NYIT Instructors: Alfred Sanabria and Rodrigo Suarez
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1 NYIT Instructors: Alfred Sanabria and Rodrigo Suarez
2 Massive stone columns, used from Stonehenge to Ancient Greece were stabilized by their own work With steel and concrete technology columns have become increasingly slender
3 Slenderness is a property that relates a column length with its cross-sectional dimensions. Slenderness is critical in the strength of columns due to buckling
4 Buckling is the sudden uncontrolled lateral displacement of a column, at which point no additional load can be supported. A sudden failure mode for members under high compressive stresses that occurs at a load that is lower than the maximum capacity of the material yield strength. BIrU
5 Short columns fail by crushing due to material failure (yield), long, slender columns fail by buckling
6 Since short columns fail by crushing (material failure) the formula for the compressive stress is simply: a P actual σ σ A allowable
7 Since short columns fail by crushing (material failure) the formula for the compressive stress is simply: Actual Axial Stress a P actual σ σ A Axial Load on Column Column Cross- Sectional Area allowable Allowable Axial Stress
8 Since short columns fail by crushing (material failure) the formula for the compressive stress is simply: A Column Cross- Sectional Area required Axial Load on Column σ P actual allowable Allowable Axial Stress
9 Buckling happens because loads aren t perfectly axial and there are material irregularities that develop some bending in the member. Euler s equation: P critical π EI L min
10 Buckling happens because loads aren t perfectly axial and there are material irregularities that develop some bending in the member. Euler s equation: Axial Load at which the column buckles P critical π EI L min Effective Length of the column
11 The radius of gyration of a column cross section is a measure of the distribution of area (mass) around the centroid of the cross section Radius of gyration r I A Moment of Inertia (smallest) Cross-Sectional Area I Ar
12 The radius of gyration of a column cross section is a measure of the distribution of area (mass) around the centroid of the cross section I Ar P critical π EI π E( Ar ) σ min critical A AL AL π ( L / E r)
13 The radius of gyration of a column cross section is a measure of the distribution of area (mass) around the centroid of the cross section σ critical π E ( L / r ) Slenderness ratio
14 For columns, the radius of gyration and moment of inertia that matter are the smaller ones. The higher the slenderness ratio (long, thin column), the lower the load at which the column will buckle
15 The most efficient columns have rxry W sections used as columns are usually boxy and 10 to 14 deep
16 Determine the critical buckling stress for a 30 long w1x65 steel column. Assume simple pin connections at the top and bottom. Fy36ksi (A36 steel) E9,000 ksi rx5.8 ry3.0
17 Determine the critical buckling stress for a 30 long w1x65 steel column. Assume simple pin connections at the top and bottom. σ L L r crit x r y Use119. π E ( L r) (30x1) in 5.8in (30x1) in 3.0in Fy36ksi (A36 steel) E9,000 ksi rx5.8 ry3.0
18 Determine the critical buckling stress for a 30 long w1x65 steel column. Assume simple pin connections at the top and bottom. σ crit π E ( L r) π (9,000ksi) (119.) 0.1ksi Fy36ksi (A36 steel) E9,000 ksi rx5.8 ry3.0
19 In the simple analysis, columns were assumed to be pinned at either end, so the column would buckle in a smooth curve. The length that is free to buckle is greatly influenced by the end supports. A K factor is used to increase or reduce the effective length of the column to take the end supports into account.
