Notes

CHAPTER 5 COMPRESSION MEMBER CONTENTS: 5.1 Introduction 5.2 Critical Load 5.3 Euler’s Theory 5.4 Column Having Various Types of Supports 5.5 Secant Formula 5.6 Perry-Robertson Formula 5.7 Rankine-Gordon Formula 5.1

Introduction ¾ Column = vertical member carries compressive axial loads. ¾ Sometimes, the compressive axial load is applied at the centroid and offset from centroid. P1 P2 P e1 e2

Axially loaded column

Eccentric column

Type of columns; a) short column = when the dimensional x-section is not very large as compared to its height. It fails due to ‘crushing’ of column material. P PP

CRUSH

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b) long/slender column = when the dimension x-section is very large as compared to its height. It fails due to ‘buckling’. P P

5.2

Critical Load Assume the column is pinned at both support; P Pcr

L

P

Pcr

¾ For the beginning, the column is considered as an ideal column :perfectly straight before loading. ¾ When the load,P is increasing until failure occurs by fracture/yielding. ¾ When the critical load, Pcr is reached, the column is becoming unstable. ¾ Any additional loading will cause the column to buckle and therefore deflect laterally as shown above. Ideal column = is initially perfectly straight, made of homogeneous material and the load is applied through the centroid of the cross section.

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Critical load, Pcr = the maximum load that a column can support when it is on the verge of buckling. This loading represents a case of neutral equilibrium. To illustrate the phenomenon of buckling, the concept of stability is applied. 5.3

Euler’s Theory ¾ Developed by the Swiss mathematician Leonard Euler on 1757 ¾ Initially developed for slender column Column with pinned-ended support; Assumptions; 1) column is perfectly straight 2) load,P is applied exactly at the centroid of column 3) no lateral loads act along the height of column 4) material behaves within elastic region or ideal rigid-plastic or elastic-plastic behaviour P

P

L

y

y x

P P Assume when the column is pinned support;

M

x P

Applying Theory of Bending; M E σ = = I R y EI 1 d2y = = curvature but R R dx 2 d2y Cut any distance with x vertically; EI 2 = M dx = M + Py d2y M = -Py EI 2 = − Py dx M =

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d 2 y − Py = ...let dx 2 EI

α2 =

P EI

d2y + α 2 y = 0..........(1) 2 dx Solve the equation (1) by using ‘complementary function’; y = A cos αx + B sin αx...........( 2)

Applying the boundary condition; a) y = 0 and x = 0 into (2); 0=A+0 A= 0 …. substitute A into (2) b) y = 0 and x = L into (2); y = A cos αx + B sin αx 0 = 0 + B sin αL B sin αL = 0 (sin αL =0 when αL = π ,2π ,3π ....)

αL = nπ ( n = 1,2,3..) from α 2 =

Sine graph (column started buckle at sin π =0) +1 0

90

180 270

360 (ºdegree)

-1

P substitute αL = n π EI

P xL = nπ EI P nπ = EI L P n 2π 2 = 2 EI L n 2π 2EI P= L2

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n=1

n=2

n=3

Buckle shape that might be happened

2

Pcr = π EI 2 L

if n =1

where; Pcr = critical or maximum axial load on the column just before it begins to buckle. This load must not cause the stress in the column to exceed the proportional limit. E = modulus of elasticity for the material I = least moment of inertia (the weakest axis) for the column’s cross sectional area. L = unsupported length of the column, whose ends are pinned For purpose of design; from equation 2

Pcr = π EI …….(1) 2 L

Pcr ....(2) and I = Ar 2 ......(3) A Substitute ( 2)and (3) ⇒ (1);

σ cr =

Pcr =

π 2 E ( Ar 2 )

L2 Pcr π 2 Er 2 = A L2

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σ cr =

π 2E (L )2 r

……… (L/r) - slenderness ratio, buckling will occur about the axis when the ratio gives the greatest value. - measurement of the column’s flexibility

where; σcr = critical stress which is an average stress in the column just before the column buckles. This stress is an elastic stress, σcr ≤ σy , (σy = yield stress of material) E = modulus of elasticity of material L = unsupported length of the column whose ends are pinned r = radius of gyration determined from r = I , where I is the least moment of A inertia of the column’s cross sectional area, A. Euler’s formula can be used to determine the buckling load since the stress in the column remains elastic. It is also important to realize the column will buckle about the principal axis of the cross section having the least moment of inertia (the weakest axis).

