Answer:
a) the friction angle of the sand is 36.87°
b) the deviator stress at failure is 450 kN/m³
Explanation:
Given the data in the question;
For a consolidated drained test
The effective major principle stresses
σ₃ = σ₃' = 80 kN/m²
and
σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 80 kN/m² + 240 kN/m² = 320 kN/m²
now
a) friction angle of the sand
σ₁' = σ₃'tan²( 45° + Ф/2' ) + 2c' tan( 45° + Ф/2 )
for sand; c' = 0
so
σ₁' = σ₃'tan²( 45° + Ф/2' )
we substitute
320 = 80 tan²( 45° + Ф/2' )
Ф' = 2 × [ tan⁻¹ (√[tex]\frac{320}{80}[/tex]) - 45° ]
Ф' = 2 × [ 63.4349° - 45° ]
Ф' = 2 × 18.4349
Ф' = 36.87°
Therefore, the friction angle of the sand is 36.87°
b) deviator stress
σ₁' = σ₃'tan²( 45° + Ф/2' )
σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' )
σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' ) 3.597
150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 150tan²( 45° + 36.87°/2 )
150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600
(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600 - 150
(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 450 kN/m³
Therefore, the deviator stress at failure is 450 kN/m³
The accompanying specific gravity values describe various wood types used in construction. 0.320.350.360.360.370.380.400.400.40 0.410.410.420.420.420.420.420.430.44 0.450.460.460.470.480.480.490.510.54 0.540.550.580.630.660.660.670.680.78 Construct a stem-and-leaf display using repeated stems. (Enter numbers from smallest to largest separated by spaces. Enter NONE for stems with no values.)
Answer:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]
Explanation:
Given
[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,[/tex]
[tex]0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,[/tex]
[tex]0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,[/tex]
[tex]0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.[/tex]
Required
Plot a steam and leaf display for the given data
Start by categorizing the data by their tenth values:
[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.[/tex]
[tex]0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,[/tex]
[tex]0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.[/tex]
[tex]0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.[/tex]
[tex]0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.[/tex]
[tex]0.78.[/tex]
The 0.3's is will be plotted as thus:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]
The 0.4's is as follows:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \ \end{array}[/tex]
The 0.5's is as follows:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}[/tex]
The 0.6's is as thus:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]
Lastly, the 0.7's is as thus:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]
The combined steam and leaf plot is:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]
3/4 + 1/2
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Ethylene glycol, the ingredient in antifreeze, does not cause health problems because it is a clear liquid
Answer:
False
Explanation:
I got it wrong picking true
A front wheel drive vehicle with four wheel disc brakes is pulling to the left. Tech A says an external kink or internal restriction in the LF brake line will result in this condition. Tech B says to use a compression fitting to repair a section of brake line. Who is correct? Tech A Tech A Tech B Tech B Both Both Neither
Answer:
Tech A is correct.
Explanation:
A front-wheel-drive pulling to the left can result from several factors. One of them is definitely a faulty break.
A correct diagnosis linking the problem to the brakes is when there is an internal restriction and the pull is constant to one side and gets worse when the brakes are applied.
To confirm this, one would need to lift the vehicle and rotate each wheel by hand to check for excessive friction.
So the restriction may be caused by:
brake calipers that are sticky to the drumtoo much brake fluid in the brake master cylinder - this prevents the caliper pistons from retracting when the brakes are released misadjusted drum brakes and or parking brakes.Cheers
A brass alloy rod having a cross sectional area of 100 mm2 and a modulus of 110 GPa is subjected to a tensile load. Plastic deformation was observed to begin at a load of 39872 N. a. Determine the maximum stress that can be applied without plastic deformation. b. If the maximum length to which a specimen may be stretched without causing plastic deformation is 67.21 mm, what is the original specimen length
Answer:
a) the maximum stress that can be applied without plastic deformation is 398.72 N/mm²
b) length of the specimen is 66.97 mm
Explanation:
Given the data in the question;
a) Determine the maximum stress that can be applied without plastic deformation
when know that; maximum stress σ[tex]_{max}[/tex] = F / A
where F is the force in the rod ( 39872 N )
A is the cross-sectional area of the rod ( 100 mm² )
so we substitute;
σ[tex]_{max}[/tex] = 39872 N / 100 mm²
σ[tex]_{max}[/tex] = 398.72 N/mm²
Therefore, the maximum stress that can be applied without plastic deformation is 398.72 N/mm²
b)
strain in the members can be calculated using the expression
ε = σ / E
where σ is the stress in the rod
E is the module of elasticity ( 110 GPa = 110000 N/mm² )
(Sl-L) / L = σ/E
where Sl-L is the change in length of the member
L is the original length of the specimen
so we substitute
(67.21 - L) / L = 398.72 / 110000
110000( 67.21 - L) = 398.72L
7393100 - 110000L = 398.72L
7393100 = 398.72L+ 110000L
7393100 = 110398.72L
L = 7393100 / 110398.72
L = 66.97 mm
Therefore; length of the specimen is 66.97 mm
why you so mean to me? leave my questions please. answer them
Answer: Why is even here then.
Explanation:
A river has an average rate of water flow of 59.6 M3/s. This river has three tributaries, tributary A, B and C, which account for 36%, 47% and 17% of water flow respectively. How much water is discharged in 30 minutes from tributary B?
Answer:
50421.6 m³
Explanation:
The river has an average rate of water flow of 59.6 m³/s.
Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:
Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s
The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken
time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds
The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³