A SHAFT COMPOSED OF SEGMENTS AC, CD, AND DB IS FASTENED TO RIGID SUPPORTS AND LOADED AS SHOWN IN FIGURE. FOR BRONZE, G = 35 GPA; ALUMINUM, G = 28 GPA, AND FOR STEEL, G = 83 GPA. DETERMΜΙNE THE ΜAXIMUM SHEARING STRESS DEVELOPED IN EACH SEGMENT. Te = 300 Nm C Aluminum D To = 700 N-m A Bronze Steel 25 mm o 25 mm ở 50 mm 2 m 2 m 2.5 m B.
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- 4 A flanged bolt coupling has ten 12-mm di ameter steel bolts on 500 mm diameter b olt circle and six 16 mm diameter aluminu m bolts on 300 mm diameter bolt circle. T he maximum shear stresses of the materi als are 60 MPa in steel and 40 MPa in al uminum. Use G = 80 GPa for steel and 3 0 GPa for aluminum. What is the maximu m required shear strength of each alumin um bolt to determine the maximum torqu e that can be applied to the system? Dra wing not included in this problem.An A36 steel plate with width of 400 mm and thickness of 12 mm is to beconnected to a plate of the same width and thickness by 34 mm diametersbolts, as shown in the figure (next slide). The holes are 2 mm larger thanthe bolt diameter. The yield strength of the steel plate is Fy = 248 MPa.Assume allowable tensile stress on net area is 0.60Fy. It is required todetermine the value of b such that the net width along bolts 1-2-3-4 isequal to the net width along bolts 1-2-4.a. Calculate the value of b in mm.b. Calculate the value of the net area for tension in plates in mm².c. Calculate the value of P so that the allowable tensile stress on net areawill not be exceeded.(Use at least 2 decimal points on your answers.)The 10-mm-diameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is 20 mm, and its inner diameter is 10 mm. If the yield stress for the steel is (ay) MPa, and for the bronze (ay)or 520 MPa, determine the magnitude of the largest elastic load P 640 that can be applied to the assembly. E 200 GPa, Ebr= 100 GPa. 10 mm 20 mm Figure 2: Steel Bolt
- (1) Given A992 steel Ag = 10 in %3D KLy- 80, ry what is $ Pa=? = 60 the' column streugth design7 Enercise:- 30 long with is subjected to a Assuming deformation is entirely elastic and manimum allowable elongation tensile 30000y that the is 0.3 mm while manimum. allowable reduction in, digmeter is 0.015mm. choose amory below only. candidate meeting requirent E (GPa) A cylindrical rod 100 mm diameter of 12 mm load of 30000 N. Aluminium alloy Bronze Titanium allow 70 100 110 U(Poiss ons vahi) 0.33 0.34 0.3H.W A Solid Steel bar of dlameter 6omm and length 350 mm is Placed inside an aluminum cylinder öf Inside diameter Femm and out side d lameter Ilomm, a comPressive load of looo lais applied on the assembly, find the stresses in the bar and the cylinder ? take Est = 200GPa and EAl = F0GPan %3D Cover Plate 0.25mm d- do 35omm A lum. cylinder Steel bar
- Determine 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 +402) Find the axial stresses of menbers FD, GD, GE State if it is tensile or Compressive. 4M 3M A LE 3m G 20 RN Go KNA 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
- H.W (2): A 37.5 mm round bar has been machined from AISI 1050 cold-drawn round bar Su =689 MPa. This part is to withstand a fluctuating tensile load varying from 0 to 72.5 kN. Because of the design of the ends and the fillet radius, a fatigue stress-concentration factor of 1.85 exists. Find the factor of safety F.S.A single unequal angle 100 x 75 x 6 is connected to a 10 mm thick gusset plate at the ends with six 16 mm diameter bolts to the 100 mm of leg to transfer tension as shown in figure. What will be the block shear strength (in kN) of the angle section assuming that the yield and the ultimate stress of steel used are 250 MPa and 410 MPa respectively? ISA (100 × 75 x 6) -T 16 mm o bolt 10 mm 40 mm 40 mm| 40 mm 40 mm 40 mm 40 mm 10 mmTopic: 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=1.32