design tensile strength
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A: Given data Steel details As = 600 mm2 Es = 200 GPa = 2 × 105 N/mm2 Bronze details AB = 300 mm2…
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A: To Determine Force at member GH, FGH
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A: Poissons ratio (μ) = 1/4 = 0.25 E = 72 GPA = 72 × 103 N/mm2 Force in x direction (Fx) = 210 KN (C)…
Q: rbitrary point, calculate the principal stresses and the бу: 2 = 100 MPa s Txy = 55.9 MPa x=200MPa
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Q: the true length of the line
A: To find: True length of Line
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Q: Q1: The C-shaped steel bar has cross section of a square of side 1.6 in. If the magnitude of…
A: Direct tensile stress=PA=21.62=0.78125ksi
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Q: Q1: The C-shaped steel bar has cross section of a square of side 1.6 in. If the magnitude of…
A: Direct tensile stress=PA=21.62=0.78125ksiDirect Bending stress=Pe(y)I=2×3.2×1.621.6412=9.375ksiso,
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A: ∑MA=012×8×4×13×4 + 10 cos36.87°×1 + 10 sin36.87°×3 + 12×6×3×4+23×3 - 33 - VD×7 - 5×4×2 -…
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A: TO FIND: BASE PERIOD
Q: Problem 2. Determine the equations for shear and bending moment for the beam shown.
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- 5. a bolted lap joint shown in the figure below. The bolts are 20m in diameter in 23 mm holes. The plates are 12m thick X350 Allowable stress of Plates: Tenaion in gross area0.60Fy Tension in net area0.50 Fu Shear on Net Area 0.30Fu Yield Strength of plate, Fy 248 Npa Ultinate tensile strength of Plate. Fu400 Mpa x, Find the safe load P based on a- Gross area yielding b. net area rapture c. block shear 2 of 3an angle bar 100x100x11 mm tension member is connected with 20-mm-dia bolts as shown in the fig. both legs of the angles are connected. use fy=248 MPa and Fu=400 MPa. 1. determine the nominal strength for tensile yielding in the gross section. a. 515.59 kN b. 429.67 kN c. 545.60 kN d. 390.10 kN 2. determine the effective net area of member a. 1573 mm^2 b. 1632 mm^2 c. 1605 mm^2 d. 1815 mm^2 3. determine the nominal strength for tensile rupture in the net area. a. 642 kN b. 730 kN c. 554 kN d. 650 kNDetermine the design tensile strength of plate (200x8 mm) connected to 10-mm thick gusset using 20 mm bolts as shown in the figure, if the yield and the ultimate stress of the steel used are 250 MPa and 410 MPa, respectively. Add 1mm around the bolt for the hole. Use LRFD method. Plate 8-mm thick 2 3 40+ 30 301 T 200 mm Gusset 10-mm thick 3af 30 2_3 *40 40+ 50,54 +40
- Two steel plate tension members have been connected using 0.72” diameter bolts arranged in an equally-spaced four by four square formation.Total plate self-weight is specified as 912 pounds.Design of the elements adhered to the set of values that are twice as much as the minimum requirements. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 54ksi and the min required edge distance as 0.125ft. a)Sketch the connection showing all the details and measurements with units. b)Find the maximum allowable service dead load(excluding the self-weight)and live load assuming live load is half as much as dead load including the self-weight of the plates.If the allowable bearing stress in the bolt is 18.855 ksi. what is the minimumrequired diameter of the bolt at B? b= .75” h= 7.874” Distances are in feet.A plate 400 x 12 mm is to be connected to a plate of the same width and thickness by 34 mm diameter bolts, as shown. The holes are 2 mm larger than the bolt diameter. The plate is A36 steel. Assume allowable tensile stress on net area is .60Fy. It is required to determine the value of b such that the net width along bolts 1-2-3-4 is equal to the net width along bolts 1-2-4. a. calculate the value of b in millimeters. b. Calculate the value of the net area for tension in plates in square millimeters. c. Calculate the value of P so that the allowable tensile stress on net area will not be exceeded. d. Calculate the nominal block shear strength based on possible failure paths
- The given plate below with width of 200 mm andthickness of 16 mm is to be connected to two plates ofthe same width with half the thickness by 20 mmdiameter rivets as shown. The rivet hole is 2 mm greaterthan the rivet diameter. Allowable tensile stress on netarea is 0.6Fy. Allowable bearing stress is 1.35Fy. Use a501 plate and a502 gr2 rivet a. Determine maximum load P without exceeding allowable tensile stress on plate b. Determine maximum load P without exceeding allowable shear stress on rivets c. Determine maximum load P without exceeding allowable bearing stress between plates and rivetsA 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 answersDetermine: Ix & ly = Moment of inertia about axis XX and YY Rx & Ry = Radius of gyration about axis XX and YY Sx & Sy = Elastic modulus about axis XX and YY steel specs are here http://www.toolsforengineer.com/steel-sections-properties-tables/ FOR WIDE FLANGE = W250X22FOR ANGLE BAR = L32X32X6.4
- What is the minimum required bolt diameter (mm) at C ifthe allowable bolt shear stress is 68 Mpa.a. 9 mm c. 11 mmb. 10 mm d. 12 mm Determine the stress (Mpa) in strut BC.a. 5.6 MPa c. 3.2 MPab. 2.8 MPa d. 6.4 MPa Calculate the bolt stress (Mpa) at A if its diameter is 20mm.a. 7.64 MPa c.11.46 MPab. 3.82 MPa d.14.92 MPapls help! Write the complete solutions and legibly. Answer in 2 decimal places. UPVOTE WILL BE GIVEN! MECHANICS OF DEFORMABLE BODIES Let Q = 0 and T = 8 (Example: 12Q°C = 120°C). Just substitute the number in letter. Thank you! For the frame shown, a 4QQT-N load is acting on member ABD at D. If the allowable material shear stress for is 4Q MPa, determine the required diameter (rounded off to the nearest 2.5 mm) of the pins at C and D. Pin C and pin D are subjected to double shear and single shear, respectively. If the thickness of member BC is 12 mm and that of member DE is 16 mm, determine the maximum bearing stress at C.Determine the nominal tensile strength of the single angle. Consider 7/8" diameter bolt and A242 steel with Fy = 50 ksi and Fu = 70 ksi. L6x6x3/4 d = 6 in. b = 6 in. t = 0.75 in. k = 1.25 in. wt./ft. = 28.8 plf. A = 8.46 in.^2 x = 1.77 in.