Q.3 The steel column shown Fig. 3 is pinned at both ends and unbraced about both x-x and braced at mid-height y-y axes. Calculate the axial capacity of a W14 x 99 section A992 (Fy = 50 ksi) to assess its the adequacy of a for the given conditions. PD = 250 k PL = 500 k L=32' Ž FIG. 3 T L=16'
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- Determine 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 21177 7a 7b 7c Alaterally supported beam was designed for flexure. The beam is safe for shear & deflection. The most economical section is structural tubing however the said section is not readily available at the time of the construction. If you are the engineer in charge of the construction what alternative section will be the best replacement? Why? The section is 8" x 8" x 7.94 mm thick: Use Fy=248 MPa: E=200,000 MPa AISC wall thickness Ix 106 S x 103 Jx 103 mm4 mm3 mm4 rx =ry Area Ag (mm2) mm Designation Weight/m 8x8 7.94 47.36 6,039 79.25 37.84 371.99 60.35 expla'n briefly your cho'ce. (transform your comparative analys's 'nto a narative form to support your cho'ce) 8x8 14.29 80.61 10,258 76.2 59.52 585.02 99.06 8x8 72.7 9,290 76.96 54.53 539.13 90.32 8x8 9.53 56.09 7,161 78.48 44.12 432.62 70.76 mm 12.7 Zx 103 mm3 437.53 714.48 650.57 512.92determine the member stresses of the truss shown. indicate if the stress in the member is in tension or compression. present complete and neat solutions. p1-30kN W-25kN/m P2=51kN AH Av Im P₂ Im P₂ Im P₂ Im Av W 1.6m
- Determine 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 DESIGNThree plates (14 mm and 16 mm in thickness) are welded to a W10 x 49 to form a built-up shape as shown. K.L= K.L= 7.6 m and F= 345 MPa, compute the design strength for LRFD. Mm x300mm WIOX49 14mm Piate PlateW3D 25 Y= 155 %31 1oow= 2500 KN 27=310mm X= 475 Z= 60 %3D 102=600mm
- Given: The T- beam cross section in Figure 3. Find: Calculate the following: a. The gross moment of inertia, Ig b. The location of the neutral axis of the cracked section and Icr. c. The effective moment of inertia, lê, assuming that Ma = 0.6M₁. (don't include the 2 # 4 bars in the top flange as compression steel when calculating M₁) f=5000 psi NWC fy = 60,000 psi Cc = 1.5 in #4 stirrups -b₁ = 50 in 20 in 10 in Figure 3-T-Beam cross section. 2 #4 2 # 10 6 in 18 inAW 12 x 65 column section shown in the figure is pinned at each end and has an additional support in the weak direction at a point 4.1 m from the top. Use X- 4.8 m K 1 for both directions. X+Y Properties of W 12 x 65 A 12323 mm2 - 134.11 mm ry 76.71 mm Fy- 248 MPa Determine the allowable compressive strength (kN).Determine the design moment capacity of the beam with given properties below. bf = 550mm bw 350mm tf = 100mm %3D d = 370mm %3D d' = 100mm %3D fc' = 34.5 MPa fy = 400 MPa %3D As = 6-32mm bars a USE NSCP 2001, Mu = kN-m b. USE NSCP 2015, Mu = kN-m
- x Determin the design Compressive 12-5 Strength of the Cross- sectiona Column shown with unbraced length kala= 35m, Kyly=2m A-118:75 I= 10872 Iy= 3799 J-247 8a e =4.13 c ? J= 247-8 %3D Tu = 9.57 C 5.746 E=2-1x1. A steel column 10 m long is fabricated from a cover plate and C section arranged as shown. Determine the safe compressive load. Fy = 248 MPa, E= 200 GPa. Use AISC/NSCP Specs. 450 mm -cover plate 'I 12 mm y2 IP d2 10 m C 310 x 37 A = 4720 mm? d = 305 mm bf = 77 mm tf = 12.7 mm tw = 9.8 mm C 310 X 37 a) Both ends of column are fixed b) Both ends of column are hinged c) One end fixed, the other end hinged Use design values of k. tw d=305- Ix = 59.9x10° mm ly = 1.85x10° mm x = 17.1 mm x=17.1P,= 500 k J. P, = 1,000 k I P,= 500 k Py=1,000 k fixed fixed Check whether a 17 ft W 14 x 211 section of A992 steel can safely carry the given applied service loads pinned 35 ft 18 ft pinned pinned Strong Axis Weak Axis