20 P critical π EI (KL) min K
21 K K1 K0.5 K0.7 K
22 Just as rigid end connections reduce the buckling length of a column, lateral bracing can also increase the column capacity by reducing the buckling length. Lateral bracing is usually provided in the weak buckling direction
23 Lb Lb P critical π EI (KL)
24 P critical π EI (KL)
25 Slenderness ratios must be calculated for both axes to determine which direction governs P critical π EI (KL)
26 Determine the critical buckling load for a 4x8 S4S Douglas Fir column that is 18 long and braced at midheight against the weak direction of buckling. E1,300 ksi
27 Determine the critical buckling load for a 4x6 S4S Douglas Fir column that is 18 long and braced at midheight against the weak direction of buckling. E1,300 ksi 6
28 Determine the critical buckling load for a 4x6 S4S Douglas Fir column that is 18 long and braced at midheight against the weak direction of buckling. E1,300 ksi P critical π EI (KL)
29 Determine the critical buckling load for a 4x6 S4S Douglas Fir column that is 18 long and braced at midheight against the weak direction of buckling. E1,300 ksi Strong Axis 3 I 1 bh 1 (1/1)(4)(6 7in 4 P critical 3 ) π EI (KL) Weak Axis 3 I 1 bh 1 (1/1)(6)(4 3in 4 3 )
30 Determine the critical buckling load for a 4x6 S4S Douglas Fir column that is 18 long and braced at midheight against the weak direction of buckling. E1,300 ksi Strong Axis 3 I 1 bh 1 (1/1)(4)(6 7in 4 P critical 3 ) π EI (KL) Weak Axis 3 I 1 bh 1 (1/1)(6)(4 3in 4 3 ) L b 18' x(1in / 16in ft) L b 9' x(1in / 108in ft)
31 Determine the critical buckling load for a 4x6 S4S Douglas Fir column that is 18 long and braced at midheight against the weak direction of buckling. E1,300 ksi Strong Axis P critical π EI (KL) Weak Axis 4 4 I 7in I 3in Lb 16 in L b 108in π EI π EI P critical ( KL) ( KL) P critical π (1,300 ksi)(7in [(1)(16 in)] 19.8k 4 ) π (1,300 ksi)(3in [(1)(108 in)] 35.k 4 )
32 Determine the critical buckling load for a 4x6 S4S Douglas Fir column that is 18 long and braced at midheight against the weak direction of buckling. E1,300 ksi Strong Axis P critical π EI (KL) Weak Axis 4 4 I 7in I 3in Lb 16 in L b 108in π EI π EI P critical ( KL) ( KL) P critical π (1,300 ksi)(7in [(1)(16 in)] 19.8k 4 ) π (1,300 ksi)(3in [(1)(108 in)] 35.k 4 )
33 What are the two conditions that need to be investigated?
34 What are the two conditions that need to be investigated? Weak Axis
35 What are the two conditions that need to be investigated? Weak Axis Strong Axis
36 In reality columns do not transition abruptly from short (crushing) to long (buckling. There is an intermediate column range in which columns fail by a combination of buckling and crushing. The analysis and design of steel columns takes this into account
37 AISC only recognizes two types of columns for design, short/intermediate and long The value for Kl/r where columns transition from short/intermediate to long is known a Cc and defined as: C c E F y
38 Cc is the theoretical value between inelastic and elastic behavior. C c E F y C c (50ksi) C c (36ksi)
39 KL C c r π σ critical F e (50ksi) KL r E KL (50ksi) r KL > (50ksi) r F cr F F y e F y Fcr F e Ω c 1.67( ASD)
40 Table 4-1 Table C-36 Table C-50 Design K-Values from AISC
41
42 Find the maximum Axial Load on your column-p(actual) Find the KLx and KLy Use Table 4-1 to find a trial section using the KLy (Tables are only for KLy) Find KLx (equivalent) by using rx/ry from table Use LARGEST KL/r for the rest of the design Using the largest KL/r, enter Table 9.1 or 9. to obtain a respective Fa Calculate the P(allowable)(Fa)x(A) of the trial section Check to see if P(allowable)>P(actual) If the P(allowable) is too low, pick a larger section and repeat If P(allowable) is larger then P(actual), check efficiency P(actual)/P(allowable) should be around 0.8 or 0.9 If your column seems overdesigned, pick a smaller section and start again Repeat until you find an ADEQUATE and EFFICIENT section.
43
44 Find the maximum Axial Load on your column-p(actual) Find the KLx and KLy Use Table 4-1 to find a trial section using the KLy (Tables are only for KLy) Find KLx (equivalent) by using rx/ry from table Use LARGEST KL/r for the rest of the design Figure out which of the two equations you need to use based on Cc Calculate F(critical) using the appropriate equation Calculate the P(allowable)(Fcr/Ωc)x(A) of the trial section Check to see if P(allowable)>P(actual) If the P(allowable) is too low, pick a larger section and repeat If P(allowable) is larger then P(actual), check efficiency P(actual)/P(allowable) should be around 0.8 or 0.9 If your column seems overdesigned, pick a smaller section and start again Repeat until you find an ADEQUATE and EFFICIENT section.
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