service dead load of 180 kips and a service live load of 320 kips? a. Use LRFD. b. Use ASD.34 - A support is a fixed, B support is a sliding joint. loads M= 30 kNm, P1= 12 kN, P2= 13 kN, q1= 6 kN ⁄ m, q2= 7 kN ⁄ m , spans are a= 2 m, b= 3 m. The support reactions in the beam whose loading condition is given in the figure will be found. Accordingly, Ax = ?A) 22.25B) 0C) None.D) 39.25E) 21/2