Select the lightest ASTM A992 W-shape column to carry an axial dead load of 200 kips and a live load of 240 kips. The column is 16 ft long and is pinned top and bottom on both axes. Limit the column size to a nominal 14-inch shape. Check the limiting Width-thickness Ratio for the flange and web.
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- Problem 1: Hanger in Tension with Block Shear The W12 X 53 tension member shown below has two rows of three 1-in-diameter A325N bolts in each flange. Assuming (Fy = 50 ksi, Fu 65 ksi) ASTM A572 steel and considering the = strength of the W12 X 53 only. 1. Determine the design tensile strength of the W12 X 53. 2. Determine the service dead load that can be supported if there is no live load. 3. If a service dead load PD 100 kips is applied, what is the maximum service live load, P₁, that can be supported? = OD Section b (a) Side View 3 -0 O O 5/2" (b) End View of BoltsCompression member.O 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
- ASSIGNMENT #5 BIAXIAL BENDING Problem: Check the beam shown for compliance with the NSCP Specification. Lateral support is provided only at the ends, and A992 steel is used, F, = 345 mPa. The 90 kN service loads are 30% dead load and 70% live load. a. Use LRFD b. Use ASD d tw by ty 1x(106) S(10³) mm mm mm mm³ mm mmª W410x100 415 10 260 16.9 398 1920 SECTION Tx mm 177 1m 90 kN 90 KN 1.5m W410x100 I Loaded Section Ly(106) Sy(10³) Ty mmª mm³ mm 49.5 381 62.4 1m Zx(10³) Zy(10³) mm³ mm³ 2130 581Refer to the beam loading diagram shown If the beam is made from A992 and is continuously braced, determine the most economical member W-shape for strength only I am using AISC Steel Construction Manual 15th Edition. For the double fixed-end bend shown, the following data applies: • Service live load, P = 30 kips Required live load factor: 1.6 Beam length, L = 30 ft Self-weight has already been included. A В C 0.5L 0.5LSITUATION 2 Determine the safe service load W permitted for this beam-column that is not part of a frame system. Assume ASTM A992 (Fy = 350 MPa; Fu = 450 MPa) as material. Use LRFD specifications. Servise koada P= 30 kips dead load 80 kips live koad Assume hinged for both principal 10- 0 directions W- 20 dead load 8OG live load 10-o Fined for both principal directions Type your final answer/s in the text box provided below.
- Prepare the Bar bending schedule of a simply supported R.C.C. Lintels from the following specification: Size of lintel 300mm widex 200mm depth. Main bars in tension zone of Fe 250(grade I) 3 bars of 16mm dia., one bar is cranked through 45 degree at 170 mm from each end 2 No. anchor bars at top 8mm dia. Two legged stirrups@150mm c/c of 6mm dia. through out. Clear span of the lintel is 1150mm. Bearing on either side is 150mmThe built-up section shown below is fastened together by passing two 10 mm diameter rivets through the top and bottom plates into the flanges of the beam. Each rivet will withstand 11.8 kN in shear. Determine the required spacing of the rivets along the length of the beam if it carries a shearing force of 199 kN. 12 mm x 180 mm plates (2) I IPE I 400x180x642.2 steel 10 mm rivetsFor the frame shown below, use the following information: • Columns are oriented so that the y axis is in plane and the x axis is out of plane. • All columns and girders are A992 steel. • The girders to column connections in the plane of the elevation are as indicated. • All out of plane girder to column connections are simple. • Base plate fixity as indicated applies to x and y axis. 1. Determine the ultimate load Pu for column 2 (i.e. along grid line 2) Pu= 2. Determine the nominal axial capacity, ¢P, for column 2 according to part 16, specifications of AISC steel manual 3. Determine the nominal axial capacity, oPn using the design tables. PPn= 4. Check if column 2 is safe or not Show all calculations and identify each step for credit. Clearly state any assumptions made. 3 FACTORED UNIFORMLY DISTRIBUTED LOAD, w = 15.0 ktf SUPLE W40x215, Ix = 16,700 W40x149, Ix = 9,780 W40x215 SPLI ERAC RACE BRACE GRACE RACE ACE FOED 50- 0" 35- 0 50- 0" Condition at Far End Sidesway Prevented,…
- Calculate the maximum, unfactored tensile dead load the angle bar can sustain based on gross yielding by LRFD if Fy = 248 MPa and that service tensile live load is 135 kN. Use resistance factor = 0.9. Write your final answer to the nearest whole number in KN. Angle bar, 150x90x8 mm, 20-mm-dia. bolts 40 40 40 O O O tät 57 63 |--50 |The tension member shown below is a PL 12 X 8 of A36 steel. The member is connected to a gusset plate with 1 1/8 in diameter bolts. It is subjected to the dead and live loads shown below. Does this member have enough strength? Assume that A₂ = An. If it does not have enough strength, how would you resolve this issue? »L 16x8 DeadThe tension member shown is an L150 × 87.5 × 8. The bolts are 19 mm in diameter. If A36 steel is used, is the member adequate for a service dead load of 138 kN and a service live load of 138 kN? a. Use LRFD. b. Use ASD. 50 37.5 37.5 37.5 56.25 62.5