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sladkih [1.3K]
4 years ago
6

The major and minor principal stresses at a certain point in the ground are 450 and 200 kPa, respectively. Determine the maximum

shear stress at this point.
Engineering
1 answer:
Elodia [21]4 years ago
8 0

Answer:

Explanation:

Diagrams of Shear force show the total shear force at each cross section of a structural member throughout the length of the beam or structural member. However, that force is not equally distributed throughout the individual cross section of the structural member. The maximum shear stress is referred to as the maximum concentrated shear force in a small area.

The Maximum Shear Stress theory states that failure occurs when the

maximum shear stress from the total of principal stresses equals

or exceeds the value obtained for the shear stress at yielding in the

uniaxial tensile test.

At yielding, in an uni-axial test, the principal stresses is given as:

σ=Sy; -

σ2 = 0 and σ3 = 0.

Therefore the shear strength at yielding

Ssy =[σ- (σ2 or σ3=0)]/2. Therefore Ssy = Sy/2

From the information given:

Tmax= σ1 - σ2/2

Tmax= 450-200/2

Tmax= 250/2

Tmax= 125 kpa

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Answer:

147.15

Explanation:

147.15 is the answer

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Spring-loaded rack guide yokes are made of ______ and have a spring that pushes on the back side of the rack to help reduce the
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Answer:

metal

Explanation:

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3 years ago
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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