Answer:
1. 1111.5MPa
2. 56.1GPa
Explanation:
1. Longitudinal tensile stress can be obtained by obtaining the strength and volume of the fiber reinforcement. The derived formula is given by;
σcl = σm (1 - Vf) + σfVf
Substituting the figures, we will have;
45(1 - 0.30) + 3600(0.30)
45(0.70) + 1080
31.5 + 1080
= 1111.5MPa
2. Longitudinal modulus of elasticity or Young's modulus is the ability of an object to resist deformation. The derived formula is given by;
Ecl = EmVm + EfVf
Substituting the formula gives;
= 2.4 (1 - 0.30) + 131 (0.30)
= 2.4(0.70) + 39.3
= 16.8 + 39.3
= 56.1GPa
Using the appropriate relation, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.
Longitudinal tensile stress can be obtained using the relation :
σcl = σm (1 - Vf) + σfVfSubstituting the values into the relation:
45(1 - 0.30) + 3600(0.30)
45 × 0.70 + 1080
31.5 + 1080
= 1111.50 MPa
2.)
Longitudinal modulus of elasticity is obtained using the relation :
Ecl = EmVm + EfVfSubstituting the values thus :
2.4 (1 - 0.30) + 131 (0.30)
= 2.4 × 0.70 + 39.3
= 16.8 + 39.3
= 56.10 GPa
Hence, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.
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What causes the charging system warning lamp to go out when the engine starts up?
Select one:
a. It turns off when ground is supplied to the lamp.
b. It turns off because voltage is applied to both sides of the lamp.
c. It turns off automatically after about 5 seconds.
d. It turns off because voltage is applied to one side of the bulb and ground to the other side.
An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft. Dead load is 20 psf. Find:
a. Basic floor live load Lo in psf
b. Reduced floor live load L in psf
c. Uniformly distributed total load to the beam in lb/ft.
d. Compare the loading in part c with the alternate concentrated load requried by the Code. Which loading is more critical for bending, shear, and deflection.?
Answer:
a. [tex]L_o[/tex] = 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, [tex]L_o[/tex] = 40 psf
b. The reduced floor live load, L in psf. is given as follows;
[tex]L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)[/tex]
For the school, [tex]K_{LL}[/tex] = 2
Therefore, we have;
[tex]L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf[/tex]
The reduced floor live load, L ≈ 30.80 psf
c. The uniformly distributed total load for the beam, [tex]W_d[/tex] = b × [tex]W_{D + L}[/tex] =
∴ [tex]W_d[/tex] = = 16 × (20 + 30.80) ≈ 812.8 ft./lb
The uniformly distributed total load for the beam, [tex]W_d[/tex] = 812.8 ft./lb
d. For the uniformly distributed load, we have;
[tex]V_{max}[/tex] = 812.8 × 26/2 = 10566.4 lbs
[tex]M_{max}[/tex] = 812.8 × 26²/8 = 68,681.6 ft-lbs
[tex]v_{max}[/tex] = 5×812.8×26⁴/348/EI = 4,836,329.333/EI
For the alternate concentrated load, we have;
[tex]P_L[/tex] = 1000 lb
[tex]W_{D}[/tex] = 20 × 16 = 320 lb/ft.
[tex]V_{max}[/tex] = 1,000 + 320 × 26/2 = 5,160 lbs
[tex]M_{max}[/tex] = 1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs
[tex]v_{max}[/tex] = 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