CMGT 372 Tulsa Community College Structural Engineering Exercise

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Engineering

CMGT 372

Tulsa Community College

CMGT

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i need help with the following exam-review and homework.

answers with explanation on them please also hand written.

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CMGT 372 – STRUCTURAL ASPECTS in CONSTRUCTION II EXAM # 2 – REVIEW 1. Given the floor plan of this 4-story building shown below, determine the lightest 10” “W shape” for Column “A” at the base (Ground Floor) if the floor to floor height is 12 feet and the superimposed Service Dead Load is 100 psf and the superimposed Service Live Load is 80 psf for floors 2 through 4. Also include the loads from the roof: Service Dead Load of 25psf and Service Live Load of 30psf. The steel for the column is to be Grade 50, it is braced in both directions at the floor/roof lines and K=1. Ignore the weight of the exterior wall. Rf 4th North 3rd 30’ 2nd Ground Plan Detail of Column A Elevation Column A 30’ 36’ 32’ Plan View Ans.: W10 x _______ 2. Using the same information in the problem above, if the unbraced length of Column “A” was 18 feet would the answer you determined in Problem #1 still be acceptable Yes or No? ______ cPn = _________ CMGT 372 Exam #2 REVIEW pg. 1 3. Using 7/8” diameter A325 bolts (threads excluded), determine the available strength based on the smaller plate. Both plates are 5/8” thick plates and are A36 steel. The smaller (top plate) is 6” wide. Consider yielding, net section rupture and block shear failures. 2” 3” Max Load Rn = _____________ kips 2” Failure mode (circle one): 1.5” P 3” Yielding P Rupture Block Shear 1.5” CMGT 372 Exam #2 REVIEW pg. 2 4. Using the figure from problem #4, determine the available shear strength of the bolts and the bearing strength of the plate. Consider both bolt bearing and tearout failures when you investigate hole failure. 2” 3” Max Load Rn = _____________ kips 2” Failure mode (circle one): 1.5” P 3” Bolt Shear P Bolt Tearout Bolt Bearing 1.5” CMGT 372 Exam #2 REVIEW pg. 3 5. Determine the net width for a 14 x ½” plate with ¾” dia bolts placed as shown in the figure below: wn = _____________ in An = _____________ in2 CMGT 372 Exam #2 REVIEW pg. 4 Chapter 10 HW
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