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- 2. A column HP 14 x 102 of A572 Gr. 55 steel has a length of 15 ft is fixed at both ends. Compute the design compressive strength for LRFD and the allowable compressive strength for ASD. (Steel section properties are provided in the next page) ASTM Designation A572 Gr. 42 Gr. 50 Gr. 55 Gr. 60⁰ Gr. 65⁰ Yield Stress (ksi) 42 50 55 60 65 Fu Tensile Stressa (ksi) 60 65 70 75 80Topic: COMBINED STRESS-AXIAL TENSION AND FLEXURE BENDING: STEEL DESIGN Please solve your Solution in a handwritten Note: It should be handwritten pleaseeee Questions A Tension member with no holes is subjected to axial loads of PD=68kN & PL=64kN.It is also subjected with bending moments of MDy=40kN-m & MLy=55kN-m. Is themember adequate? Steel is of A992 Gr 50 Specs. Use LRFD. Neglect the weight of the beam. Section properties:Lp = 1.863 md=459.99mmtw=9.02mmzx= 1835x10^3 mm^4Lr = 5.305 mLb = 4.8 mbf=191.26mmtf=16mmSx=1611x10^3 mm^4 Sy=195x10^3 mm^4Zy=303x 10^3 mm^4 Cb=1.324. Calculate the design strength (ocPn) of W24X76 with length of 12 ft. and pinned ends. A572 Grade50 steel is used. E=29x103 ksi. Show your work in detail. ASTM Classification A36 A572 Grade 50 A992 Grade 50 A500 Grade B (HSS rect, sq) A500 Grade B (HSS round) A53 Grade B Yield Strength F, (ksi) 36 50 50 46 42 35 Ultimate Strength F (ksi) 58 65 65 58 58 60
- The given angle bar L125x75x12 with Ag = 2,269 sq.mm. is connected to a gusset plate using 20 mm diameter bolts as shown in the figure. Using A36 steel with Fy = 248 MPa and Fu = 400 MPa, determine the following: 2. Determine the nominal tensile strength of the 12 mm thick, A36 angle bar shown based on: a. Gross yielding b. Tensile rupture Bolts used for the connection are 20 mm in diameter. O O O O O O O Effective net area of the tension member if the shear lag factor is 0.80. Select the correct response: 1,516.1 1,354.4 1,431.2 1,221.6The tension member shown below is C12 x 20.7 of A36 steel. Will it safely support a service dead load of 160kN and a service live load of 325kN? Use equation 3.1 for U. A. Use LRFD B. Use ASD Note: Use SI units (mm) for the following dimensions; Use CSI Steel for the Steel Properties. 12" 22" 22" 22" ооо ооо оооо 7/8-in.-diameter bolts C12 x 20.77 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 437.53 714.48 650.57 512.92
- 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'For the column shown what is the nominal compressive strength (Pn) ? Pn=? W12x50 10 12X50 THT X combilevered WITENSION MEMBERS: THE SINGLE 200 X 10 mm STEEL PLATE IS CONNECTED TO A 12 mm THICK STEEL PLATE BY FOUR 16 mm DIAMETER RIVETS AS SHOWN IN THE FIGURE. THE RIVETS USED ARE A502 GRADE 2, HOT DRIVEN RIVETS. THE STEEL IS ASTM A36 WITH Fy = 248 MPa AND Fu = 400 MPa. DETERMINE THE VALUE OF P. a. P BASED ON TENSION OF GROSS AREA b. P BASED ON TENSION OF NET AREA c. P BASED ON BEARING OF PROJECTED AREA d. P BASED ON SHEAR RUPTURE (BLOCK SHEAR)
- Situation 2. Two plates each with thicknessF16mm are bolted together with6 -22mm dimater bolts forming a lap conne ction. Bolt spacing are as follows: S1 = 40mm, S2 = 80mm, Sa = 100mm. Bolt hole diameter=25 mm 50 250mm 30 30 60 75 Allowable stress: Tensile stess on gross area of the plate=0.60 Fy Tensile stress on net area of the plate=0.5Fy Shear Stress of the bolt Fv=120MPA Bearing Stress of the bot Fp=1.2 Fu Calculate the permiss ible tensile load P under the following Conditions: 4. Based on shear capacity of bolts 5. Based on bearing capacity of bolts 6. Based on block shear strength (taking into consideration the failure path given in the figure below) 40 80 16 mm 40 180 mm 40 Bearing Failure Path #1 140 209 Bearing Failure Path #21. 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=17.1tf Sec Ag tw W16x100 29.4 0.585 0.985 Bolt (7/8 in) A structural analysis was done for -1 the steel section whose shape and characteristics are shown below, and the outputs were arranged in the table below. Choose the appropriate amount, knowing that .all numbers are in units (ksi, in) O fy ( 36 fu 0.85 20.149 584.3 635.04 58 D(bolt) 7/8 P Maximum tensile force (P) that can be exerted Effective cross- sectional area (Ae) The value of T1 Modulus M (mo) !