A W8x24 A36 column has an effective length of 10 ft about the y-axis and 22 ft a

Question

A W8x24 A36 column has an effective length of 10 ft about the y-axis and 22 ft about the tx-xis. Determine the factored compressive strength (P) using the LRFD criteria and ndicate whether the limit state is elastie or inelastie buckling. 2 A W10x60 column with an effective length of 30 ft for both axes is called upon to carry a points compressive dead load of 115 kips and a compressive live load of 160 kips. Using the LRFD criteria (and the load combinations from ASCE7-10), determine whether the column will support the load. Evaluate the strength for ) F,-50 ksi, and ()F-70 ksi. A W12x170 column of AS72 Grade 42 steel is shown here with end conditions that 3. 5 points approximate ideal conditions. Determine the factored compressive strength (.P.) for buckling about each axis. Neglect torsional buckling. i8 ft A W14x99 A992 columns is 20 ft long, pinned at each end, and braced at mid-height to S points tent buckling about the y-axis. However, this bracing is not adequate to resist torsion. Considering flexural and torsional buckling, determine the nominal strength of this compression member

Explanation / Answer

1)given section is W8x24

grade of steel=fy=36 ksi

effective length about y axis=10 ft

effective length abpout x axis=22 ft

radius of gyration about y axis = 1.61 in

radius of gyration about x axis=3.42 in

slenderness ratio about y axis=10*12/1.61=74.5

slenderness ratio about x axis=22*12/3.42=77.2

slenderness ratio about x axis will govenr the buckling strength

4.71*sqrt(E/fy)=4.71*sqrt(29000/36)=133.68 >77.2

Euler buckling load =fcr= pi2E/*kL/r)2 = pi2*29000/77.22 = 48 ksi

crtical stress = (0.65836/48)*36=26.3 ksi

area of W8x24=7.1 in2

design strength of w8x24=0.9*26.3*7.1=168 kips

since kL/r < 4.71sqrt(E/fy), inelastic buckling will happen

2)factored compressive load on column = 1.2*115 + 1.6*160=394 kips

i) effective length of column=30 ft

for grade 50 steel, compression strength can be directly obtained from Table 4-1 of AISC

design strength of column = 204 kips

The section is not adequate

ii) radius of gyration of section about minor axis = 2.57 in

slenderness ratio = 30*12/2.57=140

4.71*sqrt(E/fy)=4.71*sqrt(29000/70)=95.9<140

Euler buckling stress = pi2*29000/1402 = 14.6 ksi

crtical stress = 0.877*14.6=12.8 ksi

area of W10x60=17.6 in2

design strength = 0.9*12.8*17.6=203 kips

section is inadequate

3)effective length about minor axis=18 ft

effective length about major axis = 0.7*18=12.6 ft (<18 ft)

buckling about minor axis will govern

radius of gyration about minor axis = 3.22 in

slenderbness ratio about minor axis = 18*12/3.22=67

4.71*sqrt(E/fy)=4.71*sqrt(29000/42)=123.8 >67

Euler buckling stress = pi2*29000/672 = 63.7 ksi

crtical stress = 0.658(42/63.7)*42 = 31.9 ksi

area of W12x170=50 in2

design strength = 0.9*50*31.9=1434 kips

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