Allowable/Safety Load =

Critical Load ( Pcr ) Factor of Safety

From σcr ≤ σy; The graph below is used to identify the (L/r) for the column made of a structural steel. σcr(103)MPa

250

Structural steel (σy = 250MPa)

The graph used to represent the sritical stress versus the slenderness ratio. The curve is hyperbolic and valid only for critical stress below the material’s yield point. The yield stress of steel; (σy )s=250MPa (Es = 200GPa)

89

L/r

From the curve,the smallest acceptable slenderness ratio (L/r)s = 89

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For a steel column if (L/r)s ≥ 89 , Euler’s formula can be used to determine the buckling load since the stress in the column remains elastic. But if the (L/r)s < 89, the column’s stress will exceed the yield point and the Euler formula is not valid in this case. Or the slenderness ratio can be determined from this equation; π 2E σ cr = (L )2 r

π 2E L = σ cr r =

π 2 ( 200 x10 9 ) 250 x10 6

= 89

EXAMPLE 5.1 A 7.2m long of steel tube having the cross section shown in figure is to be used as a pinended column. Determine the maximum axial load the column can support so that it does not buckle. Check whether the Euler’s formula is appropriate or not by using the condition σcr ≤ σy .Given Es = 200GPa, σ y = 250MPa . Pcr

75mm 70mm

Pcr

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Solution; 1st Step: Identify the type of support. Pcr =

=

π 2 EI

1 @ πr 4 64 4 1 = π 754 − 70 4 4 = 6 x106 mm 4

I=

2

L

π 2 ( 200 x10 9 )(6 x10 −6 ) 7 .2 2

= 228.5kN

πd 4

[

]

= 6 x10 −6 m 4

2nd Step: Check whether the Euler’s equation is appropriate or not;

σ cr

P 228.5 x10 3 = cr = = 100.2 MPa A 2.28 x10 −3

A= =

πd 2 4

π

(0.15 4

2

− 0.14 2

= 2.28 x10 −3 m 2

Since σ cr < σ y

where

100.2 MPa < 250 MPa

So, Euler’s equation is appropriate. The maximum axial load= 228.5kN

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EXAMPLE 5.2 The steel with the size W200 x 46 shown in figure is to be used as a pin-connected at the both ends. Determine the largest axial load it can support before it either begins to buckle or the steel yields. Check whether the Euler’s formula is appropriate or not by using the condition σcr ≤ σy .Given Es = 200GPa, σ y = 250MPa .

Solution; From Appendix; A = 5890 mm 2 I x = 45.5 x10 6 mm 4 I y = 15.3 x10 6 mm 4

By inspection, the column will buckle about the principal axis of the cross section having the least moment of inertia. From the given I, Iy-y is the lowest moment of inertia. The column will buckle at this axis. Take I y = 15.3x106 mm 4 Pcr =

=

π 2 EI L2

π 2 ( 200 x10 9 )(1.53 x10 −5 )

………..For the pinned-ended support

42

= 1888.1kN

Check whether the Euler’s equation is appropriate or not; P 1888 .1x10 3 σ cr = cr = = 320 .6 N / mm 2 or 320 .6 MPa . A 5890 As we know, the yield stress of steel is σ y = 250MPa . The rule of Euler’s equation is σcr ≤ σy. ____________________________________________________________________________________________ BFC 2083-Mechanic of Materials

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From the calculation, it shows that σcr > σy where 320.6MPa > 250MPa. The stress exceeds the yield stress. Application of Euler’s equation is not appropriate. Use σy =250MPa to determine the new axial load.

