The normal stress at gage H calculated in Part 1 includes four components: an axial component due to load P, σaxial, P, a bendin
g component due to load P, σbend, P, an axial component due to load Q, σaxial, Q, and a bending component due to load Q, σbend, Q. Calculate σaxial, P and σbend,P. Use the sign convention for normal stresses. If a normal stress component is tension, it is positive, and if it is compression, it is negative.
1 answer:
Answer:
hello your question has some missing information attached to the answer is the missing component
Answer : αaxial,p = -6.034 ksi ( compressive )
αbend,p = 19.648 ksi ( tensile )
Explanation:
αaxial, p = equation 1
αbend, p = equation 2
P = load = 35 kips
A = area of column = 5.8
d = column cross section depth = 9.5 in
= 55.0
Hence equation 1 becomes
αaxial,p = -35 / 5.8 = - 6.034 ksi ( compressive )
equation 2 becomes
αbend, p = = + 19.648 ksi ( tensile )
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Answer:
(a) The force sustained by the matrix phase is 1802.35 N
(b) The modulus of elasticity of the composite material in the longitudinal direction Ed is 53.7 GPa
(c) The moduli of elasticity for the fiber and matrix phases is 124.8 GPa and 2.2 GPa respectively
Explanation:
Find attachment for explanation
Answer:
N_c = 3.03 N
F = 1.81 N
Explanation:
Given:
- The attachment missing from the question is given:
- The given expressions for the radial and θ direction of motion:
r = 0.5*θ
θ = 0.5*t^2 ...... (correction for the question)
- Mass of peg m = 0.5 kg
Find:
a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.
b) Determine the magnitude of the normal force of the slot on the peg.
Solution:
- Determine the expressions for radial kinematics:
dr/dt = 0.5*dθ/dt
d^2r/dt^2 = 0.5*d^2θ/dt^2
- Similarly the expressions for θ direction kinematics:
dθ/dt = t
d^2θ/dt^2 = 1
- Evaluate each at time t = 2 s.
θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°
r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2
- Evaluate the angle ψ between radial and horizontal direction:
tan Ψ = r / (dr/dθ) = 1 / 0.5
Ψ = 63.43°
- Develop a free body diagram (attached) and the compute the radial and θ acceleration:
a_r = d^2r / dt^2 - r * dθ/dt
a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2
a_θ = r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)
a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2
- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:
Radial direction: N_c * cos(26.57) - W*cos(24.59) = m*a_r
θ direction: F - N_c * sin(26.57) + W*sin(24.59) = m*a_θ
Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:
N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5
N_c = 3.03 N
F - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5
F = 1.81 N
