Question

Problem 1 Floor Systems Self Weight, Tributary Load and Load Path (25 pts.) The attached structure supports a floor system which will be used for office space. This structure is located inside a larger structure and it is independently supported both laterally and vertically Essentially, it stands by itself. It is call a mezzanine system. Lateral bracing on all four walls is not shown for clarity. The sections provided below are preliminary. Using the information provided on the sketch determine the following loads . Is the system one or two way action? 2. Interior joist 2 A-B uniformly distributed w_dead- 3. Edge joist 3 A-B uniformly distributed w dead 4, Interior girder 2-3 B uniformly distributed w-dead- 5. Edge girder 2-3 A uniformly distributed w dead- 6. Interior column 2B P dead 7. Edge column 2A P dead 8, Corner column 3AP dead- Draw free bod numerically explain your answers. This problem deals with self-weight determination. y diagrams of each case with the calculated dead loads. Be as clear as possible to Conrere ps PS sTeel Are No Te s n mere VN 12x2D ←w.shape dis na pon depth

Answer 1

25 The ratio of length to the width of the slab bay is3.57 > 2. The system is having a one-way slab action. Calculate the self-weight of the slab Self weight -ftx150 pcf 12 15 psf Self-weight of the beam W8x21 for the interior joist 2 A-B is equal to 21lb/ft Calculate the tributary width of the beam. Width -+ -6.5ft Calculate the uniformly distributed self-weight dead load on the interior joist 2 A-B 1kip +211bjft 1,0001b 1,0001b lkip Load 6.5ft x75psf x 0.5085 kip/ft Therefore, the uniformly distributed self-weight dead load on the interior joist 2 A-B is 0.5085 kipfft

Calculate the tributary width of the edge joist 3 A-B Width + 1 -4.5ft Self-weight of the beam W8x21 for the edge joist 3 A-B is equal to 21lb/ft Calculate the uniform load on the edge joist 3A-B. 1kip Load-4.5x75 Dsf X-Ikip+211b/ftx IK'P_ 1,0001b 0.3585 kip/ft Therefore, the uniformly distributed self-weight dead load on the edge joist 3 A-B is 0.3585 kip/ft Span of the interior girder 2-3 B is equal to 14 feet Calculate the tributary width of the interior girder 2-3 B as below. widtht 30 25 - 27.5ft

Self-weight of the beam W12x 30 for the interior girder 2-3 B is equal to 301Ib/ft Calculate the uniform load on the interior girder 2-3 B 1kip _IK'P. 1,0001b Load-27.5 x75 psfX-1Kdp +301b/ftx 1,0001b 2.0925 kip/ft Therefore, the uniformly distributed self-weight dead load on the interior girder 2-3 B is 2.0925 kip /ft Span of the edge girder 2-3 A is equal to 14 feet. Calculate the tributary width of the edge girder 2-3 A as below. 30 width-+1 - 16ft Self-weight of the beam W10x 26 for the edge girder 2-3 A is equal to 26lb/ft Calculate the uniform load on the edge girder 2-3 A. Load -16x 75 psf x1,0001Ib Ikip_ +261b/ft×10001b = 1.226 kip/ft

Therefore, the uniformly distributed self-weight dead load on the edge girder 2-3 A is 1.226 kip/ft 6. Tributary area for load distribution on the interior column 2B is 6 25 7 256 30 7 30 22丿(22丿(22)(22, Load on the column is equial Load = 178.75×75 psf x 1kip 1,0001b = 14 kips Therefore, the point dead load on the interior column 2B is 14 kips

Tributary area for load distribution on the edge column 2A is 6 (30 7 30 Load on the column is equal to 1kip 1,0001b Load-104x75 psf x-1Kp -7.8 kips Therefore, the point dead load on the edge column 2A is 7.8 kips Tributary area for load distribution on the c column 3A is 30 ++172ft Load on the column is equal to Load - 72x75 psf x IXiP 1kip 1,000lb 5.4 kips Therefore, the point dead load on the corner column 3A is5.4 kips