σ cr = 250 x106 =

Pcr A P 5890 x10 −6

P = 1472.5kN

EXAMPLE 5.3 A steel column has a length of 9m and pinned at both ends. If the cross sectional area has the dimensions shown in figure, determine the critical load. Check whether the Euler’s formula is appropriate or not by using the condition σcr ≤ σy .Given Es = 200GPa, σ y = 250MPa . 200

10 10

150

10

Solution; The cross section is symmetry due to the same dimensions. Centroid or neutral axis is x , y = (100 ,85 ) mm .

y 1 d1 2 d2

y = 85 3

x

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PART

1

2

3

bh 3 AREA(mm ) 12 3 (10 ) (mm4) 200(10 3 ) 200(10)=2000 12 =16.67 10(150 3 ) 150(10)=1500 12 =2812.5 200(10 3 ) 200(10)=2000 12 =16.67 2

Ix =

d (mm)

Ad2(103)(mm4)

85-5= 80

12800

0

0

80

12800

∑ I xx = [Ix + Ad2]1 + [Ix + Ad2]2 +[Ix + Ad2]3 = [(16.67 + 12800)x2 + 2812.5] x103 = 28.45 x106 mm4

PART

1

2

3

b 3h AREA(mm ) 12 3 (10 ) (mm4) 200 3 (10) 200(10)=2000 12 =6666.67 10 3 (150 ) 150(10)=1500 12 =12.5 200 3 (10) 200(10)=2000 12 =6666.67 2

Iy =

s (mm)

As2(103)(mm4)

0

0

0

0

0

0

∑ I yy = [Ix + As2]1 + [Ix + As2]2 +[Ix + As2]3 = [6666.67+12.5+6666.67] x103 = 13.35 x106 mm4 To determine the critical load, pick the value of lowest I=13.35 x106 mm4

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Pcr =

=

π 2 EI L2

π 2 ( 200 x10 9 )(1.34 x10 −5 )

………..For the pinned-ended support

92

= 326.63kN

Check whether the Euler’s equation is appropriate or not; P 326 .63.1x10 3 σ cr = cr = = 59.4 N / mm 2 or 59.4 MPa . A 5500 As we know, the yield stress of steel is σ y = 250MPa . The rule of Euler’s equation is σcr ≤ σy where 59.4MPa 100. Given E=200 kN/mm2 and factor of safety=2.0. b) the allowable load for the column.

84 20 96

10

6m

20 (all units in mm) [Ans : L/r > 100 = slender column, allowable load = 54.4kN]

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EXERCISE 5.2 The 4m steel pipe column has an outer diameter of 75mm and a thickness of 6mm. Determine the critical load if the ends are assumed to be pin connected. a) Check whether the Euler’s formula is appropriate or not by using the b) condition σcr ≤ σy .Given Es = 210GPa, σ y = 250MPa .

[Ans : Pcr = 101.1kN]

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5.4

Column Having Various Types of Supports

In the previous section, we derived the Euler load for a column that is pin connected of free to rotate at its ends. However, column may be supported in some other way with the different type of supports. Cantilever supports EXAMPLE 5.4

P

P

P δ y L

L

δ y P

Mx

The diagram shows the column is fixed at its base and free at the top. The internal moment at arbitrary section is Mx = P(δ – y). Prove the critical load 2 occurs at the column is Pcr = π EI . 2 4L Solution; = − M X + P (δ − y ) d2y = Mx dx 2 d2y EI 2 = P (δ − y ) dx d 2 y Pδ Py = − dx 2 EI EI EI

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d 2 y Py Pδ + = .....let dx 2 EI EI

α2 =

P EI

d2y + α 2 y = α 2δ ..........(1) 2 dx Solve the equation (1) by using ‘complementary function’; y = A cos αx + B sin αx + δ ...........( 2)

Applying the boundary condition; a) x = 0 and y = 0 into (2); 0=A+0+δ A= -δ …. substitute A into (2) b) x = 0 and

dy = 0 into (2); dx

dy = − Aα sin αx + Bα cos αx dx 0 = 0 + Bα Bα = 0 B=0

A= -δ and B=0, substitute into (2); y = A cos αx + B sin αx + δ ...........( 2) y = −δ cos αx + δ

