MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months)
MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months)
5th Edition
ISBN: 9781305581159
Author: Nicholas J. Garber; Lester A. Hoel
Publisher: Cengage Learning US
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Chapter 6, Problem 20P
To determine

(a)

The maximum queue length that will be formed, total delay, number of vehicles that will be affected by the incident, the average individual delay for the expected demand flow of 70%of the capacity of the highway.

Expert Solution
Check Mark

Answer to Problem 20P

For 70 % expected demand flow:

Themaximum queue length that will be formed is qmax=400Veh.

The total delay is dr=445hr.

The number of vehicles that will be affected by the incident is 10800Veh.

The average individual delay is 0.0412hr.

Explanation of Solution

Given:

We have been given the following information:

Total number of lanes = 3,

  V=90%of6000,

  tinc=2hr,

Mean free flow speed of the highway = 55 mi/h,

Jam density = 135 veh/mi/ln,

  Cr=4000.

Following is the lay out of the given highway section:

  MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months), Chapter 6, Problem 20P , additional homework tip  1

Calculation:

For the expected 70 % demand flow:

We have the following formula for the determination of maximum queue length that will be formed:

  qmax=(VCr)tinc

Where, qmax is the maximum queue length,

  V is the total volume,

  Cr is the reduced capacity,

  tinc is the duration of the incident.

Considering 70 percent of the flow and that the capacity of each lane is 2000 Veh/ h

Substituting the values in the following equation, we have

  qmax=(VCr)tincqmax=(( 70%of6000)4000)2qmax=(( 6000× 70 100 )4000)2

  qmax=(( 60×70)4000)2qmax=(42004000)2qmax=400Veh.

The maximum queue length that will be formed is qmax=400Veh.

Now, the total delay, we have the following formula

  dr=tinc2(VCR)(CCR)2(CV)

Where, CR is the reduced capacity and can be found as follows:

  CR=Numberoflanesinuse×capacityofeachlane.CR=2000×(31).CR=2000×2=4000Veh/h.

And C is the total capacity and can be found as

  C=Totalnumberoflanesinuse×capacityofeachlane.C=2000×3.C=6000Veh/h.

Now, substituting the values in the required equation, we have

  dr=t inc2( V C R )( C C R )2( CV)dr=22( 70%of60004000)( 60004000)2( 600070%of6000)

  dr=4(42004000)(2000)2(60004200)

  dr=4( 100)( 2000)(1800)dr=40009

  dr=444.44=445hr.

The total delay is dr=445hr.

The number of vehicles that will be affected by the incident.

To calculate the number of vehicles that will be affected by the incident can be found using the following formula:

Number of vehicles affected = V×tinc.

Substituting the values, we have

  =5400×2.=10800Veh.

The number of vehicles that will be affected by the incident is 10800Veh.

To calculate the average individual delay, we have the following formula:

Average individual delay =drNumber of vehicles affected 

Substituting the values, we have

  Averageindividual delay=drNumber of vehicles affected =44510800.=0.0412hr.

The average individual delay is 0.0412hr.

Conclusion:

Therefore, for 70 % expected demand flow:

Themaximum queue length that will be formed is qmax=400Veh, the total delay is dr=445hr,

The number of vehicles that will be affected by the incident is 10800Veh and the average individual delay is 0.0412hr.

To determine

(b)

The maximum queue length that will be formed, total delay, number of vehicles that will be affected by the incident, the average individual delay for the expected demand flow of 75%of the capacity of the highway.

Expert Solution
Check Mark

Answer to Problem 20P

For 75 % expected demand flow:

Themaximum queue length that will be formed is qmax=1000Veh.

The total delay is dr=1334hr.

The number of vehicles that will be affected by the incident is 10800Veh.

The average individual delay is 0.1235hr.

Explanation of Solution

Given:

We have been given the following information:

Total number of lanes = 3,

  V=90%of6000,

  tinc=2hr,

Mean free flow speed of the highway = 55 mi/h,

Jam density = 135 veh/mi/ln,

  Cr=4000.

Following is the lay out of the given highway section:

  MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months), Chapter 6, Problem 20P , additional homework tip  2

Calculation:

For the expected 75 % demand flow:

We have the following formula for the determination of maximum queue length that will be formed:

  qmax=(VCr)tinc

Where, qmax is the maximum queue length,

  V is the total volume,

  Cr is the reduced capacity,

  tinc is the duration of the incident.

