Design the most economical W14 column to carry a load of 1650k if Lx = 24' and Ly = 16'. Flx-pin supports and Fy = 50ksi:
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A: Given W10x33 Dead load=20kips Live load=50kips B=12inch A36 steel Fc=3ksi
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A: at section a-a b = 350 mmh = 600 mmfc' = 35 MPafy = 420 MPaArea of steel at bottom = 5- dia 25…
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A: The FBD of the given structure can be drawn as shown below
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Q: 1. Design the most economical A36 steel base plate to resist a dead load = 1,400 kN and a live load…
A: Solving the first problem: Given: Live Load = 1700 kN; Dead Load = 1400 kN fc = 27 MPa; Column…
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A: Apply method of section: cut section is :
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A: Given: Section = W 14 x 120 fy=36 ksi fu=58 ksi Ag = 35.30 in2 tf = 0.94 in To find: The yielding…
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Q: 2. Given: A cover plate (3/4" x 10") is welded to one flange of a W12x45 as shown. Fy= 50 ksi The…
A: Size of plate = 3/4" x 10" W12 x 45 Fy = 50 ksi L = 22 ft or 264 in
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A: Given- A36 steel. Now, For A36 steel: fy=36 ksi=248.211 MPafu=58 ksi=400 MPa Factored load: LRFD:…
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A: As per Bartleby's guidelines we are supposed to answer only first question. Kindly repost other…
Q: From the Truss member shown, design all the tension members for ASD and LRFD, if the connection for…
A: Solution: A36 steel: fy=36 ksi=248.211 MPafu=58 ksi=400 MPa Factored load: LRFD:…
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A: fy =275MPa = 275N/mm2 Mmax = 600KNm
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- Situation 4: A W12x106 simply supported beam carries a uniformly distributed load. The properties of the section relevant to the problem are as follow: Use A50 steel and Cb - 1.14 d - 328 mm bf-310 mm tf = 25.1 mm tw = 15.5 mm kdes - 40.4 mm ry - 79.0 mm lx = 388 x 10^6 mm^4 ly - 125 x 10^6 mm^4 Sx - 2380 x 10^3 mm^3 Zx=2690 x 10^3 mm^3 J-3800 x 10^3 mm^3 Cw=2870 x 10^9 mm^6 1. Which of the following nearly gives the Design Strength Moment (LRFD) if the span of the beam is 2.5m? 878 KN-m 793 KN-m 836 KN-m 811 KN mload, 2 wat O: A fixed and beam is subjected to a load, 1/302 span from the left support as shown in the figure. The collapse load of the bean is 2W *4/3- X₂ Me 2mp71 Y K4/₂4/₂ Date: ZDetermine the type and amount of stress in each member of the truss shown below due to wind load using the method of joints. a.) Draw FBD b.) Solve the stress of all the members c.) Tabulate your answers (Length, Force, Compression/Tension, etc.)
- The beam shown in Figure is laterally braced at A, B, C, and D. Is a W18 × 119 adequate forFy = 50 ksi?a. Use LRFDb. Use ASD.Please answer the problem attached image using nscp 2015. ASAP thank you very much.ure B 2m2 m C -2 m D 4 m 500 N 700 N 2 of 2 Part B Identify the zero-force members in the truss shown in (Figure 2). Check all that apply. 000 CD DE CG CF FG BC BG DF AG EF AB Submit Provide Feedback Request Answer
- A W12x72 is used as a column and is supported by a concrete pier with top surface dimensions of 16 inch x 16 inch. f’c of column is 4,000 psi and is to carry a total service load of 250 kips with a Live load / Dead load = 3/2. Use LRFD solution. If B=N=14 inches, what is the minimum required computed thickness of the column base plate in inch? Express your answer in three decimal places. Use A36 for base plate.Situation 3: A W12x22 simply supported beam carries a uniformly distributed load. The properties of the section relevant to the problem are as follow: Use A50 steel and Cb - 1.14 d=312 mm bf - 102 mm tf 10.8 mm tw = 6.60 mm kdes 18.4 mm ry-21.5 mm Ix- 64.5 x 10^6 mm^4 ly- 1.94 x 10^6 mm^4 Sx416 x 10^3 mm^3 Zx 480 x 10^3 mm^3 J-122 x 10^3 mm^3 Cw-44.0 x 10^9 mm^6 1. Which of the following nearly gives the Design Strength Moment (LRFD) if the span of the beam is 2.5m? 89 KN-m O 114 KN-m O 123 KN-m O 101 KN-mO 88 130% v - + I Annotate T| Edit Trial expired Unlock Full Version ENGR 263 + A A A T O 4.10 Member AB is the beam under consideration. As shown in the illustration of the loading condition, member AB is an overhanging beam that supports a uniformly distributed roof load of 500 lb/ft. It also carries concentrated loads from a rooftop HVAC unit (4000 lb), an interior hanging display support (2000 lb), and a marquee sign (3000 lb). Marquee overnang Displau SLIPPort II Raof Kiots Ol I W = 500 Ib/Ft A B 4000 b 2000 000 I Steel beam !! I1 3 FE 5 FE - Ft RoOFtop HVAC unit 10 FE 4 Ft Marquee sign< R R2 Steel beam (negligible weight) Free-body diagram Dispiay Support Loading condition
- Situation 3: A W12x22 simply supported beam carries a uniformly distributed load. The properties of the section relevant to the problem are as follow: Use A50 steel and Cb - 1.14 d-312 mm bf - 102 mm tf = 10.8 mm tw = 6.60 mm kdes - 18.4 mm ry-21.5 mm Ix - 64.5 x 10^6 mm^4 ly-1.94 x 10^6 mm^4 Sx - 416 x 10^3 mm^3 Zx 480 x 10^3 mm^3 J-122 x 10^3 mm^3 Cw-44.0 x 10^9 mm^6 1. Which of the following nearly gives the maximum safe uniform load (Wu) that the beam could if the span of the beam is 0.800m? Neglect weight of the beam. 960 KN/m 933 KN/m 922 KN/m 975 KN/mCalculate the quantity of steel required for stirrups for the following beam. Y -2 - 12 Dia Anchor bars 4 - 16 Dia 8 mm Dia at 200mm C/C 2- 12 & 2 -16 Dia 650 mm 650 mm X Y 8 mm Dia at 350 mm C/C Clear Span 4000mm Clear Cover Bottom= 25mm Top= 25mm Ends=32mm Sp.Wt of Steel =7860kg/Cum 2- 12 mm Dia 2- 12 mm 2-16 mm Dia 2- 16 mm Dia 4- 16 mm Dia 550 mm 230 mm 1944.7 mm 550 mm -Q3/ A 7.0 m simply supported rectangular beam has a width of 450 mm and a depth of 850 mm. Supports, in addition to its own weight, a 40 kN/m dead load and a 35 kN/m live load. Knowing that fc' = 21 MPa and fy = 420 MPa, Find the required area of steel to support the applied loads