Since the deflection occur at the top of the column; At x = L, y = δ δ = −δ cos αL + δ

cosines graph (column started buckle at sin 0.5π =0) +1

0 = −δ cos αL -90

0 =δ cos α L 0 = cos αL

0

90

180

270

360 (ºdegree)

-1 cos 0.5π, 1.5 π, 2.5π = 0

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Known that α 2 =

P or α = EI

P EI

P π xL = EI 2 P π = EI 2 L P ⎛π ⎞ =⎜ ⎟ EI ⎝ 2 L ⎠

π2 P = EI 4 L2

2

…. Its proven

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Table 5: Effect of support conditions Pinned-pinned

Fixed-fixed

Cantilever

Fixed-pinned

Le

Pcr =

Basic Formula L

Le Pcr

Pcr =

π 2 EI (L) 2

π 2 EI Le

2

0.5L Pcr =

π 2 EI (0.5L) 2

2L Pcr =

π 2 EI ( 2 L) 2

0.7L Pcr =

π 2 EI ( 0. 7 L ) 2

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EXERCISE 5.3 The steel with size W310-129 has a length of 4m. If its bottom end is fixed supported while its top is free, a) determine the effective length of column b) the slenderness ratio c) the allowable axial load it can support. Use a factor of safety with respect to buckling of 1.75. Given Es = 210GPa, σ y = 250MPa .

[Ans: Pallow = 1850.44kN]

P

EXAMPLE 5.5 The aluminum column is fixed at its bottom and is braced at its top by cables in order to prevent movement at the top along x axis. If it is assumed to be fixed at its base; a) Check whether the Euler’s formula is appropriate or not by using the condition σcr ≤ σy . b) Determine the largest allowable load, P that can be applied.

Use a factor of safety for buckling = 3.0. Given Eal = 70GPa, σy=215MPa, A = 7.5(10-3) m2, Ix = 61.3 (10-6) m4, Iy = 23.2(10-6) m4.

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Solution; Due to this condition of bracing, the column will buckle with different axis namely are x and y axis. In y axis, the column will behave as the fixed at its bottom and pinned at the π 2 EI . top end. So, the critical load is Pcr = ( 0. 7 L ) 2

π 2 EI

Pcr = =

(0.7 L) 2

π 2 (70 x109 )( 23.2 x10 −6 ) (0.7 x5) 2

= 1308.8kN

In x axis, the column will behave as the fixed at its bottom and free at the top end. So, the π 2 EI critical load is Pcr = . ( 2 L) 2

Pcr = =

π 2 EI ( 2 L) 2

π 2 (70 x109 )(61.3x10 −6 ) ( 2 x5) 2

= 423.6kN

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By comparison the column will buckle to the lowest P which is x axis. Pcr = 423.6kN a) And check whether the Euler’s equation is appropriate or not; P 423.6(10 3 ) σ cr = cr = = 56.48 MPa A 7.5(10 −3 ) Given, the yield stress of aluminum is σ y = 250MPa . The rule of Euler’s equation is σcr ≤ σy where 56.48MPa < 215MPa. The application of Euler’s equation is appropriate. b) The allowable load is; Pallow =

Pcr 423 .6 x10 3 = = 141 .2 kN F .O.S 3

The critical load = 423.6kN The allowable load = 141.2kN

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5.5

Secant Formula

The Euler formula was derived with the assumption; i) the load,P is always applied through the centroid of the column’s cross sectional area and; ii) the column is perfectly straight. This is actually quite unrealistic since manufactured columns are never perfectly straight. In actual condition, column never suddenly buckle, instead they begin to bend slightly upon the application of the load. The actual criterion for load application will be limited either to a specified deflection of the column or by not allowing the maximum stress exceed the allowable stress in the column. To investigate this effect load, P is applied to the column at a short eccentric distance, e from the centroid of the cross section.

y

y

L y

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From Theory of Bending;

M E σ = = I R y EI but R d2y EI 2 = M dx

M=

1 d2y = = curvature R dx 2

From figure above, internal moment column is; M = − P( e + y ) EI

d2y = − P( e + y ) dx 2 d2y − P = (e + y ) dx 2 EI d 2 y − Pe Py = − dx 2 EI EI

d 2 y Py Pe + =− 2 dx EI EI d2y + α 2 y = −α 2 e 2 dx

.....α 2 =

P EI

............(1)

Solve the equation (1) by using ‘complementary function’;

y = A cos αx + B sin αx − e ...........(2) where A and B are constants and can be determined by using boundary condition.