Considering 75 percent of the flow and that the capacity of each lane is 2000 Veh/ h

Substituting the values in the following equation, we have

  qmax=(VCr)tincqmax=(( 75%of6000)4000)2qmax=(( 6000× 75 100 )4000)2qmax=(( 60×75)4000)2qmax=(45004000)2qmax=1000Veh.

Themaximum queue length that will be formed is qmax=1000Veh.

Now, the total delay, we have the following formula

  dr=tinc2(VCR)(CCR)2(CV)

Where, CR is the reduced capacity and can be found as follows:

  CR=Numberoflanesinuse×capacityofeachlane.CR=2000×(31).CR=2000×2=4000Veh/h.

And C is the total capacity and can be found as

  C=Totalnumberoflanesinuse×capacityofeachlane.C=2000×3.C=6000Veh/h.

Now, substituting the values in the required equation, we have

  dr=t inc2( V C R )( C C R )2( CV)dr=22( 75%of60004000)( 60004000)2( 600075%of6000)dr=4( 45004000)( 2000)2( 60004500)dr=4( 500)( 2000)3000dr=8( 500)3dr=1333.33=1334hr.

The total delay is dr=1334hr.

The number of vehicles that will be affected by the incident.

To calculate the number of vehicles that will be affected by the incident can be found using the following formula:

Number of vehicles affected = V×tinc.

Substituting the values, we have

  =5400×2.=10800Veh.

The number of vehicles that will be affected by the incident is 10800Veh.

To calculate the average individual delay, we have the following formula:

Average individual delay =drNumber of vehicles affected 

Substituting the values, we have

  Averageindividual delay=drNumber of vehicles affected =133410800.=0.1235hr.

The average individual delay is 0.1235hr.

Conclusion:

For 75 % expected demand flow:

Themaximum queue length that will be formed is qmax=1000Veh, the total delay is dr=1334hr, the number of vehicles that will be affected by the incident is 10800Veh and the average individual delay is 0.1235hr.

To determine

(c)

The maximum queue length that will be formed, total delay, number of vehicles that will be affected by the incident, the average individual delay for the expected demand flow of 80% of the capacity of the highway.

Expert Solution
Check Mark

Answer to Problem 20P

For 80 % expected demand flow:

Themaximum queue length that will be formed is qmax=1600Veh.

The total delay is dr=2667hr.

The number of vehicles that will be affected by the incident is 10800Veh.

The average individual delay is 0.247hr.

Explanation of Solution

Given:

We have been given the following information:

Total number of lanes = 3,

  V=90%of6000,

  tinc=2hr,

Mean free flow speed of the highway = 55 mi/h,

Jam density = 135 veh/mi/ln,

  Cr=4000.

Following is the lay out of the given highway section:

  MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months), Chapter 6, Problem 20P , additional homework tip  3

Calculation:

For the expected 80 % demand flow:

We have the following formula for the determination of maximum queue length that will be formed:

  qmax=(VCr)tinc

Where, qmax is the maximum queue length,

  V is the total volume,

  Cr is the reduced capacity,

  tinc is the duration of the incident.

Considering 80 percent of the flow and that the capacity of each lane is 2000 Veh/ h

Substituting the values in the following equation, we have

  qmax=(VCr)tincqmax=(( 80%of6000)4000)2qmax=(( 6000× 80 100 )4000)2qmax=(( 60×80)4000)2qmax=(48004000)2qmax=1600Veh.

The maximum queue length that will be formed is qmax=1600Veh.

Now, the total delay, we have the following formula

  dr=tinc2(VCR)(CCR)2(CV)

Where, CR is the reduced capacity and can be found as follows:

  CR=Numberoflanesinuse×capacityofeachlane.CR=2000×(31).CR=2000×2=4000Veh/h.

And C is the total capacity and can be found as

  C=Totalnumberoflanesinuse×capacityofeachlane.C=2000×3.C=6000Veh/h.

Now, substituting the values in the required equation, we have

  dr=t inc2( V C R )( C C R )2( CV)dr=22( 48004000)( 60004000)2( 60004800)dr=4( 800)( 2000)2( 1200)dr=4( 800)( 20)24dr=2666.66hr.dr=2667hr.

