a load p is supported by a steel pin that has been inserted in a short wooden member hanging from the ceiling. the ultimate strength of the wood used is 60 MPa in tension and 7.5 MPa in shear, while the ultimate strength of the steel is 145 MPa in shear. knowing that b= 40 mm, c= 55mm, and d=12mm, determine the load P if an overall factor of safety of 3.2 is desired. [ans : 10.25 ]
To answer your question, I would need to see the figure that defines b, c and d, and explains how the pin is supported by the wood.
To determine the load P, we need to analyze the tensile and shear strength of both the wooden member and the steel pin, considering a factor of safety. Let's go step by step:
1. Tensile strength analysis:
The load P will create tension in the wooden member due to its weight. We need to ensure that the tensile stress in the wooden member is below its ultimate strength.
Tensile stress (σ_t) = Load P / Cross-sectional area of the wooden member
Cross-sectional area (A) = b * c
To maintain a factor of safety, we'll use the allowable stress, which is the ultimate strength divided by the factor of safety.
Allowable tensile stress (σ_allow_t) = Ultimate tensile strength of wood / Factor of safety
Therefore, σ_t ≤ σ_allow_t
2. Shear strength analysis:
The steel pin in the wooden member supports the load P, creating shear stress. We need to ensure that the shear stress in the steel pin and in the wooden member is below their respective ultimate strengths.
Shear stress (τ) = Load P / (Area of shear plane in wooden member or steel pin)
Area of shear plane for the wooden member (A_shear_wood) = b * d
Area of shear plane for the steel pin (A_shear_steel) = π * d^2 / 4
Allowable shear stress in wood (τ_allow_wood) = Ultimate shear strength of wood / Factor of safety
Allowable shear stress in steel (τ_allow_steel) = Ultimate shear strength of steel / Factor of safety
Therefore, τ ≤ τ_allow_wood and τ ≤ τ_allow_steel
3. Solve for load P:
By substituting the values and equations from the above steps, we can solve for the load P.
Calculations:
Given:
b = 40 mm
c = 55 mm
d = 12 mm
Ultimate tensile strength of wood = 60 MPa
Ultimate shear strength of wood = 7.5 MPa
Ultimate shear strength of steel = 145 MPa
Factor of safety = 3.2
1. Tensile strength analysis:
Cross-sectional area (A) = b * c = 40 mm * 55 mm = 2200 mm^2
σ_allow_t = 60 MPa / 3.2 = 18.75 MPa
σ_t = P / A
Therefore, P / A ≤ σ_allow_t
P ≤ σ_allow_t * A
2. Shear strength analysis:
Area of shear plane for the wooden member (A_shear_wood) = b * d = 40 mm * 12 mm = 480 mm^2
Area of shear plane for the steel pin (A_shear_steel) = π * d^2 / 4 = π * (12 mm)^2 / 4 = 113.097 mm^2
τ_allow_wood = 7.5 MPa / 3.2 = 2.34375 MPa
τ_allow_steel = 145 MPa / 3.2 = 45.3125 MPa
Therefore, τ ≤ τ_allow_wood and τ ≤ τ_allow_steel
3. Solve for load P:
P ≤ σ_allow_t * A (from tensile strength analysis)
τ ≤ τ_allow_wood and τ ≤ τ_allow_steel (from shear strength analysis)
The smallest load P that satisfies all the conditions is the maximum allowable load.
P = min(σ_allow_t * A, τ_allow_wood * A_shear_wood, τ_allow_steel * A_shear_steel)
Substitute the respective values and calculate P.
P = min(18.75 MPa * 2200 mm^2, 2.34375 MPa * 480 mm^2, 45.3125 MPa * 113.097 mm^2)
P = min(41,250 N, 1,125 N, 5,114.628 N)
Therefore, the load P is approximately 1,125 N.
The answer you provided, 10.25, does not match the calculated values based on the given information and calculations above. Please verify the values and calculations to determine if there was an error in transcription or calculation.