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Applying the boundary condition; a) y = 0 and x = 0 into (2) 0=A+0- e A= e subs A into (2) y = e cos αx + B sin αx − e ...........(3) dy = − eα sin αx + Bα cos αx dx

b)

.......( 4)

dy = slope = θ = 0 and x = L/2 into (4) dx

From

( 4) →

dy = −eα sin αx + Bα cos αx dx

eα sin αx = Bα cos αx sin αx = e tan αx cos αx ⎛ αL ⎞ = e tan⎜ ⎟ ..............(5) ⎝ 2 ⎠

B=e

Substitute (5) into (3); y =e cos αx + B sin αx − e

⎛ αL ⎞ =e cos αx + e tan⎜ ⎟ sin αx − e ⎝ 2 ⎠

...........(6)

Equation (6) is a general equation to find deflection anywhere within the column height. But there will be one position, y becomes maximum.

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For this pin-ended column, ymax ocuurs at mid-height; At x = L/2 into (6) P M=P.e

⎛ αL ⎞ (6) → y = e cos αx + e tan ⎜ ⎟ sin αx − e ⎝ 2 ⎠

From

ymax

⎡ ⎛ αL ⎞ ⎛ αL ⎞ ⎛ αL ⎞⎤ y max = ⎢e cos ⎜ ⎟ + e tan ⎜ ⎟ sin ⎜ ⎟⎥ − e ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠⎥⎦ ⎢⎣ ⎡ ⎛ αL ⎞ ⎤ = e ⎢sec ⎜ ⎟ − 1⎥ ⎝ 2 ⎠ ⎦ ⎣

M=P.e P

............(7)

deflection max occurs at x = L/2 Resubstitute α 2 =

P into (7); EI

⎡ ⎛ αL ⎞ ⎤ y max = e ⎢sec ⎜ ⎟ − 1⎥ ⎝ 2 ⎠ ⎦ ⎣ ⎡ = e ⎢sec ⎣

P L ⎤ . − 1⎥ EI 2 ⎦

.......(8)

Equation (8) indicates that ymax infinite when

L 2

π P = EI 2

While deflection does not actually become infinite and nevertheless becomes unacceptably large. P should not be allowed to reach the critical value, Pcr. For this support condition; π 2 EI Pcr = 2 L Pcr L2 EI = 2

π

⎡ π ymax = e ⎢sec 2 ⎣

…into (8)

⎤ P − 1⎥ …….(9) Pcr ⎦

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To find maximum stress; σmax The column will reached compressive stress and σmax occurs when moment is maximum;

y

M = P(e + y ) M max = P (e + y max )

Substitute equation (8) into Mmax;

⎡ ⎛ M max = P ⎢ e + e ⎜⎜ sec ⎝ ⎣⎢ ⎡ = Pe ⎢1 + ⎣⎢ ⎛ = Pe ⎜⎜ sec ⎝

⎛ ⎜ sec ⎜ ⎝

⎞⎤ P L . − 1 ⎟⎟ ⎥ EI 2 ⎠ ⎥⎦ ⎞⎤ P L . − 1 ⎟⎟ ⎥ EI 2 ⎠ ⎥⎦

P L⎞ . ⎟.......... (10 ) EI 2 ⎟⎠

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From Theory of Bending; P M .y σ max = + max A I P Pe. y ⎛ P L⎞ ⎜ sec . ⎟ = + A I ⎜⎝ EI 2 ⎟⎠

