(c) Fig. Q3 shows the S-N curve of a metal rod with a diameter of 14 mm. The metal rod is subjected to a cyclic loading about a mean stress of zero. If the maximum cyclic tensile load applied is 4730 N, would the metal rod fail by fatigue and why? Stress Amplitude, σ (MPa) 20 10 50 100 90 80 0 100 1000 10000 100000 1e6 1e7 1e8 Number of Cycles to Failure, N₁ Fig. Q 3 S-N curve of the metal rod [8 marks]

1- Spiral fractures of the tibia are common in ski accidents and are caused by torsional load on bone. Ski binding are designed to release the boot when a desire torque is applied to the leg. The bone has a shear modulus of G-3.5 GPa. The bone can be modelled as a hollow cylinder with inner diameter of 1.5 cm and outer diameter of 5cm. If the bone will fracture (failure) at the shear stress of 17MPA and the design of binding must not allow the bone to reach to the maximum shear stress of 60% of the failure shear stress. a) Determine the value for the maximum torque that the release mechanism should act to prevent bone fracture b) Determine the angle of twist when torque reach to the value obtain from part a 45cm (Answer: a) 248.31 N.m; b) o = 0.052 rad ) %3D A R

1- Rankin's theory of failure is applicable for following type of materials (a) Brittle (b) ductile (c) elastic (d) plastic 2- Stress concentration is caused due to (a) Variation in properties of material from point to point in a member (b) Pitting at 'points or areas at which loads on a member are applied (c) Abrupt change of section (d) all of the above (e) none of the above 3- Yield point in fatigue loading as compared to static loading is (a) Same (b) higher (c) lower (d) depends on other factors (e) none of the above. 4- Transverse fillet welded joints are designed for (a) Tensile strength (b) compressive strength (c) Shear strength (d) bending strength (e) torsional strength. 5. The stress represented by cos (t) belongs to which category? a) Fluctuating Stresses b) Alternating Stresses c) Repeated Stresses d) Reversed Stresses 6. If the mean stress value for a sinusoidal stress function is zero, then this type of stress falls in which category? c) Repeated Stresses b) Alternating…

1- Rankin's theory of failure is applicable for following type of materials (a) Brittle (b) ductile (c) elastic (d) plastic 2- Stress concentration is caused due to (a) Variation in properties of material from point to point in a member (b) Pitting at 'points or areas at which loads on a member are applied (c) Abrupt change of section (d) all of the above (e) none of the above 3- Yield point in fatigue loading as compared to static loading is (a) Same (b) higher (c) lower (d) depends on other factors (e) none of the above. 4- Transverse fillet welded joints are designed for (a) Tensile strength (b) compressive strength (c) Shear strength (d) bending strength (e) torsional strength. 5. The stress represented by cos (t) belongs to which category? a) Fluctuating Stresses b) Alternating Stresses c) Repeated Stresses d) Reversed Stresses 6. If the mean stress value for a sinusoidal stress function is zero, then this type of stress falls in which category? a) Fluctuating Stresses d) Reversed…

For a given homogeneous, isotropic, linearly elastic material, E = 15e6 psi and v = 0.3. Solve for the shear modulus. 2.1.1 Homogeneous, isotropic, linearly elastic materials For specimens undergoing small deformations, the stress-strain diagram often ex- hibits a linear behavior. Although this is a very crude approximation to the behavior of actual materials, it is a convenient assumption that is often used for preliminary evaluation. A linear relationship between stress and strain can be expressed as 01 = E €1, (2.1) where the coefficient of proportionality, E, is called Young's modulus or modulus of elasticity. Since strains are non-dimensional quantities, this coefficient has the same units as stress quantities, i.e., Pa. This linear relationship is known as Hooke's law. The elongation of a bar in the direction of the applied stress is accompanied by a lateral contraction that is also proportional to the applied stress. The resulting defor- mations for this uniaxial state of stress…

Two forces are applied to a flywheel 50 cm in diameter. Each force is 500 N. The shaft is fixed to the wall using three grade 5 steel bolts. The distance between the bolts and the axis of rotation is 10 cm.    (a) What moment of resistance must develop the three bolts together    Answer: 250 Nm    (b) What is the resistance force F of each bolt?   Answer: 833.3 N    (c) What is the diameter of the bolts if their factor of safety is 10?    Answer: 4.14mm

nt 2 poa page 25 ol 40) - Google Orome nline.gmitie mod/quz/attempt.phplattempt-514100omid-4251718pages24 GMIT 25 A strel -200 Ga rod with a circular cross section k long Determine the minimum diameter reguired if the rod must transmit a tensile force of 50N withut eceeding an alowable stress of 180 M State your answer in mm without including the units and correct to two decimali places Answer ous page Nt pg 9ype here e ah

7:Y. O %Y. O 36 ll ll く ロ quiz 2 elearning.uowa.edu.iq Marked out of 1.00 P Flag question quiz 2: flexural stress which fluctuates between + 250 MN/m2 and - 115 MN/m² applied on a machine component. Calculate the ultimate strength value according to Goodman's relation. Take yield strength = 0.55 Ultimate %3D strength; Endurance strength = 0.5 Ultimate strength; and factor of safety = 1.5. A- В I D II

From the shear stress - shear strain diagram shown below, all the following are true except: Shear Stress (MPa) 500 450 400 350 300 250 200 150 100 50 0. 0.2 0.4 0.6 0.8 1.2 1.4 1.6 1.8 Shear Strain (rad) Shear Stress-Strain Curve for Brass Select one: O a. Shear stress and Shear strain are 430 MPa and 1.72 rad at fracture. O b. Shear stress and Shear strain are 250 MPa and 0.05 rad at Yielding. Oc. The elastic zone ends at a shear stress of 400 MPa and shear strain of 1 rad Od. The ultimate Shear stress and Shear strain are 430 MPa and 1.72 rad.

A tensile test was conducted on a mild steel& the following data was obtained as follows. Diameter of the steel bar = 3 cm,Gauge length of the bar = 20 cm. Load at elastic limit = 250 kN . Extension at a load of 150kN = 0.21 mm . Maximum load = 380 kN . Total extension = 60 mm . Diameter of the rod at failure = 2. 25 cm Determine: (a) Young’s modulus (b) stress at elastic limit (c)the percentage of elongation (d) Percentage decrease in area.