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The document provides an example of calculating the instantaneous value of a sine wave passing through zero at 30 degrees. It gives the amplitude as 15V, frequency as 50Hz, and time as 15ms. Through the calculations, it finds the instantaneous value to be -12.99V.
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This document provides the solution to calculating the maximum stress under an existing load on a structure. It first establishes the coordinate axes and calculates the reactions at points A and B. It then draws the shear and moment diagrams. The maximum moment is found to be 16 kN-m where the two loads are applied. Finally, it calculates the section modulus and finds the maximum stress to be 10.24 MPa using the relationship that stress is equal to the maximum moment divided by the section modulus.
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The document describes calculating the forces acting on a beam attached to a hinge by a cable and balanced force. It finds that the cable exerts 88N of tension to balance the 50N force on the beam and no net torque. It then calculates the x and y components of the hinge force as 30N rightward and 97N upward to balance all the forces acting on the beam.
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Stress Mohr's circle
1. The principal stresses are σ1=164MPa and σ2=-24MPa. The maximum in-plane shear stress is τm,ip=-94MPa.
2. The normal and shear stresses in the wood grain direction are σx'=0MPa and τx'y'=63MPa.
3. Mohr's circle is used to determine the stresses by rotating the stress element to principal planes and the plane of maximum shear.
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The document contains examples of solving word problems involving complementary angles, supplementary angles, and mixed problems involving both. Several word problems are presented and solved involving an angle (x), its complement (90-x), and supplement (180-x). Additional word problems at the end involve finding the measure of an angle given a relationship to its complement or supplement.
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This document provides the solution to calculating the maximum stress under an existing load on a structure. It first establishes the coordinate axes and calculates the reactions at points A and B. It then draws the shear and moment diagrams. The maximum moment is found to be 16 kN-m where the two loads are applied. Finally, it calculates the section modulus and finds the maximum stress to be 10.24 MPa using the relationship that stress is equal to the maximum moment divided by the section modulus.
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This document provides calculations to determine the resistance (R%) and reactance (X%) percentages for a 1600 kVA trihal standard transformer. It first calculates R% as 0.594% based on the transformer's full load and no load losses. It then uses the impedance relation and R% to calculate X% as 5.97%. The final X/R ratio is determined to be 10.05.
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The document describes calculating the forces acting on a beam attached to a hinge by a cable and balanced force. It finds that the cable exerts 88N of tension to balance the 50N force on the beam and no net torque. It then calculates the x and y components of the hinge force as 30N rightward and 97N upward to balance all the forces acting on the beam.
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This document contains calculations for the trunnion support of a pipe. It provides input data on the pipe dimensions and material properties, as well as loading conditions. The calculations determine the linear loads, bending stresses, primary stresses, and combined stresses on the pipe. The results indicate that the circumferential stress of 163.03 MPa and longitudinal stress of 170.73 MPa exceed the allowable stress of 148.94 MPa, meaning the current trunnion design is inadequate for the given pipe and loading.
Stress Mohr's circle
1. The principal stresses are σ1=164MPa and σ2=-24MPa. The maximum in-plane shear stress is τm,ip=-94MPa.
2. The normal and shear stresses in the wood grain direction are σx'=0MPa and τx'y'=63MPa.
3. Mohr's circle is used to determine the stresses by rotating the stress element to principal planes and the plane of maximum shear.
Solving word problems ~ complementary and supplementary angles
The document contains examples of solving word problems involving complementary angles, supplementary angles, and mixed problems involving both. Several word problems are presented and solved involving an angle (x), its complement (90-x), and supplement (180-x). Additional word problems at the end involve finding the measure of an angle given a relationship to its complement or supplement.
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2) It provides examples of the trig values of the six functions at 30°, 45°, 60°, 90°, 135°, 150°, 225°, 300°, and 315° by relating x and y to the radii of 30-60-90 and 45-45-90 triangles on the unit circle.
3) The values of the trig functions are reflected across the axes and change signs depending on the quadrant of the unit circle.
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To better understand the stress state for a new tractor, a structural analysis computer program was used to determine stresses in the tractor's front-end bucket. For the location shown, it was determined that the stresses are σx = 30 psi, σy = 10 psi, and τxy = 20 psi
Using Mohr's circle, what is
1) the principal direction and principal normal stresses
2) maximum shear direction and the maximum shearing stress.
APPROACH:
Construct the basic Mohr's circle for the given stress state.
Determine the principal direction from the Mohr's circle diagram, and then the principal stresses.
Find the maximum shear stress direction from the Mohr's circle diagram.