I = r2 A

=

P Pe. y ⎛ P L⎞ . ⎟ known that + 2 ⎜⎜ sec A r A⎝ Er 2 A 2 ⎟⎠

=

P Pe. y ⎛ P L⎞ . ⎟ + 2 ⎜⎜ sec A r A⎝ EA 2 r ⎟⎠

=

⎛ L P ⎡ e. y ⎢1 + 2 sec⎜⎜ A ⎢⎣ r ⎝ 2r

1 1 = r2 r

P ⎞⎤ ⎟⎥.........(11) EA ⎟⎠⎥⎦

where; σmax = maximum elastic stress in the column which occurs at the inner concave side at the column’s midpoint. This stress is compressive. P

= vertical load applied to the column. P < Pcr unless e = 0, then P = Pcr.

e

= eccentricity of the load P measured from the neutral axis of the column’s cross sectional area to the line of P action.

y

= distance from the neutral axis to the outer fiber of the column where the maximum compressive stress σmax occurs.

A = cross sectional area of the column.

⎛ L ⎜ ⎜ 2r ⎝

L

= unsupported length of the column. For supports other than pins, the effective length Le should be used.

E

= modulus of elasticity for the material.

r

= radius of gyration, r = bending axis.

ey r2

= eccentricity ratio

I / A where I is computed about the neutral or

P ⎞ ⎟ = Euler angle EA ⎟⎠

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Summary of Secant Formula

Maximum Deflection; ⎡ y max = e ⎢sec ⎣ y

P L ⎤ . − 1⎥ EI 2 ⎦

Maximum Stress; ⎛L P ⎡ e. y σ max = ⎢1 + 2 sec⎜⎜ A ⎢⎣ r ⎝ 2r

P ⎞⎤ ⎟⎥ EA ⎟⎠⎥⎦

EXAMPLE 5.6 The steel column shown in figure is assumed to be pin connected at its top and base. a) Determine the allowable eccentric load, P that can be applied. Due to bracing assume buckling does not occur about the y axis. b) Determine the maximum stress and deflection of the column due to this loading.

Given Es = 200 GPa and σ y = 250MPa . z

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Solution; a)

Determine the allowable eccentric load, P that can be applied. y 50mm x 150mm

bh 3 50(150 3 ) = = 14.06 x10 6 mm 4 12 12 3 3 b h 50 (150) Iy = = = 1.56 x106 mm 4 12 12 Ix =

A = 50(150) = 7500 mm2 e = 25mm Refer to the question; Due to bracing assume buckling does not occur about the y axis. Take Ix.

r=

14.06x106 = 43.3mm 7500

In the secant formula, the curves has been established for Es = 200 GPa and σ y = 250MPa , this curves can be used to determine the value of P/A to avoid a trial and error solution of the secant formula. From the curves L/r is the slenderness ratio and the type of support is pinned connected at the both ends, L/r is 4500/43.3 = 104. If the type of support is cantilever L/r will become as 2L/r and so on. y is the distance from the neutral axis to the outer fiber of the column where the maximum compressive stress occurs. So, the distance is 75mm.

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ey = 0.1 r2

ey =0 r2

Using the curves, the eccentricity ratio,

ey 25(75) = =1 r2 43.32

The value of P/A approximate to be, P ≈ 83MPa A

P = 83(7500) = 622.5kN The allowable eccentric load, P = 622.5kN

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b)

the maximum stress and deflection of the column due to this loading.

To calculate the maximum stress, σmax;

σ max = =

⎛ L P ⎡ e. y ⎢1 + 2 sec⎜⎜ A ⎢⎣ r ⎝ 2r

P ⎞⎤ ⎟⎥ EA ⎟⎠⎥⎦

⎛ 4500 622.5(10 3 ) ⎡ 25(75) 622.5(10 3 ) ⎞⎟⎤ ⎜ sec ⎢1 + ⎥ ⎜ 2( 43.3) ( 200 x10 3 )7500 ⎟⎥ 7500 ⎢ 43.32 ⎝ ⎠⎦ ⎣

