Figure 1 shows a composite shaft ABCD acted upon horizontal forces. The
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A: dear Student i have explained in detail . kindly refer the below solution
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- Torsional T is applied to the hollow axis (G=80 GPa). Ewha Womans University, the maximum shear strain is γmax = 640 × 10-6rad Igo, inside andThe outside diameters are 120mm and 150mm, respectively.a. What is the maximum axial strain?b. Maximum axial stress?c. What is the maximum torque (T) meeting the above conditions?K=17 so 107 MPa strength of material question.If the shear modulus of a material is 120GPa & the bulk modulus is 140 GPa, then Young’s modulus is------ a. 820 GPa b. 280 Pa c. 280 GPa d. 820 Pa
- – The weight is suspended from structural A–36 steel alloy and 2014-T6 aluminum alloy wires, eachhaving the same initial length of 10 ft and cross-sectional area of 7×10–3 in2 (Figure Q1). If the weight is increasedgradually, determine which wire yields first? Explain which material model and yield criterion you will use? Justifyyour choicesPlease answer the problem completely. Provide a free body diagram. Final answer: • Pac = 8.84 kN (Compression) • Normal Stress = 8.84 kN (Compression)Find the amount of depression in mm at the midpoint of the bar AB. Mo= 1kNm, L = 1m, E = 200 GPa, I=1x106mm4
- BD rod made of G1 = 78 GPa with a diameter of 27 mm attached to the CA tube at point B, the tube made of G2 = 133 GPa had an outer diameter of 59 mm and a wall thickness of 10 mm. If T1 = 935 Nm and T2 = 1683 Nm, answer the following questions: 1- It is the polar inertial moment of the bar BD. 2- Maximum shear stress BD. 3- It is the polar moment of inertia of the CA tube. 4- It is the maximum shear stress BD. 5- It is the maximum shear stress of the CA tube. 6- It is the maximum shear stress of the assembly. 7- is the torsion angle between D and B. 8- It is the torsion angle between C and A. -9 is the torsion angle at D. 10- is the torsion angle between C and B. 11- is the reaction at point A.The cross sectional area of the bar is 5cm^2. Use E = 70GPa for aluminum, and E = 200 GPa for steel What is the elongation in segment AB (Aluminum)? a. 7.8571 x10^-3 m b. 6.8571 x10^-3 m c. -7.8571 x10^-3 m d. -6.8571 x10^-3 m What is the elongation in segment BC (Steel)? a. -3x10^-3 m b. 3x10^-3 m c. 2 x10^-3 m d. -2 x10^-3 m What is the elongation in segment CD (Aluminum)? a. -6.8571 x10^-3 m b. 6.8571 x10^-3 m c. -2.8571 x10^-3 m d. 2.8571 x10^-3 m What is the total elongation of the bar? a. 5 x10^-3 m b. 6 x10^-3 m c. 4 x10^-3 m d. -6 x10^-3 m -Draw and label the diagram correctly, No diagram in the solution will be marked wrong. -Shortcut solution will be marked wrong.- Direction of the assumption of the equilibrium equation must be shown, no direction will be marked wrong.Find the amount of depression in mm at the midpoint of the bar AB.Mo = 1kNm, L = 1m, E = 200 GPa, I=1x106 mm4
- Find the amount of slump at the midpoint of the bar AB, in mm. Mo= 1kNm, L = 1m, E = 200 GPa, I=1x106mm4(6) A key is to be designed for a 12.7cm shaft which will transmit power of 150kW @ 360 rpm. If the allowable compressive stress is 1,200kg/cm2, determine: (a) cross-sectional dimensions of the square key to be used; (b) length of the key.mechanics of rigid bodies The shaft assembly is made from a steel tube 1 in. in diameter, which is bonded to a solid brass core 0.5 in. in diameter. If a torque of 250 lb-ft is applied at its end, determine the maximum shear stress in psi experienced by the assembly. Take G for steel at 11.4x103 ksi and G brass at 5.2 x 103 ksi. with diagram if necessary.