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The document provides calculations to determine:
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2) Reactions for a frame with multiple forces. Vertical and horizontal reactions are calculated.
3) Internal forces - axial forces, shear forces, and bending moments - for sections of the frame.
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3. Design of the longitudinal rebar, including positive moment, negative moment, and temperature rebar. Rebar sizes, amounts, and spacings are calculated.
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2) The first solution uses the center-of-mass frame, where the maximum φ is given by sinφ = m/M.
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2) It provides examples of the trig values of the six functions at 30°, 45°, 60°, 90°, 135°, 150°, 225°, 300°, and 315° by relating x and y to the radii of 30-60-90 and 45-45-90 triangles on the unit circle.
3) The values of the trig functions are reflected across the axes and change signs depending on the quadrant of the unit circle.
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To better understand the stress state for a new tractor, a structural analysis computer program was used to determine stresses in the tractor's front-end bucket. For the location shown, it was determined that the stresses are σx = 30 psi, σy = 10 psi, and τxy = 20 psi
Using Mohr's circle, what is
1) the principal direction and principal normal stresses
2) maximum shear direction and the maximum shearing stress.
APPROACH:
Construct the basic Mohr's circle for the given stress state.
Determine the principal direction from the Mohr's circle diagram, and then the principal stresses.
Find the maximum shear stress direction from the Mohr's circle diagram.
Ejericio analisis
The document provides calculations to determine:
1) The design strength of an angle section under LRFD using nominal strength and fracture strength. Fracture strength controls at 388 lbs.
2) Reactions for a frame with multiple forces. Vertical and horizontal reactions are calculated.
3) Internal forces - axial forces, shear forces, and bending moments - for sections of the frame.
4) Design of longitudinal stiffeners on a plate with specified steel grades and loads. Stiffener size is calculated to be 2.12 lbs/ft2.
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This document presents the solution to determining the vortex strengths (Γ1, Γ2, Γ3) for a problem involving 3 point vortices. It sets up a system of equations relating the vortex strengths to the velocities at 3 control points, but notes this system has infinite solutions. It then explains that applying the Kutta condition, which sets the velocity around one point vortex to zero, allows determining a unique solution by setting Γ3 = 0 and solving the reduced system.
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This document provides the design calculations for a reinforced concrete staircase. It includes:
1. Preliminary sizing and load calculations for the inclined ramp and landing sections. Applied loads include self-weight, finishes weight, and live loads.
2. Structural analysis to calculate bending moments and shear forces.
3. Design of the longitudinal rebar, including positive moment, negative moment, and temperature rebar. Rebar sizes, amounts, and spacings are calculated.
4. Check of shear capacity to ensure the concrete can resist the shear forces without rebar.
So in summary, this document performs the structural design of a reinforced concrete staircase under gravity and temperature loads. It sizes the rebar
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1. The document discusses different types of symmetry in periodic waves: even symmetry, odd symmetry, and half-wave symmetry.
2. Even symmetry waves have the property that the wave has the same shape on both sides of the y-axis. Odd symmetry waves have the opposite shape on each side of the y-axis.
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https://youtu.be/ILZThGyKV7s
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2) The r's complement of a number N is defined as rn - N, where r is the base and n is the number of digits. This includes the 1's complement for binary and 9's complement for decimal.
3) To subtract N from M, the r's complement of N is added to M. If the sum is greater than or equal to the base, the r's complement is taken. Otherwise the negative sign is placed in front of the sum.
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The document summarizes the propagation of waves in a vacuum and free space. It defines constants like frequency, conductivity, permittivity and permeability. It then calculates wave properties like period, phase and group velocity, impedance, wavelength and amplitude based on these constants. Graphs are presented showing the decay of the electric and magnetic field amplitudes over distance as the wave propagates.
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2) The first solution uses the center-of-mass frame, where the maximum φ is given by sinφ = m/M.
3) The second solution uses conservation of momentum and energy in the lab frame, arriving at the same result that the maximum φ satisfies m/M ≥ sinφ.
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The document summarizes the solution process for determining the deflection, bending moment, and stress of a simply supported beam subjected to an external forcing function. Natural frequencies are calculated as ωn = n^2π^2/L^2(EI/ρA). Mode shapes are sinusoidal functions of nπx/L. Forced response is determined using mode superposition, yielding a deflection proportional to sin(Ωt) and sin(ω1t) with spatial dependence of sin(πx/L). Bending moment and stress are derived from deflection and have the same functional forms.
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1) SHM is motion where the acceleration is proportional to and directed towards the displacement from a fixed point.
2) Common examples include a vibrating tuning fork, weight on a spring, boy on a swing.
