A 7.3-m long simply supported beam is subjected to a concentrated live load of 71.1 kN at midspan and to a uniformly distributed dead load of 26.2 kN/m (excludes self-weight). A W360X91 of A992 steel is used on the beam. Lateral bracing is provided at the ends only. Calculate the design (LRFD) moment strength (kN-m) of the beam. Next
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- 5. The beam-column has a height of 7.5m and is a member of a braced frame. It is subjected to an axial load of DL-65KN and LL= 1OOKN. It is pinned at both ends. A W12X35 section was used and was made up of A-36 steel with Fy-250MPA. Compute the interaction value for beam- column based on NSCP 2015 code provisions. Ngelect the weight of the section. A = 6645 mm2 Zx- 839x10^3 mm3 d- 317.50mm Sx- 747x10^3 mm3 tw= 7.62 Rx= 133.35 bf= 166.62 ly= 10x10^6 mm4 tf- 13.21 Sy= 122x10^3 mm3 Ix= 119x10^6 mm4 Ry3 39.12mm 3.5 Note: Use C=D1.00%--0.2The beam shown has the following properties: steel cover=60mm, f'c=21 MPa, and fy=420 MPa. Determine the ultimate moment capacity of the beam in kN-m. Determine also the type of failure and the beam condition (under-reinforced, over-reinforced or balanced condition) 250mm 2-Ø20mm 3-025mm h = 450mm 50mm1. Check the beam shown for compliance with the AISCS. Lateral support is provided only at the ends, and A992 steel (E 345 MPa and Fy= 450 MPa). The only uniform service dead load is the weight of the beam. The 70 KN service loads are 30%DL and 70%LL. Use LRFD. 70 KN 70, KN 70 KN W 14 x 68 0.90m 1.20m 0.90m WI4 x68 IR, R2
- A built up section of A992 steel, F, = 345 mPa is made from plates fully welded together. The flanges consist of PL16X380 and PL12X500 for the web. Use the NSCP 2015 specifications. PL16x380 WL = 1.5wp %3D A Wu = 3.6wp 10m Which of the following best gives the maximum service dead load, wp? 40.2 kN/m 26.8 kN/m O 10.8 kN/m O 96.6 kN/m PL12x500DESIGN FOR TENSION ONLY (ACI METHOD fe=21mPa fy=276 MPa trength Design 2015 NSCP use ba 24 TL I SPAN 1: DL= 8 KN LL=12 KW/m L = 4.50m 2: DL= 10x/m LLIG W/m L=450m Ⓒ: DL=10 b/m LL 15/m L=4.70 m (4): DL = 8 KN/ 1 =12xx/m L = 4.60m TL ITSelect an A992 (E= 345 MPa and Fu= 450 MPa) W-shape for the floor beam AB of the floor system shown. In addition to the weight of the beam, the service dead load consists of 125 mm thick reinforced concrete slab (we= 24 KN/m³). The service live load is 3.80 KPa and there is 0.95 KPa movable partition load. The total deflection (DL + LL) must not exceed L/240, where L is the length between supports. Use ASD tairder Flear Beam - Hax column Im overhang 4@1.5 m = 6 m
- Bearing plate is design to distribute the reaction of W12x65 with a span length of 12 ft center-to-center of supports. The total service load including the beam weight is 15 kips/ft with Live load / Dead load ratio equals to 2. The beam is to be supported on RC walls with f’c =4,000 psi. For the beam, Fy = 50 ksi and Fy = 36 ksi for the plate. Use LRFD. If ℓb = 8 inch and B = 14 inch, what is the minimum computed thickness of the bearing plate in inch? Express your answer in three decimal places.Bearing plate is design to distribute the reaction of W12x65 with a span length of 12 ft center-to-center of supports. The total service load including the beam weight is 15 kips/ft with Live load / Dead load ratio equals to 2. The beam is to be supported on RC walls with f'c =4,000 psi. For the beam, Fy = 50 ksi and Fy = 36 ksi for the plate. Use LRFD. What is the minimum bearing length, {, to prevent web yielding in inch? Express your answer in two decimal places.A bracket of length 1.5m is rigidly attached to the simply-supported beam at its end. The beam is subjected to the combined effect of M,, M, and compressive force P caused by the factored loads as shown. There are lateral supports only at the ends. The member is made of HE300A section and steel grade is S275 (Fy-275MPA, Fu= 430 MPa). The loads acting are determined from LRFD combinations. Determine whether the member can safely carry the applied loads. Multiply first-order moment diagrams by a factor of 1.2 to account for second-order effects. No need to check shear limit state. You do not need to additionally consider the beam self-weight. E Use the following simple expression to calculate the value of L:: L, = T*rts 0.7*Fy y P = 400kN 1.5 m F = 150kN 2 т 4 т X: Lateral support НЕЗ00А
- AW460×60 steel beam is loaded as shown. Assume w - 50 kN/m, Mo - 95 kN-m, Las - 3.8 m, Lec - 1.6 m, E- 200 GPa, and I- 255 x 10° mm. (a) Determine the reaction force at roller B. Forces are positive upwards. (b) Determine the maximum positive bending moment in the span. (c) If the allowable bending stress is 210 MPa, determine the minimum acceptable section modulus for the steel beam. Mo LAB LBC Answer: (a) Ву - kN (b) Mpos kN-m (c) Smin x 10° m3An 80-foot long plate girder (see below) is fabricated from a ½-inch x 78-inch web and two 3inch x 22-inch flanges. Continuous lateral support is provided. The steel is A992. The loading consists of a uniform service dead load of 1.0 kip/ft (including the self-weight), a uniform service live load of 2.0 kips/ft, and a concentrated service live load of 500 kips at midspan. Stiffeners are placed at each end and at 4 feet, 16 feet, and 28 feet from each end. Once stiffener is placed at midspan. Determine whether the flexural strength is adequate using LRFD.Verify the adequacy of column AB, part of a sway frame structure, to carry the loads shown in the figure. The column is made of ASTM A36 steel (Fy = 250 MPa). Use ASD specifications. For determining its effective length factor, neglect inelastic effect. %3D W21X62 C W21X62 L=5 m L=5 m Flanges of Columns and web of girders are in the plane of the frame NOTE: Take note of the moment of inertia to be used (ly) for columns. W18X50 W18X50 L=5 m L=5 m P(DL) = 40 kN NM 001 = (10)d NM 00S = (1)d NY SZI = (1)d W21X62 L-4.2m W21X62 L=4.2 m