= 83[1 + 1 sec(1.0585rad )] = 83[1 + 1 sec(60.65°)] = 252.3N / mm

1.0585 x

2

180

π

= 60.64°

σy > σmax where 250MPa > 252.3 MPa Use σy =250 MPa because σmax can not be exceed than σy To calculate the maximum deflection, ymax ; ⎡ y max = e ⎢sec ⎣ ⎡ = 25⎢sec ⎣⎢

[

P L ⎤ . − 1⎥ EI 2 ⎦ 622.5(103 ) 4500 ⎤ − 1⎥ . 3 6 ( 200 x10 )(14.06 x10 ) 2 ⎦⎥

]

= 25 sec 1.0585rad − 1 = 25(sec 60.65° − 1) = 26mm

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EXERCISE 5.4 The steel column shown in figure below is assumed to be pin-connected at its base and top. a) Determine the allowable eccentric load, P that can be applied. b) Calculate the maximum deflection of the column due to this loading? 3

Due to bracing, assume buckling does not occur about the y-axis. Take E=200(10 ) MPa and σy=250MPa. C,centroid

60mm P 30mm 120mm

4.5m

[Ans: Peccentric = 360kN, ymax = 28mm]

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5.6

Perry-Robertson Formula

The formula used for structural steelwork is the Perry-Robertson formula that represented as the average end stress to cause yield in a strut.

where; σp = stress based on Perry-Robertson formula. σE = Euler’s stress σyield = yield stress depending on the yield strength of material η = constant depending on the material. For a brittle material η = 0.015L/r, for a ductile material η = 0.3 (L/100r)2 5.7

Rankine-Gordon Formula

σR =

where a =

Material

Mild steel Cast iron Timber

σy 1 + a( L / r)2

σy π 2E

theoretically but is usually found by experiment. Typical values are :

Compressive yield stress (MN/m2) 315 540 35

a Pinned ends Fixed ends

1/7500 1/1600 1/3000

1/30000 1/6400 1/12000

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Appendix A American Standard Channels or C Shapes

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TUTORIAL 5

1. A steel column has a length of 9m and is fixed at its bottom and free at its top. If the cross-sectional area has the dimensions shown; a) determine the critical load. b) also check whether the Euler’s Equation is appropriate or not. Given Est= 200GPa, σy= 250 MPa. 200mm 10mm 10mm

150mm 10mm

[Ans: a) Pcr = 82.25kN, b) σcr = 14.95MPa] 2. a) The 4m steel column has an outer diameter of 80mm and a thickness of 6mm. Determines the critical load if the bottom is fixed and the top are pinned. Also check whether the Euler’s Equation is appropriate or not. Given E= 210GPa, σy= 250 MPa. b) A steel strut is 6.5m long and constructed from circular tube with an outside diameter of 120mm as shown in figure below. The type of strut is fixed at the bottom and pinned at the top. The strut must resist an axial load of 150kN from buckling. Using a factor of safety of 2.0, determine the required tube thickness and the critical buckling stress. Given E = 200GPa.

[Ans: a) Pcr = 254kN, σcr = 183MPa, b) t = 5.3mm, σcr = 157MPa]

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2

3. The steel angle has a cross sectional area of 1550mm . A radius of gyration about the x axis is 32mm and about y axis is 22mm. The smallest radius of gyration occurs about z axis is 16mm. If the angle is to be used as a pin-connected with 3m long column, determine the largest axial load that can be applied through its centroid, C without causing it to buckle, E= 200GPa. y z x

C y

x z

[Ans: Pcr =87.03kN] 4. a) Define the following terms; i. short column ii. slender column iii. Euler buckling load b) List 5 assumptions or limitation of Euler’s formula. c) A column is 8m high and is constructed from circular hollow section as shown in figure. The column is fixed at one end and is free at the other end. Determine the critical buckling load. Given E = 2x105 N/mm2.

[Ans: c) Pcr = 36.86kN]

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5. a) The 4m steel pipe column has an outer diameter of 100mm and a thickness of 8mm. Determine the critical load if the ends are assumed to be pin connected. Also check whether the Euler’s Equation is appropriate or not. Given E=210GPa , σy=250MPa. b) Determine the Euler’s buckling load for an I-section of a column as shown in figure with length of 10m and fixed at both ends. Given E = 200GPa.

