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german
2 years ago
14

A cross beam in a highway bridge experiences a stress of 14 ksi due to the dead weight of the bridge structure. When a fully loa

ded tractor-trailer crosses over the bridge, however, the stress in the beam increases to 45 ksi. The beam is fabricated from steel with an ultimate tensile strength of 76 ksi, a yield strength of 50 ksi, and an endurance limit of 38 ksi. Find the safety factor for an infinite fatigue life:
a. if the effect of mean stress on fatigue strength is ignored
b. when the effect of mean stress on fatigue strength is considered.
Engineering
1 answer:
zlopas [31]2 years ago
8 0

Answer:

a) 2.452

b) 1.256

Explanation:

Stress due to dead weight. = 14 Ksi

Stress due to fully loaded tractor-trailer = 45Ksi

ultimate tensile strength of beam = 76 Ksi

yield strength = 50 Ksi

endurance limit = 38 Ksi

Determine the safety factor for an infinite fatigue life

a) If mean stress on fatigue strength is ignored

β = ( 45 - 14 ) / 2

  = 15.5 Ksi

hence FOS ( factor of safety ) = endurance limit / β

                                                 = 38 / 15.5 = 2.452

b) When mean stress on fatigue strength is considered

β2 = 45 + 14 / 2

    = 29.5 Ksi

Ratio  = β / β2 = 15.5 / 29.5 = 0.5254

Next step: applying Goodman method

Sa =  [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]

     = 19.47 Ksi

hence the FOS ( factor of safety ) = Sa / β

                                                      = 19.47 / 15.5 = 1.256

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The electric motor exerts a torque of 800 N·m on the steel shaft ABCD when it is rotating at a constant speed. Design specificat
kodGreya [7K]

Answer:

d= 4.079m ≈ 4.1m

Explanation:

calculate the shaft diameter from the torque,    \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}

Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).

r = Radius of the shaft.

T = Twisting Moment or Torque.

J = Polar moment of inertia.

C = Modulus of rigidity for the shaft material.

l = Length of the shaft.

θ = Angle of twist in radians on a length.  

Maximum Torque, ζ= τ ×  \frac{ π}{16} × d³

τ= 60 MPa

ζ= 800 N·m

800 = 60 ×  \frac{ π}{16} × d³

800= 11.78 ×  d³

d³= 800 ÷ 11.78

d³= 67.9

d= \sqrt[3]{} 67.9

d= 4.079m ≈ 4.1m

3 0
2 years ago
Read 2 more answers
The critical resolved shear stress for a metal is 39 MPa. Determine the maximum possible yield strength (in MPa) for a single cr
damaskus [11]

Answer:

78 MPa

Explanation:

Given that the critical resolved shear stress for a metal is 39 MPa, the maximum possible yield strength for a single crystal of this metal is twice the critical resolved shear stress for the metal. The maximum yield yield strength for a single crystal of this metal that is pulled in tension (\sigma_y) is given as:

\sigma_y=2*critical\ resolved\ shear\ stress(\tau_{css})\\\\\sigma_y=2*\tau_{css}\\\\\sigma_y=2*39\\\\\sigma_y=78\ MPa

4 0
2 years ago
6.28 A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 4 ft from the right
dimulka [17.4K]

Answer:

Assume Base free flow speed (BFFS) = 70 mph

Lane width = 10 ft

Reduction in speed corresponding to lane width, fLW = 6.6 mph

Lateral Clearance = 4 ft

Reduction in speed corresponding to lateral clearance, fLC = 0.8 mph

Interchanges/Ramps = 9/ 6 miles = 1.5 /mile

Reduction in speed corresponding to Interchanges/ramps, fID = 5 mph

No. of lanes = 3

Reduction in speed corresponding to number of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 70 – 6.6 – 0.8 – 3 – 5 = 54.6 mph

Peak Flow, V = 2000 veh/hr

Peak 15-min flow = 600 veh

Peak-hour factor = 2000/ (4*600) = 0.83

Trucks and Buses = 12 %

RVs = 6 %

Rolling Terrain

fHV = 1/ (1 + 0.12 (2.5-1) + 0.06 (2.0-1)) = 1/1.24 = 0.806

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 2000/ (0.83*3*0.806*1.0) = 996.54 ~ 997 veh/hr/ln

Vp < (3400 – 30 FFS)

S = FFS

S = 54.6 mph

Density = Vp/S = (997) / (54.6) = 18.26 veh/mi/ln

7 0
3 years ago
3.3 Equation (2) for VCPP is rather difficult to prove at this time. Take it as a challenge to derive it as you learn increasing
podryga [215]

Answer:

For an RC integrator circuit, the input signal is applied to the resistance with the output taken across the capacitor, then VOUT equals VC. As the capacitor is a frequency dependant element, the amount of charge that is established across the plates is equal to the time domain integral of the current. That is it takes a certain amount of time for the capacitor to fully charge as the capacitor can not charge instantaneously only charge exponentially.

