PRIN.OF HIGHWAY ENGINEERING&TRAFFIC ANA.
7th Edition
ISBN: 9781119610526
Author: Mannering
Publisher: WILEY
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Chapter 2, Problem 17P
To determine
The difference in minimum theoretical stopping distances with and without aerodynamic resistance considered.
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15'-0"
15'-0"
Architect specifications:
1) Superimposed dead load (including finishes, partition, mep etc) for each floor are shown next page.
2) The facade weight is 200 plf and it's all around the building, the parapet weight at the roofs is 100 plf. Don't forget to add these loads at the perimeter beams.
3) Wind Load is 30 psf
4) Use steel or LVL for beams, TJI of common wood for joists.
5) Submit your calc package by Sunday May 11th
ALL STRUCTURAL ITEMS IN RED ARE PART OF THE GRAVITY SYSTEM
ALL STRUCTURAL ITEMS IN BLUE ARE PART OF THE LATERAL SYSTEM
I
10'-0"
2
10'-0"
10'-0"
B I
I
I
LAYOUR FOR OFFICES
Things Required
Dead and Live Load Schedule
Gravity design for office floor -
Load calculations (DL + LL + facade load if on perimeter)
Beam sizing
Joist spacing/type (TJI or LVL)
Provide calculations for:
Reaction forces
Maximum moments
Shear
•
Beam sizes (steel or LVL)
Key Map of elements (your office layout plan with beam/joist marks)
I
please solve the problem in the picture make sure you show all of your work, thank you so much for your help!
Please solve the question in the picture make sure you show all of your work and show every step you do please. Thank you so much for your help!
Chapter 2 Solutions
PRIN.OF HIGHWAY ENGINEERING&TRAFFIC ANA.
Ch. 2 - Prob. 1PCh. 2 - Prob. 2PCh. 2 - Prob. 3PCh. 2 - Prob. 4PCh. 2 - Prob. 5PCh. 2 - Prob. 6PCh. 2 - Prob. 7PCh. 2 - Prob. 8PCh. 2 - Prob. 9PCh. 2 - Prob. 10P
Ch. 2 - Prob. 11PCh. 2 - Prob. 12PCh. 2 - Prob. 13PCh. 2 - Prob. 14PCh. 2 - Prob. 15PCh. 2 - Prob. 16PCh. 2 - Prob. 17PCh. 2 - Prob. 18PCh. 2 - Prob. 19PCh. 2 - Prob. 20PCh. 2 - Prob. 21PCh. 2 - Prob. 22PCh. 2 - Prob. 23PCh. 2 - Prob. 24PCh. 2 - Prob. 25PCh. 2 - Prob. 26PCh. 2 - Prob. 27PCh. 2 - Prob. 28PCh. 2 - Prob. 29PCh. 2 - Prob. 30PCh. 2 - Prob. 31PCh. 2 - Prob. 32PCh. 2 - Prob. 33PCh. 2 - Prob. 34PCh. 2 - Prob. 35PCh. 2 - Prob. 36PCh. 2 - Prob. 37PCh. 2 - Prob. 38PCh. 2 - Prob. 39PCh. 2 - Prob. 40P
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- 8. A prestressed concrete beam is subjected to the following stress distributions: Pi is the initial prestressing force, Pe is the effective prestressing force, M, is the bending moment due to self- weight, Ma and M, are the dead load and live load bending moment, respectively. The concrete has the following properties: fr = 6000 psi and fri = 4200 psi +250 -85 -2500 +550 Pe+ Mo+Ma+Mi P alone P₁+ Mo -2450 -3500 Stress at midspan +210 +250 P, alone Pe alone -2500 -3500 Stress at ends Using Table 22.1, evaluate whether the stresses at the center of the span and the end of the span comply with the permissible stress limits. The beam is classified as U-class. Provide justifications for each condition listed in the table. Note: Calculated stresses are to be taken from the above diagram, and permissible stresses are to be calculated using Table 22.1. Compressive stresses immediately after transfer Tensile stresses immediately after transfer Compressive stresses under sustained and total…arrow_forward10. A short column is subjected to an eccentric loading. The axial load P = 1000 kips and the eccentricity e = 12 in. The material strengths are fy = 60 ksi and f = 6000 psi. The Young's modulus of steel is 29000 ksi. (a) Fill in the blanks in the interaction diagram shown below. (2pts each, 10pt total) Po Pn (1) failure range H 3" 30" Ast 6 No. 10 bars = P 22" I e H 3" (4) e = e small Load path for given e Radial lines show constant (2) eb (3) e large failure range Mn (5) e= Mo (b) Compute the balanced failure point, i.e., P and Mb.arrow_forwardNo chatgpt plsarrow_forward
- 11. The prestressed T beam shown below is pretensioned using low relaxation stress-relieved Grade 270 strands. The steel area Aps = 2.5 in². The tensile strength is fpu = 270 ksi, and the concrete compressive strength is fr = 6000 psi. (a) Calculate the nominal moment strength Mn with hr = 6 in. 22" 15" T hf (b) Since this beam is a T-beam, the nominal moment strength M₁ increases with a thicker hf. However, M, stops increasing if he reaches a value. Determine the minimum thickness hy that can achieve the maximum nominal moment strength Mr. Also, calculate the corresponding maximum nominal moment strength Mn with the computed hf.arrow_forward10. A short column is subjected to an eccentric loading. The axial load P = 1000 kips and the eccentricity e = 12 in. The material strengths are fy = 60 ksi and f = 6000 psi. The Young's modulus of steel is 29000 ksi. (a) Fill in the blanks in the interaction diagram shown below. 30" Ast 6 No. 10 bars = Pn (1) Po (4) e = e small Load path for given e failure range Radial lines show constant (2) eb (3) e large failure range Mn (5) e= Mo (b) Compute the balanced failure point, i.e., P and Mb. H 3" P 22" I e H 3"arrow_forward10. A short column is subjected to an eccentric loading. The axial load P = 1000 kips and the eccentricity e = 12 in. The material strengths are fy = 60 ksi and f = 6000 psi. The Young's modulus of steel is 29000 ksi. (a) Fill in the blanks in the interaction diagram shown below. 30" Ast 6 No. 10 bars = Pn (1) Po (4) e = e small Load path for given e failure range Radial lines show constant (2) eb (3) e large failure range Mn (5) e= Mo (b) Compute the balanced failure point, i.e., P and Mb. H 3" P 22" I e H 3"arrow_forward
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