Q.1 A W14 x 120 section A992 (Fy= 50 ksi) is specified for the column shown in Fig. 1 Calculate the available strength of the column if it is pinned at both ends and unbraced about both axes. (a) b) Using tables determine a suitable column section to safely support the applied loads if the end conditions are fixed at the base and pinned at the top about both axes. PD=250 k PL=450 k L=32' FIG. 1
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- Answer the following for the section at Point D Only Calculate the distributed load "w" that: Will cause the section crack Will cause the reinforcement to yield. Material Properties: F'c = 5000 psi Fy = 60000 psi Es = 29000000 psi Ln = 27 ft L wl₂² 16 wl,² 14 CD L wl,2 vl₁² 10 11 win² 16 h: 28 in A=4 in² b=14 in n d: 25 inDetermine the safe load of the column section shown, if it has a yield strength of 25 MPa. E = 200000 MPa. Use NSCP Specifications. Fyz248 mpa Properties of Channel Section d = 305 mm t₂ = 7.2 mm A = 3929 mm² t₁ = 12.7 mm Ix=53.7 x 10mm¹ x = 117 mm Properties of W 460 x 74 A = 9450 mm² b = 190 mm ly= 1.61 x 10 mm x = 17.7 mm tw = 9.0 mm rx = 188 mm ry = 41.9 mm d = 457 mm tr = 14.5 mm Ix = 333 x 10 mm Iy = 16.6 x 10mm* 7.21 When the height of column is 6 m. When the height of column is 10 m. Assume K= 1.0 457 CIVIL ENGINEERING- STEEL DESIGN504.2 Tensile Strength The design tensile strength, $¿Pns and the allowable tensile strength, Pn/N, of tension members, shall be the lower value obtained according to the limit states of tensile yielding in the gross section and tensile rupture in the net section. 1. For tensile yielding in the gross section: Pn = F,A. (504.2-1) P: = 0.90 (LRFD) N, = 1.67 (ASD) 2. For tensile rupture in the net section: Pn = FµAe (504.2-2) Pe = 0.75 (LRFD) h = 2.00 (ASD) %3D %3D An angle bar 100x100x11 mm tension member is connected with 20-mm-diameter bolts as shown in the figure. Both legs of the angles are connected. Use Fy = 248 MPa and Fu = 400 MPa. 6 @ 50 mm 63.5 mm
- The beam is constructed from three boards. (Figure 1) The allowable shear stress for the wood is τallow=465 psiτallow=465 psi. Each nail can resist a shear force of 395 lblb. -Determine the maximum loads P that it can support. -What is the maximum allowable spacing ss of the nails used to hold the top flange to the web? (Consider only regions AC and BD.) -What is the maximum allowable spacing ss of the nails used to hold the bottom flange to the web? (Consider only regions AC and BD.)A 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0. Determine the design tensile strength of the section based on yielding of the gross area. Answer: |0 O O О оDetermine the force of the members for the space truss shown in the figure. Joints A and B are supported by ball-and-socket while Joint C is supported by short link along z-axis. Indicate whether member is tension (T) or compression (C). 2m y Sm Az By BX Bz PAR Ax CSXScanned with Ca canner a SISU KN 15m 1 2117
- A 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0. a) Determine the allowable tensile strength of the section based on tensile rupture of the net area.b) Determine the allowable tensile strength of the section based on yielding of the gross area.For the column shown what is the nominal compressive strength (Pn) ? Pn=? W12x50 10 12X50 THT X combilevered WITHE BEAM SHOWN BELOW HAS THE FOLLOWING PROPERTIES. f’c =30 MPa, fc=13.5 MPa, fy=400 MPa, fs=200MPa. DETERMINE THE FOLLOWING. b)THE UNCRACKED SECTION PROPERTIES KdAND IU.
- Question 4 Complete the table below for the steel sections shown. All sections are bending about their horizontal axis. Section Location hep Ney Slenderness (show calculations) of element bf Flange te ld ++ tw be = 800 mm Web tf = 20 mm d = 1440 mm tw = 16 mm Grade 300, heavily 入。= Whole Asp= Asy= welded plates section br Flange tr d Web bf = 800 mm te = 20 mm d = 1420 mm tw = 16 mm Grade 300, heavily welded plates 入。= Whole Asp= Asy= sectionQ1: For the simply supported beam of the cross sectional dimensions and load shown in the figure, determin 3- The spacing required for nails A and B if they stand a force of 2 kN. S = 1- The max. bending stress in the beam. 2- The max. shear stress in the beam. F Fblt vQ 9 = ON 나오 50 Fail A w= 10 kN m 150 2 m 4 m 150 50 Nail B All dimensions in mm 50 150 50Problem 3: Method of Sections Inputs: ● ● h = 12 ft b = 7 ft F₁ = 2100 lb F₂ = 800 lb 0 = 25° You must use the method of sections as described below. You will not receive credit if you use the method of joints! Take a horizontal section cut through the middle of the truss and keep the top portion for your FBD. That way you don't have to bother with the reactions at A and D. Using the FBD of the top portion of the truss, perform the following steps: F₁ B a) Use EFX = 0 to compute BD. Indicate tension vs. compression. b) Use MD = 0 to compute AB. Indicate tension vs. compression. c) Use [MB = 0 to compute CD. Indicate tension vs. compression. d) Use [Fy = 0 to check all values D F₂