Question

You are the project manager's assistant on a construction site. The project manager is off site for the day and you are responsible for the ongoing progress of the site in her absence. Two similar beams are delivered to a construction site. Each beam is a hollow box beam, constructed of A36 structural steel, (E=29.0 x 10ʻpsi, density of 490 1bm/ft), each beam having outside dimensions of 16" base and 18" height, 18 feet long. Beam A has top and bottom wall thicknesses of 0.5" and side wal thickness of 0.375". Beam B has top and bottom wall thicknesses of 0.4375" and side wall thickness of 0.3125". The structure to be constructed calls for a beam that will support a point load of 55,000 lbs at the mid- point of the span of the beam, 16 feet between supports, with a maximum allowable deflection at the mid-point of 0.20" 2. The Contractor places Beam B into position according to the construction plans. Will this beam support the design load without exceeding the allowable deflection? If so, what is the factor of safety of the beam as installed? If not, what is the actual deflection of beam at the design load? What is the weight of Beam B as installed? Document your work. If a problem exists based on your analysis, describe the problem and present a recommended solution to the problem, including any effects this may cause to the construction of the project (i.e., additional costs, delays to the schedule, performance of the structure compared to its design and intended purpose, etc.). Document this in a memo to your project manager and be prepared to make an oral presentation (2-3 minutes) to the project manager if needed to defend your solution.

Answer 1

Given outside dimension = 16 base and 18" height, 18 feet long conceitratea Loaa (W) = 55000 lbf = 55 kips E =1sxloopsi Aluminium Density = 160 lb/ft² 17.125" [Alumini 1811 K 15.3751 The moment of inertia of given section about X-X axis is given by :- -1611 12 I = bd² - b, d,3 12 16x183_15.375 X17-1253 12 - 1341.33 in 4 We know that WE W = 3.05k/ft ㄷ Also Actual deflection /we" 384EI (3.05 x (1 8X1234 384X15000X1341133 = 0.8591 =0085911

We know that Allowable deflection -0.611 (i) No, this beam will not support the design load without exceeding the allowable deflection because actual deflection is greater than allowable aeflection. 0.859">006! (ii) Actual deflection = 0.85911 We know that Pii) A=w14 384 EI 0-2 =wXC18X1234 384 X1 5000 X1341:33 wo 0.709 Kiplfeet w=12777 ugebf ..] he maximum allowable load beam (B) can support = 12777.49 lbf (iv) We know that Area = (bc-bidi) (16"x18'' – 15.375"x170125') = 24.70 in? volume = 24.70X18X12) = 5335.87 in3 Density = 160 lb/ft3 = 0.0925 ebling .. weight of beam (B)=(5335.87X0.0925) = 493.56 lb Answer

Reccomended solution to the problem.

If we have time then we should wait for the steel beam because steel beam is more efficient.

Explanation

Since the elastic modulus of Aluminium is about 1/3 that of Steel, for the same beam shape properties, for the same loading, for the same span and support conditions, the Aluminium beam deflection will be three times more.

Density of steel is coincidentally about three times more than that of aluminium and so the same beam in steel will be three times heavier.

If subjected to same loading and having same moment of inertia of the cross section, aluminum beam will deflect more.

The reality is that the relative cost of selecting one or other material depends on the required application. However, as a little background:
The cost of steel varies according to its grade. By weight commercial grades can cost anything from about 1/2 the cost of . However, aluminium is about 1/3rd the density of steel, which means it breaks even with even the cheapest steel .

Since the havier section is requred to meet the requirement so more time will be consumed for fabrication leading to the delay in project completion.