To become familiar with the general equations of plane strain used for determining in-plane principal strain, maximum in-plane s
hear strain, and average normal strain. The state of strain at a point has components of ϵx=380.0×(10−6), ϵy=−250.0×(10−6), and γxy=180.0×(10−6).a. Equivalent in-plane strains on the oriented element Determine the equivalent in-plane strains on an element rotated counterclockwise at an angle of θ = 60.0 ∘ . Find ϵx',ϵy', γxy' Express your answers, separated by commas, to three significant figures.b. In-plane principal strains on the oriented element Determine the in-plane principal strains on the oriented element. Find ϵ1, ϵ2. Express your answers, separated by a comma, to three significant figures.c. Maximum in-plane shear strain and average normal strain on the oriented element Determine the maximum in-plane shear strain and the average normal strain on the oriented element. Express your answers, separated by a comma, to three significant figures.
1 answer:
Answer:
a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4
b) 3.926 x 10∧-4, -2.626 x 10∧-4
c) 6.552 x 10∧-4, 6.5 x 10∧-5
Explanation:
a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4
b) 3.926 x 10∧-4, -2.626 x 10∧-4
c) 6.552 x 10∧-4, 6.5 x 10∧-5
The explanation is shown in the attachment. I hope i have been able to help.
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Answer:
See explanations for step by step procedures to get answer.
Explanation:
Given that;
Determine the deflection at the center of the beam. Express your answer in terms of some or all of the variables LLL, EEE, III, and M0M0M_0. Enter positive value if the deflection is upward and negative value if the deflection is downward.
Answer:
a. = 40 psf
b. L ≈ 30.80 psf
c. The uniformly distributed total load for the beam = 812.8 ft./lb
d. The alternate concentrated load is more critical to bending , shear and deflection
Explanation:
The given parameters of the beam the beam are;
The span of the beam = 26 ft.
The width of the tributary, b = 16 ft.
The dead load, D = 20 psf.
a. The basic floor live load is given as follows;
The uniform floor live load, = 40 psf
The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²
Therefore, the uniform live load, = 40 psf
b. The reduced floor live load, L in psf. is given as follows;
For the school, = 2
Therefore, we have;
The reduced floor live load, L ≈ 30.80 psf
c. The uniformly distributed total load for the beam, = b ×
=
∴ = = 16 × (20 + 30.80) ≈ 812.8 ft./lb
The uniformly distributed total load for the beam, = 812.8 ft./lb
d. For the uniformly distributed load, we have;
= 812.8 × 26/2 = 10566.4 lbs
= 812.8 × 26²/8 = 68,681.6 ft-lbs
= 5×812.8×26⁴/348/EI = 4,836,329.333/EI
For the alternate concentrated load, we have;
= 1000 lb
= 20 × 16 = 320 lb/ft.
= 1,000 + 320 × 26/2 = 5,160 lbs
= 1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs
= 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI
Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load