The total delay is dr=2667hr.

The number of vehicles that will be affected by the incident.

To calculate the number of vehicles that will be affected by the incident can be found using the following formula:

Number of vehicles affected = V×tinc.

Substituting the values, we have

  =5400×2.=10800Veh.

Therefore, the number of vehicles that will be affected by the incident is 10800Veh.

To calculate the average individual delay, we have the following formula:

Average individual delay =drNumber of vehicles affected 

Substituting the values, we have

  Averageindividual delay=drNumber of vehicles affected =266710800.=0.247hr.

The average individual delay is 0.247hr.

Conclusion:

For 80 % expected demand flow:

Themaximum queue length that will be formed is qmax=1600Veh, the total delay is dr=2667hr, the number of vehicles that will be affected by the incident is 10800Veh and

the average individual delay is 0.247hr.

To determine

(d)

The maximum queue length that will be formed, total delay, number of vehicles that will be affected by the incident, the average individual delay for the expected demand flowof 85% of the capacity of the highway.

Expert Solution
Check Mark

Answer to Problem 20P

For 85 % expected demand flow:

Themaximum queue length that will be formed is qmax=2200Veh.

The total delay is dr=4889hr.

The number of vehicles that will be affected by the incident is 10800Veh.

The average individual delay is 0.453hr.

Explanation of Solution

Given:

We have been given the following information:

Total number of lanes = 3,

  V=90%of6000,

  tinc=2hr,

Mean free flow speed of the highway = 55 mi/h,

Jam density = 135 veh/mi/ln,

  Cr=4000.

Following is the lay out of the given highway section:

  MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months), Chapter 6, Problem 20P , additional homework tip  4

Calculation:

For the expected 85 % demand flow:

We have the following formula for the determination of maximum queue length that will be formed:

  qmax=(VCr)tinc

Where, qmax is the maximum queue length,

  V is the total volume,

  Cr is the reduced capacity,

  tinc is the duration of the incident.

Considering 85 percent of the flow and that the capacity of each lane is 2000 Veh/ h

Substituting the values in the following equation, we have

  qmax=(VCr)tincqmax=(( 85%of6000)4000)2qmax=(( 6000× 85 100 )4000)2qmax=(( 60×85)4000)2qmax=(51004000)2qmax=2200Veh.

Therefore, the maximum queue length that will be formed is qmax=2200Veh.

Now, the total delay, we have the following formula

  dr=tinc2(VCR)(CCR)2(CV)

Where, CR is the reduced capacity and can be found as follows:

  CR=Numberoflanesinuse×capacityofeachlane.CR=2000×(31).CR=2000×2=4000Veh/h.

And C is the total capacity and can be found as

  C=Totalnumberoflanesinuse×capacityofeachlane.C=2000×3.C=6000Veh/h.

Now, substituting the values in the required equation, we have

  dr=t inc2( V C R )( C C R )2( CV)dr=22( 85%of60004000)( 60004000)2( 60005100)dr=4( 51004000)( 2000)2( 900)dr=4( 1100)( 20)18dr=4888.88hr.dr=4889hr.

Therefore, the total delay is dr=4889hr.

The number of vehicles that will be affected by the incident.

To calculate the number of vehicles that will be affected by the incident can be found using the following formula:

Number of vehicles affected = V×tinc.

Substituting the values, we have

  =5400×2.00.=10800Veh.

The number of vehicles that will be affected by the incident is 10800Veh.

To calculate the average individual delay, we have the following formula:

Average individual delay =drNumber of vehicles affected 

Substituting the values, we have

  Averageindividual delay=drNumber of vehicles affected =488910800.=0.453hr.

the average individual delay is 0.453hr.

The graph of average individual delay versus the expected demand flow is as follows:

  MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months), Chapter 6, Problem 20P , additional homework tip  5

Conclusion:

For 85 % expected demand flow:

Themaximum queue length that will be formed is qmax=2200Veh, the total delay is dr=4889hr, the number of vehicles that will be affected by the incident is 10800Veh and

the average individual delay is 0.453hr.

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Traffic and Highway Engineering
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ISBN:9781305156241
Author:Garber, Nicholas J.
Publisher:Cengage Learning