Answer:
Explanation:
Given that,
Average fiber diameter is 0.01mm
d = 0.01mm = 1 × 10^-5m
The average fiber length is 2.5mm
L = 2.5mm = 0.0025m
Volume of the fraction of fibers is 0.40
Vf = 0.40
Fiber matrix bond strengths is 75MPa
τ = 75 MPa
The fraction strength of the fibers is 3500 Mpa
σf = 3500 MPa
The matrix street fiber is 8 MPa
σm = 8 MPa
We need to find the critical fiber length and compare it to original fiber length
Ic = σf•d / 2τ
Ic = 3500 × 0.01 / 75 × 2
Ic = 0.233 mm
Since the critical fiber length of 0.233 mm is much less than the provided length of the fiber (2.5mm) , so we can use the following equation to find the longitudinal tensile strength
σcd = σf•Vf(1—Ic / 2L) + σm(1—Vf)
σcd = 3500×0.4[1—0.233/(2 × 2.5)] + 8(1—0.4)
σcd = 1400(1—0.0467) + (8 × 0.6)
σcd = 1334.67 + 4.8
σcd = 1339.47 MPa
The longitudinal tensile strength of the aligned glass fiber-epoxy matrix composite is 1339.47 MPa
In signal processing, a filter is a device or process that removes some unwanted components or features from a signal. Filtering is a class of signal processing, the defining feature of filters being the complete or partial suppression of some aspect of the signal. Most often, this means removing some frequencies or frequency bands. However, filters do not exclusively act in the frequency domain; especially in the field of image processing many other targets for filtering exist.
Explanation:
Answer and Explanation:
The main objective that the torsion test serves is the determination of the material behavior or the behavior of the test sample when subjected to torsional stresses or forces due to the application of moments that results in shear stress along the axis.
Plasticity is the property of elastic material and tension or shear stresses leads to plasticity in a material where these links are the weakest, that gives torsion test a major advantage over the tension test.
Torsion tests are performed on materials to deduct properties like the shear modulus of elasticity, the torsional strength, and the MOR, i.e., Modulus of Rupture.
This test can be used to obtain larger strain values of strain without any complexity as that in tension test.
This test provides a curve of shear-stress-shear strain which is more significant in determining the plasticity as compared to the curve of stress-strain in tension test.
Maximum torque for a given value of maximum stress will be 2 times higher in torsion as that of tension.
In torsion, for plastic flow, the threshold value of shear stress is achieved before the threshold value of normal stress for fracture whereas in tension the critical value of normal stress is achieved sooner than the critical shear for plastic flow.
Answer and Explanation:
The computation is shown below
Total average cost + total variable cost = total cost
Let number of output be x
So,
Total fixed average cost = x × $30
Total variable cost = x × $15
Total cost = $2,500
Therefore,
$20 × x + $30 × x = $2,500
50 × x = $2,500
x = 50
Now the total variable cost is
= 50 × $20
= $1,000
And, the fixed cost is
= 50 × $30
= $1,500
Answer:
12 4
Explanation:
because the production average is variable
Answer:
ask someone u fool hhhhhhhh
• Use the reference on "Practical Considerations for Rectangular Reinforced Concrete Beams"
• Include references to ACI code – see slides from second class
• Include references to Tables from Appendix A
• Draw a sketch of the reinforced concrete beam showing all dimensions, number and size of rebars, including stirrups.
Answer:
Beam of 25" depth and 12" width is sufficient.
I've attached a detailed section of the beam.
Explanation:
We are given;
Beam Span; L = 20 ft
Dead load; DL = 0.50 k/ft
Live load; LL = 0.65 k/ft.
Beam width; b = 12 inches
From ACI code, ultimate load is given as;
W_u = 1.2DL + 1.6LL
Thus;
W_u = 1.2(0.5) + 1.6(0.65)
W_u = 1.64 k/ft
Now, ultimate moment is given by the formula;
M_u = (W_u × L²)/8
M_u = (1.64 × 20²)/8
M_u = 82 k-ft
Since span is 20 ft, it's a bit larger than the average span beams, thus, let's try a depth of d = 25 inches.
Effective depth of a beam is given by the formula;
d_eff = d - clear cover - stirrup diameter - ½Main bar diameter
Now, let's adopt the following;
Clear cover = 1.5"
Stirrup diameter = 0.5"
Main bar diameter = 1"
Thus;
d_eff = 25" - 1.5" - 0.5" - ½(1")
d_eff = 22.5"
Now, let's find steel ratio(ρ) ;
ρ = Total A_s/(b × d_eff)
Now, A_s = ½ × area of main diameter bar
Thus, A_s = ½ × π × 1² = 0.785 in²
Let's use Nominal number of 3 bars as our main diameter bars.
Thus, total A_s = 3 × 0.785
Total A_s = 2.355 in²
Hence;
ρ = 2.355/(22.5 × 12)
ρ = 0.008722
Design moment Capacity is given;
M_n = Φ * ρ * Fy * b * d²[1 – (0.59ρfy/fc’)]/12
Φ is 0.9
f’c = 4,000 psi = 4 kpsi
fy = 60,000 psi = 60 kpsi
M_n = 0.9 × 0.008722 × 60 × 12 × 22.5²[1 - (0.59 × 0.008722 × 60/4)]/12
M_n = 220.03 k-ft
Thus: M_n > M_u
Thus, the beam of 25" depth and 12" width is sufficient.
Answer:
Explanation:
Given that
Load is uniformly distributed load and beam is simply supported.
Ra + Rb= wL
Ra = Rb =wL / 2
Lets x is measured from left side,then the deflection of beam at any distance x is given as
The maximum deflection of beam will at x = L/2 (mid point )