(2) 3#9 12" 21" I 3" fé=3000 psi Grade 60 steel NWC. & Mn = ?
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- Derive the maximum steel ratio and the recommended rteel ratio fe' O.856 fy Po-004 3 0.85 fc' pooos= 0-85In the connection shown, 1/4-in. side and end fillet welds are used to connect the 3-in.-by-1 in. tension member to the plate. The applied load is 60,000 lb. Find the required dimension L. The steel is ASTM A36 and the electrode used is an E70. - Fillet welds L 3" Activate Windows Go to Settings to activate Wir TABLE 19.2 Allowable stresses in ksi (MPa) 12: 12/1 DELL end F12 insert prt sc F10 home F11 F6 F8 F9 F3A L 3 x 2 x ¼ is connected to a gusset plate via six bolts. The nominal diameter of the bolt is 0.25 inches, the pitch spacing is 1.75 inches, the gage spacing is 2 inches, and the thickness of the connection is (¼) inch. The yield stress is 50 Ksi and the ultimate stress is 60 Ksi. Consider section line (a-a') for the analysis. . a. What is the effective net area (Ae) of the angle section in inches2 [a-a']?(The shear lag factor is "0.80") b. What is the design tensile yielding strength in Kips for the steel member? c. What is the design tensile rupture strength in Kips for the steel member? d. What is the minimum Factor of Safety if a tensile load of 23 Kips is applied to the angle section? please make sure the answer is correct 100% I only need the final answers
- GIVEN: Solid alumınum bar shown: 4 in. by4 in. E = 10 x 10° psi oy = 36 ksi 4000 2000 1b 4000 16 B 12' REQ’D: A) Load in each section of the bar (state whether C or T) PAB = Pac PCD = B) Maximum axial stress in the bar. In which section does this occur? C) Final length of the barWhat 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 areaDetermine the stress and total load on each stringers if the web is negligible. Estainless steel = 28000 ksi, Esteel = 29000 ksi, EMagnesium = 6500 ksi, EAlum. Ally = 10500 ksi 6" Steel 10 .10 a Stainless Steel My = 5000"# 10" Mx = 10000"# Magnesium .1 Alloy Area of Each 10 Alum. Alloy Stringer = 1 sq. in. 4"
- Question j=169" R=2.1" ;f 4- 2.81 d $ 100 08 96 07 h £=0.5" * # >P d=7.3" Material is ASTM A36. Plate is fixed 511 by 4 screws of {"$. using the information provided, calculate the ultimate stress (factored by LRFD) for shear fracture and tensile rupture of block pattern passing through points the shear 3-10-9-4..Subject: Steel design- Bolted Steel Connection *Use NSCP 2015 formula/guide to solve this problem *Use Handwritten A plate having a width of 400 and a thickness of 20mm is to be connected to two plates with thickness of 12mm by four rivets. The allowable shear stress for the rivets is 150 MPa and for bearing is 1.35 Fy. If the yield strength and the ultimate strength is 248 MPa and 400 MPa respectively. Determine the Following: a.) Required diameter of rivets w/o exceeding the aloowable stress in the shear. b.) Required diameter of rivets w/o exceeding the allowable stress in bearing. c.) Required diameter of rivets w/o exceeding the allowable stress in tension.STRENGTH OF MATERIAL(SIMPLE STRESS) PLEASE INCLUDE FBD 100% GUARANTEE UPVOTE
- 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 opyWhat is the axial stress in the bronze rod (in MPa)? 2m 3m 6o kN mm 15mm banzerodCentroid + Inertia 200 x 20 mm Plate The strength of the rolled section shown is increased by combining S section, channel, plate and then welded. 1. Determine the location of its centroid; and 5310 x 47.3 C310 x 30.8 2. Determine the moments of inertia of the composite section with respect to its centroidal x and y Semi-circle Weld, 10mm radius axes.