Therefore the capacitor current can be written as:

 

his basic equation above of iC = C(dVc/dt) can also be expressed as the instantaneous rate of change of charge, Q with respect to time giving us the following standard equation of: iC = dQ/dt where the charge Q = C x Vc, that is capacitance times voltage.

The rate at which the capacitor charges (or discharges) is directly proportional to the amount of the resistance and capacitance giving the time constant of the circuit. Thus the time constant of a RC integrator circuit is the time interval that equals the product of R and C.

Since capacitance is equal to Q/Vc where electrical charge, Q is the flow of a current (i) over time (t), that is the product of i x t in coulombs, and from Ohms law we know that voltage (V) is equal to i x R, substituting these into the equation for the RC time constant gives:

We have seen here that the RC integrator is basically a series RC low-pass filter circuit which when a step voltage pulse is applied to its input produces an output that is proportional to the integral of its input. This produces a standard equation of: Vo = ∫Vidt where Vi is the signal fed to the integrator and Vo is the integrated output signal.

The integration of the input step function produces an output that resembles a triangular ramp function with an amplitude smaller than that of the original pulse input with the amount of attenuation being determined by the time constant. Thus the shape of the output waveform depends on the relationship between the time constant of the circuit and the frequency (period) of the input pulse.

By connecting two RC integrator circuits together in parallel has the effect of a double integration on the input pulse. The result of this double integration is that the first integrator circuit converts the step voltage pulse into a triangular waveform and the second integrator circuit converts the triangular waveform shape by rounding off the points of the triangular waveform producing a sine wave output waveform with a greatly reduced amplitude.

RC Differentiator

For a passive RC differentiator circuit, the input is connected to a capacitor while the output voltage is taken from across a resistance being the exact opposite to the RC Integrator Circuit.

A passive RC differentiator is nothing more than a capacitance in series with a resistance, that is a frequency dependentTherefore the capacitor current can be written as:

 

 

device which has reactance in series with a fixed resistance (the opposite to an integrator). Just like the integrator circuit, the output voltage depends on the circuits RC time constant and input frequency.

Thus at low input frequencies the reactance, XC of the capacitor is high blocking any d.c. voltage or slowly varying input signals. While at high input frequencies the capacitors reactance is low allowing rapidly varying pulses to pass directly from the input to the output.

This is because the ratio of the capacitive reactance (XC) to resistance (R) is different for different frequencies and the lower the frequency the less output. So for a given time constant, as the frequency of the input pulses increases, the output pulses more and more resemble the input pulses in shape.

We saw this effect in our tutorial about Passive High Pass Filters and if the input signal is a sine wave, an rc differentiator will simply act as a simple high pass filter (HPF) with a cut-off or corner frequency that corresponds to the RC time constant (tau, τ) of the series network.

Thus when fed with a pure sine wave an RC differentiator circuit acts as a simple passive high pass filter due to the standard capacitive reactance formula of XC = 1/(2πƒC).

But a simple RC network can also be configured to perform differentiation of the input signal. We know from previous tutorials that the current through a capacitor is a complex exponential given by: iC = C(dVc/dt). The rate at which the capacitor charges (or discharges) is directly proportional to the amount of resistance and capacitance giving the time constant of the circuit. Thus the time constant of a RC differentiator circuit is the time interval that equals the product of R and C. Consider the basic RC series circuit below.

Explanation:

3 0
3 years ago
If a system of pulleys results in a force of 25% of the load, how far will the rope need to move to pull the load a distance of
GaryK [48]

Answer:

  40 ft

Explanation:

Assuming no loss of energy in the system of pulleys, the work done is the same whether you move the load directly or through the pulleys.

  W = Fd . . . . . . . . work is the product of force and distance

  F(10 ft) = (0.25F)(d) . . . . . where d is the distance we want to find

  d = 10F/(0.25F) = 40

The rope will need to move 40 feet.

8 0
2 years ago
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