Assignment 1: Structural Design of Rectangular Reinforced Concrete Beams for Bending Perform structural design of a rectangular reinforced concrete beam for bending. The beam is simply supported and has a span L=20 feet. In addition to its own weight the beam should support a superimposed dead load of 0.50 k/ft and a live load of 0.65 k/ft. Use a beam width of 12 inches. The depth of the beam should satisfy the ACI stipulations for minimum depth and be proportioned for economy. Concrete compressive strength f’c = 4,000 psi and yield stress of reinforcing bars fy = 60,000 psi. Size of stirrups should be chosen based on the size of the reinforcing bars. The beam is neither exposed to weather nor in contact with the ground, meaning it is subjected to interior exposure.
• 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.
Answers
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.
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. Were you able to observe ???? = 0 in the circuit you constructed during lab? Why or why not? Hint: What value of resistance would be needed for ???? = 0? 2. What feature in the time response of an RLC circuit distinguishes a critically damped response from an underdamped response? 3. Why must an op-amp be powered to be used in a circuit? 4. If you were handed a parts kit with an unknown op-amp, what information would you need to find prior to using it in a circuit?
A 1-m3 tank containing air at 10°C and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35°C and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 19.5°C. Determine the volume of the second tank and the final equilibrium pressure of air. The gas constant of air is R = 0.287 kPa·m3/kg·K.
An input voltage of 9.2 V is to be converted into its digital counterpart using an analog-to digital converter. The voltage range is 0 to 16 V. The ADC has 4-bit capacity. Determine: (a) What are the number of quantization levels, resolution, and the maximum quantization error of this ADC
Answers
Answer:
C)185,500 KJ
Explanation:
Given that
Latent heat fusion = 333.23 KJ/kg
Latent heat vaporisation = 333.23 KJ/kg
Mass of ice = 100 kg
Mass of water = 40 kg
Mass of vapor=60 kg
Ice at 0°C ,first it will take latent heat of vaporisation and remain at constant temperature 0°C and it will convert in to water.After this water which at 0°C will take sensible heat and gets heat up to 100°C.After that at 100°C vapor will take heat as heat of vaporisation .
Sensible heat for water Q
For water
Q=4.178 x 40 x 100 KJ
Q=16,712 KJ
So total heat
Total heat =100 x 333.23+16,712 + 60 x 2257 KJ
Total heat =185,455 KJ
Approx Total heat = 185,500 KJ
So the answer C is correct.
Answers
Answer:
the relative compaction is 105.88 %
Explanation:
Given;
dry unit weight of field compaction, = 18 kN/m³
maximum dry unit weight measured, = 17 kN/m³
Relative compaction (RC) of the site is given as the ratio of dry unit weight of field compaction and maximum dry unit weight measured
Relative compaction (RC) = dry unit weight of field compaction / maximum dry unit weight measured
substitute the given values;
RC (%) = 105.88 %
Therefore, the relative compaction is 105.88 %
Answers
Answer:
Option C..Farmers saught new technology to help with the workload
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The normal melting point and boiling point of nickel are 1453°C and 2730°C, respectively, and the density and atomic weights of Ni are 8.91 g.cm^-3 and 58.71 atomic mass units respectively. Calculate the energy in joules needed to evaporate the required quantity of nickel.
Answers
Answer:
Check the explanation
Explanation:
Let’s take for instance, when an object with a mass of 10 kg (m = 10 kg) is moving at a 5 meters per second (v = 5 m/s) velocity rate, the kinetic energy is equal to 125 Joules
Kindly check the attached images below to get the step by step explanation to the question above.
Answers
Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.
Answers
Answer:
A) Ductility = 11% EL
B) Radius after deformation = 4.27 mm
Explanation:
A) From equations in steel test,
Tensile Strength (Ts) = 3.45 x HB
Where HB is brinell hardness;
Thus, Ts = 3.45 x 250 = 862MPa
From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.
Also, from image 2,at CW of 27%,
Ductility is approximately, 11% EL
B) Now we know that formula for %CW is;
%CW = (Ao - Ad)/(Ao)
Where Ao is area with initial radius and Ad is area deformation.
Thus;
%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100
%CW = [1 - (rd)²/(ro)²]
1 - (%CW/100) = (rd)²/(ro)²
So;
(rd)²[1 - (%CW/100)] = (ro)²
So putting the values as gotten initially ;
(ro)² = 5²([1 - (27/100)]
(ro)² = 25 - 6.75
(ro) ² = 18.25
ro = √18.25
So ro = 4.27 mm