3) The motion can be defined by the force equation F = -kx, where k is the spring constant.
4) Kinematics equations for SHM include the position equation x = x0sin(ωt + φ) and related equations for velocity and acceleration.
WavesExamplesAnswers.pdf
1. The document provides answers to example problems involving wave propagation and hydraulics. It analyzes wave characteristics such as wavelength, phase speed, and acceleration for different water depths.
2. Methods like iteration of the dispersion relationship are used to determine wave numbers and properties for scenarios with and without current.
3. Key wave parameters like height and wavelength are calculated from pressure readings using linear wave theory and shoaling equations. Different cases consider deep, intermediate, and shallow water conditions.
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OConnor_SimulationProject
This document summarizes simulations of a mass-spring-damper system representing a bus. Simulation 1 examines the system's response to step inputs without and with increased force. Adding damping is found to significantly reduce displacement peaks. Simulation 2 models a 2-degree-of-freedom system for a bus traveling on a bumpy road. Initially, large displacement peaks occur. By adding damping with a ratio of 0.3, displacements are reduced by 30-40% and the system quickly compensates to match the road input.
Standing wave lo5
This document discusses standing waves, which are formed by the superposition of two waves traveling in opposite directions with the same amplitude, wavelength, and frequency. It defines nodes as points of zero amplitude and antinodes as points of maximum amplitude. Equations are provided for calculating the position of nodes and antinodes. An example problem demonstrates using these equations to find the location of the first node and antinode for a given wave function.
Newton's Laws of Motion L2.1
This document discusses motion in a straight line and key concepts related to velocity and acceleration, including:
- The differences between average and instantaneous velocity and acceleration. Average values are calculated over a time interval, while instantaneous values are the limit as the time interval approaches zero.
- Velocity is a vector quantity that involves displacement over time, while speed is a scalar quantity involving distance over time.
- Acceleration is the rate of change of velocity with respect to time. It can be calculated as the change in velocity over a time interval for average acceleration or the derivative of velocity with respect to time for instantaneous acceleration.
- Examples are provided to demonstrate calculating average and instantaneous values and interpreting velocity-
Solb98mc 090504233342-phpapp02
1. The document provides solutions to multiple choice questions from an AP Physics B exam in 1998. It explains the basic ideas and solutions for 27 multiple choice questions covering topics like kinematics, forces, energy, momentum, circuits, fields, and more. The solutions are concise and directly address the key principles or calculations required to solve each problem.
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7.4.6 slip
This document describes how to calculate the rotor frequency of a two-pole, 50 Hz induction motor given the rotor speed of 2850 rpm. It shows that the slip speed is 3000 rpm, the slip percentage is 5%, and using the formula fs x %slip / 100, the rotor frequency is calculated to be 2.5 Hz.
4.2.1 Better Load
The document describes the circuits and loads for an electrical installation. It lists 19 circuits with various lighting, power outlet, motor, and appliance loads. It calculates the demand current in amps for each phase based on adding the full load or percentage of full load for each circuit based on rating and number of devices. The highest calculated demand is 153.25 amps on phase L2. With a 10% allowance for future additions, the total recommended maximum current is 168.575 amps.
4.1.13 Light Demand 3
This short document does not provide any substantive information to summarize in 3 sentences or less. It only notes that an explanation is not contained in a workbook, but provides no other context or details.
4.1.12 Light Demand 2 Table Part2
The document outlines the demand calculations for 19 different circuit load groups across 3 phases. It lists the load description, current allowance calculation method, and resulting demand current for each phase. The total demand current per phase is calculated at the bottom, with values of 143.8 amps for phase 1, 153.25 amps for phase 2, and 145.05 amps for phase 3.
4.1.11 Light Demand 2 Table
The document contains a table that calculates the current demand per phase for various circuit load groups in an electrical system. It lists 19 load groups categorized by letters A through D, describing each load. It shows the current allowance calculation method and resulting demand in amps for each phase. The total demand current calculated per phase is 143.8 amps for L1, 153.25 amps for L2, and 145.05 amps for L3.
4.1.10 Light Demand 2
This 3 sentence document provides instructions to refer to a specific table on a specific page of a particular standard for additional explanatory information not contained in the current workbook. The instructions direct the reader to Table C2 on page 359 of AS/NZA 3000:2007 for an explanation that is not included in the current document.
4.1.9 Light Demand 1
The document describes the functions of 19 electrical circuits in a building. It lists the types of equipment connected to each circuit such as fluorescent lighting, outlets, motors, and appliances. It also indicates which of the 3 electrical phases (L1, L2, L3) each circuit is connected to.
