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ratelena [41]
3 years ago
10

Interlocking stacked material is done by​ _____.

Engineering
2 answers:
VikaD [51]3 years ago
7 0

Answer:

B. Placing the objects at right angles

Explanation:

Workers are required to follow the safety guidelines when handling stacking materials. A material can fall or collapse causing injuries or death. According to OSHA guidelines, worker must do the following but not limited to:

  • when using hands, stacks should not exceed 16 feet high and 20 feet high if using a forklift
  • remove nails from used lumber before stacking
  • stacks should be stable and self-stable
  • material stored in racks should face away from main aisles
Dmitry [639]3 years ago
3 0

Answer:

B

Explanation:

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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.
saul85 [17]

Answer:

a. L_o  = 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, L_o  = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)

For the school, K_{LL} = 2

Therefore, we have;

L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, W_d = b × W_{D + L} =

∴  W_d =  = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, W_d = 812.8 ft./lb

d. For the uniformly distributed load, we have;

V_{max} = 812.8 × 26/2 = 10566.4 lbs

M_{max} =  812.8 × 26²/8 = 68,681.6 ft-lbs

v_{max} = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

P_L = 1000 lb

W_{D} = 20 × 16 = 320 lb/ft.

V_{max} = 1,000 + 320 × 26/2 = 5,160 lbs

M_{max} =  1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

v_{max} = 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

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