A single-degree-of-freedom mass-spring-damper system is observed during its free vibration and the displacement amplitude decays
1 answer:
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Answer:
a) The ductility = -30.12%
the negative sign means reduction
Therefore, there is 30.12% reduction
b) the true stress at fracture is 658.26 Mpa
Explanation:
Given that;
Original diameter = 12.8 mm
Final diameter = 10.7
Engineering stress = 460 Mpa
a) determine The ductility in terms of percent reduction in area;
Ai = π/4( )² ; Ag = π/4(
)²
% = π/4 [ ( ( )² - (
)²) / ( π/4 (
)²) ]
= ( ( )² - (
)²) / (
)² × 100
we substitute
= [( (10.7)² - (12.8)²) / (12.8)² ] × 100
= [(114.49 - 163.84) / 163.84 ] × 100
= - 0.3012 × 100
= -30.12%
the negative sign means reduction
Therefore, there is 30.12% reduction
b) The true stress at fracture;
True stress =
( 1 +
)
is engineering strain
= dL / Lo
= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49
= 49.35 / 114.49
= 0.431
so we substitute the value of into our initial equation;
True stress = 460 ( 1 + 0.431)
True stress = 460 (1.431)
True stress = 658.26 Mpa
Therefore, the true stress at fracture is 658.26 Mpa
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