4.1.7 Example Domestic 3
The document calculates the electrical load of communal services in an apartment building. It shows that 24 lighting points will draw 240 watts and 6 10A sockets will draw up to 12A, for a total demand of 18A per phase. The total demand current per phase for communal services is 18A.
4.1.6 Example Domestic 2
The document calculates the electrical load and demand for 6 living units per phase. It lists the types of loads in each unit, the quantity and allowance per unit, and uses this to calculate the total demand current for phases L1, L2 and L3, which is 154.4 amps for each phase. The key loads included are lighting, power outlets, cooking ranges, air conditioners and hot water systems.
4.1.5 Example Domestic 1
This document discusses the number of living units per phase of a project. It calculates that for 18 total living units divided into 3 phases, there would be 6 living units per phase.
4.1.4 Step 2
This document summarizes the electrical load calculations for 11 circuits in a home. It lists the load type and description for each circuit, the current allowance per unit, and calculates the demand in amps for circuits 1-10. The largest demands are 17.7 amps for an air conditioner and 15 amps for an off-peak hot water system. The total calculated demand current for each phase is 40.7, 39.5, and 46.5 amps respectively.
4.1.3 Example Solution Three
The document calculates the electrical load and demand current for 11 circuits in a home. It groups the loads into categories like lighting, outlets, appliances, and assigns each a description, allowance, and demand calculation. The total demand current per phase is summarized at the bottom, with the highest draw of 46.5 amps on phase L3.
4.1.2 Example Three
The document outlines the functions and ratings of 12 electrical circuits. Circuit 1 provides power for 13 indoor lighting points. Circuit 10 powers an air conditioner rated at 23.6 amps per phase and can draw power from circuits L1, L2, and L3. Circuits 11a and 11b each power a controlled load hot water unit rated at 3.6 kW.
4.1.1 Example Single
This document calculates the total demand current for an electrical installation consisting of:
- 21 lighting points and 12 double sockets, contributing 5A and 15A respectively
- 15 single sockets contributing 10A
- A 6kW oven contributing 0.5A
- A 2.4kW water heater contributing 0.33A
The total demand current calculated is 45.83A.
5.4.4 LR Circuits
The document discusses the time constant and final current value for an RL circuit. It states that:
1) The time constant for the circuit is 0.17 seconds based on the given inductance and resistance values.
2) It will take approximately 0.85 seconds (5 time constants) for the current to reach its final value.
3) Using Ohm's law, the approximate final current after 0.85 seconds is 2 amps.
14.5.4 Example 1 Part 1
This document calculates the apparent power, power factor, and phase angle for a circuit. It finds that the apparent power is 2.308 kVA by multiplying the current of 9.615 by the voltage of 240. This apparent power is larger than the actual power of 1.5 kW, indicating a poor power factor of 0.65 or a 49.46 degree phase angle between the current and voltage.
14.1.2 TRA Power
Reactive power (Q) and true power (P) combine to form apparent power (S). Apparent power is the combination of true power, which is the usable energy in a circuit, and reactive power, which is stored energy that results from the combination of voltage and current out of phase.
14.5.8 Example 1 Part 5
This document calculates the true power, apparent power, and total current for a circuit. It determines that the true power is 1.5 kW, the apparent power is 1.5009 kVA, and the power factor is 0.99994. It then calculates that with an apparent power of 1.5009 kVA at 240 Volts, the total current would be 6.25 amps.
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8.1.6example2
- 1. Example 2 Determine the instantaneous value of a sine wave that has a positive slope passing through zero at 30 . The sine wave has the following values. Am = 15 V f = 50 Hz t = 15 mS Note : Multiply ω t by 57.2958 to convert Radians to Degrees. Working e = E m sin ( ω t + 30) = 15 x sin (( 2 π f t )+ 30) = 15 x sin (( 2 x 3.141593 x 50 x 0.015 x 57.2958) + 30 ) = 15 x sin (270 + 30) = 15 x sin 300 = 15 x -0.866 Answer= - 12.99 V
- 2. Example 2 Determine the instantaneous value of a sine wave that has a positive slope passing through zero at 30 . The sine wave has the following values. Am = 15 V f = 50 Hz t = 15 mS Note : Multiply ω t by 57.2958 to convert Radians to Degrees. Working e = E m sin ( ω t + 30) = 15 x sin (( 2 π f t )+ 30) = 15 x sin (( 2 x 3.141593 x 50 x 0.015 x 57.2958) + 30 ) = 15 x sin (270 + 30) = 15 x sin 300 = 15 x -0.866 Answer= - 12.99 V