[Ans: a) Pcr = 19.31kN, σcr = 138.8MPa, b) Pcr = 459kN] 6. For the column fixed at the base and the top, P

P

M P

L

y

L

P

Mx Cut section

M

Cut any section vertically.

M The internal moment at arbitrary section is; = M x + Py − M . M x = − Py + M The critical load acted is Pcr =

4π 2 EI . Prove the equation. L2

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7. a) Write the relationship between Euler’s buckling stress with slenderness ratio. b) If a pin-ended solid circular strut is 3m long and 60mm diameter, determine the strut’s slenderness ratio. c) If the same material on 7(b) is reshaped into a square bar with the length and same width of 60mm, determine the percentage reduction in the slenderness ratio. [Ans: b) L/r = 200, c) 13.4%] 8. A steel strut with total length of 5m is constructed of circular tube. The dimension of tube is diameter, Doutside = 100mm and thickness t = 12mm. The type of strut is partially restrained which is fixed at the bottom and pinned at the top. Determine; a) slenderness ratio. b) Euler buckling load. c) ratio of axial stress under the action of the buckling load to the elastic strength of the material. Given E=200GPa , σy=250MPa. d) safety load the column can support with a factor of safety of 1.5. [Ans: a) L/r = 111.46, b) Pcr = 527kN, c) 0.635, d) Pallow = 351.3kN] 9. a) For a pin-ended solid rectangular column with a dimension 120cm x 50cm and 5m in length. Given E = 200GPa. Determine; i. slenderness ratios at both axes. ii. Euler’s buckling load. iii. allowable load with a factor of safety of 2.5. b) Determine the Euler’s buckling load for an I-section of length 6m when fixed at both ends as shown in figure.

[Ans: a) i) (L/r)x = 34.65, (L/r)y = 14.43, ii) Pcr = 987MN, iii) Pallow =394.5MN, b) Pcr =26.4MN]

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10. a) State all Euler’s buckling load formulas for different end support conditions. b) Determine the Euler’s buckling load for a C-section of column as shown in figure with length of 10m and fixed at both ends. Given E = 200GPa.

[Ans: Pcr =436.63kN] 11.

a) The cross sectional area of column which is fixed at the bottom and pinned at the top is shown in figure. The length of column is 1.5 meter. Given E = 200GPa. Determine; i) slenderness ratio. ii) Euler’s buckling load. iii) the allowable load with a factor of safety of 2.5.

b)

For a fixed and free ends column of length 2m with a cross section as shown in figure, determine; slenderness ratio of the column. i) Euler’s buckling load. ii) allowable load that the column can carry by assuming a factor of iii) safety is 2.5. Given E = 200GPa.

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[Ans: a) i) L/r = 20.21, ii) Pcr = 200.13MN, iii) Pallow =80MN, b) i) L/r = 128, ii) Pcr = 414.5kN, iii) Pallow = 165.8kN] 12.

a)

A steel column has a hollow circular cross sectional area with an outside diameter of 120mm and inside diameter of 100mm. The column is 6.0m length and is pin ended at both ends. Given E = 200GPa. Determine; i. slenderness ratio ii. Euler’s Buckling load iii. Axial stress in the column when the column is subjected to the Euler’s Buckling load.

b)

The cross sectional area of the column given in 12(a) is reshaped with the new radius of gyration r = 39.9mm and area = 11 355mm2. Determine the current value of slenderness ratio, Euler Buckling load and axial stress.

c)

A column of 3m height has a cross section as shown in figure. Both of its end are fixed. Determine the safe load when a factor of safety equals to four. Check whether the Euler’s Equation is appropriate or not. Given E= 200GPa, σy= 250 MPa.

[Ans: a) i) L/r = 153.8, ii) Pcr = 288.96kN, iii) σcr = 83.6MPa, b) i) L/r = 150.4, ii) Pcr = 992.4kN, iii) Pallow = 87.4MPa] ____________________________________________________________________________________________ BFC 2083-Mechanic of Materials

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