As shown in figure, a homogeneous 50-kg rigid block is suspended by the three springs whose lower ends were originally at the same level. Each steel spring has 24 turns of 10-mm-diameter on a mean diameter of 100 mm, and G = 83 GPa. The bronze spring has 48 turns of 20-mm-diameter wire on a mean diameter of 150 mm, and G=42 GPa. Compute the maximum shearing stress in each spring using approximate equation.

Q: 5 - An aluminum circular shaft, fixed at both ends, is loaded by torques applied at points B and C,…

A: Given Data  Shear modulus (G) = 85 GPa  Diameter of shaft (d) = 40mm  Length of AB (LAB) = 0.5m =…

Q: Each bar of the truss shown in the figure below has a diameter of 20 mm Suppose that P= 36 kN…

A:

Q: f the shaft?A

A: We knowμ=Lateral strainLongitudinal strainLateral strain=δddLongitudinal strain=δLL

Q: The beam shown in the figure below is subject to a moment of M = 17 kN.m as illustrated. Calculate…

A: Given:-  M = 17 kNm

Q: Refer image below

A: Given data: Steel:L=2 mDi=70 mmDe=100 mmBronze:L=1.5 mD=70 mm

Q: A solid aluminum shaft 2 in. in diameter as shown is subjected to two torques in the figure.…

A:

Q: . What is the normal stress in member BC?

A:

Q: Determine the magnitude of the pin force at A. Assume W = 640 Ib, a = 3.5 ft, b = 2.6 ft, r = 8 in.…

A: Step 1

Q: allowed.

A: Given data- Diameter of the solid segment 1 d1 = 41 mm= 0.041mDiameter of the solid segment 2 d2 =…

Q: For the beam shown, calculate the bending stress at a point 70 mm. from the top of the beam. P = 59,…

A:

Laws of Equilibrium

A body is said to be in equilibrium if the system of forces acting on the body tends to keep the body at rest, in other words, there is no translation motion or net torque on the body. Under such conditions, the equilibrium conditions are to be applied to analyze the condition. The branch of mechanics that deals with this condition is known as 'statics'. Under equilibrium, there are no net forces on the body.