A built up l-section has a total unsupported length of 6m. It is made up of 2 - 400mm x 12mm plates as flanges and 450mm x 12 mm plate as web. The plates are of Grade 40 steel (fy = 276MPa). Considering the local buckling of the stiffened and unstiffened slender elements, determine the ultimate axial load capacity of the built-up column for both LRFD and ASD. Use NSCP 2015 Specification and effective length factor k = 1.0. Properties of the built-up section: A = 15,000 mm² ly = 128.06 x 106 mm4 Ix = 6.035 x 108 mm4 rx = 200.58 mm ry = 92.40 mm What is the Slenderness Ratio? KL/r Using LRFD, what is the ultimate load capacity of the column? Express your answer in KN Using ASD, what is the allowable load capacity of the column? Express your answer in KN

A built up l-section has a total unsupported length of 6m. It is made up of 2 - 400mm x 12mm plates as flanges and 450mm x 12 mm plate as web. The plates are of Grade 40 steel (fy = 276MPa). Considering the local buckling of the stiffened and unstiffened slender elements, determine the ultimate axial load capacity of the built-up column for both LRFD and ASD. Use NSCP 2015 Specification and effective length factor k = 1.0. Properties of the built-up section: A = 15,000 mm² ly = 128.06 x 106 mm4 Ix = 6.035 x 108 mm4 rx = 200.58 mm ry = 92.40 mm What is the Slenderness Ratio? KL/r Using LRFD, what is the ultimate load capacity of the column? Express your answer in KN Using ASD, what is the allowable load capacity of the column? Express your answer in KN

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

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%--0.2
4.7-2 An HSS 10 x 6 x 6 with F 46 ksi is used as a column. The length is 15 feet. Both ends are pinned, and there is support against weak axis buckling at a point 6 feet from the top. Determine a. the design strength for LRFD. b. the allowable stress for ASD.
A reinforced concrete simply supported beam has a span of 20 ft and supports service dead load of 2.5 kip/ft and a uniform service live load of 2.5 kip/ft in addition to its self-weight. Design a beam section as per ACI 318-14 to resist the factored external bending moment. The concrete compressive strength at 28 days is(4 + 15)ksi and yielding strength of steel is (60+ 15)ksi.
Principle of superposition The beam is indeterminate to the first degree. Force in member BD is chosen as the redundant. This member is therefore sectioned to eliminate its capacity to sustain a force. 60 kN/m 3 m 1.8 m 60 45 actual structure (a) 160 kN/m B FaD redundant рrimary structure ARD FRp applied Fap D
For the given loaded truss, the allowable tensile stress in any member is up to 800 MPa while the allowable compressive stress is up to 650 MPa. The reduced stress in compression is specified to reduce the danger of buckling. Determine the minimum cross-sectional area of member CE. Solve with complete solution and proper FBD.
I need the answer as soon as possible
Solve it.
SOLVE please Determine the design bending strength of ISLB 350 @ 486 N/m, considering the beam to be laterally supported. Assume factored design shear force will not exceed 0.6 times the design shear strength of the cross-section. If it is used as simply support beam of unsupported length 5 m then determine the factored distributed load the beam can carry. Use Fe410 grade steel For ISLB 350 @ 486 N/m, section properties are as follows: h = 350 mm; b= 165 mm; t₁ = 11.4 mm; t = 7.4 mm; R₁ = 16 mm; Izz = 13158.3 × 104 mm²; I = 631.9 × 10¹ mm²; Zpz = 851.11 × 10³ mm³; Zez = 751.9 x 10³ mm³
Figure 1 shows the cross-section of a heavily welded I-section steel column under axial compression. The yield stress of the steel section is 300 MPa. (a) Check the slenderness of the flanges and webs. (b) Determine the effective and gross areas of the cross-section. (c) Determine the design axial capacity of the cross-section. 380 E 10 425 -8 Figure 1. Cross-section of the rectangular hollow section steel column (Units: mm)
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
-This Engineering Mechanics - Static of Rigid Bodies : Method of Sections - [ ] Solution must include FBDs #4
The floor structural system shown in Fig 1 comprise W-section steel beams and girders supporting a 5-inch thick concrete slab which provides full restraint to the flanges of the beams. The density of concrete is 150 lbs. /ft³. and the uniformly distributed live load L=50 lbs/ft². L-22¹ B B. Girder G1: - [82] 12' Gl GI [BI] FIG. 1-Steel Floor Beam a) Strength - Bending and Shear 12' + I 1) Produce a sketch of the floor plan showing the tributary load distribution areas for the beams and girders [zul