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From See. 3-12, the maximum shear stress in a solid round bar of diameter. d, due to an applied torque. T, is given by
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Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
- A strain gauge mounted at a potentially critical point in a steel part has recorded the stress history shown below during 20 seconds of typical use. Identical parts have been fatigue tested under constant amplitude loading with R = -1 to give an endurance limit of 60 ksi and a fatigue strength of 140 ksi at N = 1000 cycles. The steel used has an ultimate strength of 165 ksi. Estimate the fatigue life of the part under typical use. How much will the fatigue life in problem 3 (above) be reduced if a mean stress of 10 ksi is added to the stress history given?arrow_forwardA= A specimen of magnesium having a rectangular cross section of dimension 3.2)mm x 19.1/mm and a gauge length of 63.5 is pulled in tension. Use the load-clongation characteristics tabulated as follows to: Load 0.0 1.38 2.78 5.63 7.43 8.14 9.87 12.85 14.10 14.34 13.83 12.59 6.99 (KN) Al 0.0 0.03 0.06 0.12 0.20 0.25 0.64 1.91 3,18 4.45 5.72 (mm) a) Plot the data as engineering stress versus engineering strain. b) Compute the modulus of elasticity.-M. L. 445010 c) Determine the tensile strength of this alloy.arrow_forwardA material presents S-N fatigue behaviour shown on the right with the fatigue limit being 100 MPa. If a cylindrical bar made from the material has a diameter of 20 mm is subjected to repeated tensile stress along its axis, calculate the maximum allowable tensile load to ensure no occurrence of fatigue failure. Stress amplitude, S Fatigue limit 103 104 105 106 107 Cycles to failure, N (logarithmic scale) 108 10⁹ 1010arrow_forward
- gauges system mounted at a point on a structural member that is subjected to multiaxial stresses, where we have the following readings: Ex = 6.686 x104 ;&, 3.0 x10-4; & = - 6.857 x10-5; Yxy = 2.580 x10-3; %3D %3D Determine the stress components if the member is made of an isotropic material with linear elastic behavior having the following elastic properties: Young's modulus E = 70 Gpa and Poisson's ratio v = 0.29; Select one: Ox = 70 Mpa; Oy = 50 Mpa; o, 3 30 Мра; Тҳу 35 Mра; %3D 25 Mpa; 0z = 30 Mpa; Txy = 70 Mpa; %3D Ох %3D 140 Мра; бу %3D Ox = 90 Mpa; oy = 50 Mpa; o, = 60 Mpa; Ty = 35 Mpa%3; Ox = 70 Mpa; oy = 50 Mpa; 0z = 30 Mpa; Txy = 70 Mpa; = 70 Mpa3; = 100 Mpa; Oy = 50 Mpa; 0z = 80 Mpa; Txyarrow_forward1 2 A hollow circular steel strut with its ends position- fixed, has a length of 3 m, external diameter 12.5 cm and interne diam. eter 10 cm. Before loading, the strut is bent with a maximum deviation of 0.4 cm. Assuming the central line to be sinusoidal, determine (a) the maximum stress due to a central compressive end load of 8 kN. (b) If the load has an eccentricity of 1-5 cm, then find the maximum stress in- duced. Take E = 200 GPa.arrow_forwardIf load P = 20.1 kips, determine the normal force in bar (1). (1) A 6 ft 42.0 kips 28.2 kips 33.5 kips 45.4 kips 30.9 kips B 4 ftarrow_forward
- Q1: Find the Diameter of the shaft macle of A cold-drawn steel. Assume hat ultimate stress is 370 MPa and the factor of the safety is 3. For pulley A consider 250 Dia. the loose belt tension is 15 percent of the tension on the tight side. *All dimensions units are in mm. 300 Tz A 400 T₂ 50 N. 270 N. 45° 150. 300 Dia. T P x 60 2 KN T = (T₁ - T₂) Rc Tw M z 16 + T² xxx d³arrow_forwardQ.7. A cylindrical specimen of aluminium having a diameter of 12 mm and gauge length of 48.000 mm is pulled in tension. Use the load-elongation characteristics tabulated below to complete problems a through h. Load (N) Length (mm) Load (N) 48,00 48,05 48,10 48,15 48,20 Length (mm) 49,93 50,89 51,85 52,81 53,77 54,49 55,21 42336 6927 43659 14270 44699 21830 44888 28728 43565 32508 36288 42336 48,25 48,49 48,97 a) Plot the data as stress versus strain 40257 39029 34398 55,93 b) Compute the modulus of elasticity c) Determine the yield strength (Hint: Yield point not apparent)arrow_forwardA mild steel bar went through a tensile test conducted at the University of San Carlos Mechanical Engineering Laboratory. The following data was obtained from the test: i. Diameter of the steel bar = 30 mm ii. Gauge length of the bar = 200 mm iii. Load at the elastic limit = 250 kN iv. Extension at a load of 150 kN = 0.21 mm v. Maximum load = 380 kN vi. Total extension = 60 mm vii. Diameter of the rod at the failure = 22.5 mm Determine the Young’s modulus, the stress at the elastic limit, the percentage elongation, and the percentage decrease in area.arrow_forward
- Table Q4 belowindicates results on a tensile test of a mild steel specimen. 95 kN Proportional Limit load Original diameter Original ga uge length 30 mm 50 mm Yield Point Load 100 kN Extension at proportional limit 0.050 mm Maximum recorded load 160 kN Final length between gauge points 58.8 mm Final minimum diameter 17.9 mm Table Q4 Calculate the following: a. Modulus of Elasticity for the mild steel. b. The proportional limit stress. C. The ultimate tensile strength. d. The percentage elongation. e. The percentage reduction in area.arrow_forward1. A safety valve must start opening when the force is equal to 777.48N, and be fully open when (20mm spring compression) at a force of 812.82N. The spring should not be more than 90mm outside diameter, with the maximum shear stress under 1100MPA. The material used has a shear modulus of (G = 77.2 GPa). Determine the springs Mean Diameter, Wire diameter, Number of active coils and unloaded Length. D C = d 4С — 1 0.615 8FD Kw T = K,- nd³ + 4С — 4 C - 8FD³N y = 8FD³N dªG 1+ 2C² dªGarrow_forwardThe induced nominal stress in a machine member is 28 N/mm? and the value of maximum stress near the irregularities is 61N/mm?, the value of stress concentration factor is SCF, K = Bearings are designated by a standard number internationally, the first digit in the bearing number 6409 represents type of the bearing. Select one: O True O False The value of curvature correction factor k; will be . if the spring index is 6. O a. 1.33 O b. 1.08 O c.0.,92 O d. 4.00 The figure below shows the components of normal forces, the value of radial force on the gear wheel due to a tangential force of 642 N is . N, Take pressure angle as 1420. Pitch point P. Gear driven 02 The radial force isarrow_forward
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