Chapter 5 Compression Member This chapter starts with the behaviour of columns, general discussion of buckling, and determination of the axial load needed to buckle. Followed b the assumption of Euler s Theor and the calculation with the different tpes of support in the column. t the end of the chapter, Secant formula will be discussed when the axial load acted at the offset from centroid. fter successfull completing this chapter ou should be able to: Determine the tpe of failure in compression member Determine the shape of buckling in compression member nalse the compression member using Euler s theor and Secant formula 5.1 Introduction The selection of the column is often a ver critical part of the design of structure because the failure of the column usuall has catastrophic effects. If a column is long compared to its width, it ma fail b buckling (bending and deflection laterall). The buckling ma be either elastic or inelastic depends upon the slenderness of the column. 5. Critical load and Euler Theor In this section we discuss a theor of a straight column that is simpl supported at either side. This theor was first developed b eonard Euler and is named after him. The assumptions for this theor are: a) The column is perfectl straight; b) The cross section of the column is uniform; c) The column material is homogeneous; d) The column behave elasticall; e) The compression force acted on the centroid of the section. 14
x x M Figure 5.1 Figure 5. Figure 5.1 shows a simpl supported column that is axiall loaded with force. et the bending deflection at an location x be given b as in Figure 5.. B balancing the moment at oint, we obtain M + = 0 The differential equation for moment curvature relationship is given b d dx M Therefore Where d 0 dx d dx 0 B substitution d 0 dx 15
The solution to the differential equation is: = kos λx + B sin λx With a boundar condition x = 0 = 0 we obtain = 0, thus = B sin λx nd with x =, = 0; we obtain B sin λ = 0 If B = 0, than we obtain a trivial solution, For a nontrivial solution, the sun function mest equal to zero Therefore sin λ = 0 is called buckling equation λ = nπ ( n = 1,,..) λ = n π n The critical buckling load is cr n (5.1) cr the critical buckling load is also caller Euler load. Buckling will occur about the axis that has minimum area of moment of inertia. The importance of each buckled mode shape is shown in Figure 5. as n is increased; the deflection curve has more and more inflection points. 16
1 1 First mode of buckling 1 Second mode of buckling 4 Third mode of buckling Figure 5.: The value of n defines the buckling mode shape. 9 For case 1 where n = 1, the value of that should be used is depends on the support condition at both ends of the column. So the equation 5.1 can be written as: cr (5.) e Where i) If both ends are pinned; e = ii) If one end fixed, other end free; e = iii) If both end are fixed; e = / iv) If one end fixed, other end pinned; e = 0.7 17
Figure 5.: Effect of support condition EXME 1 m long pin ended column of square cross section is to be made of wood. ssuming E = 1 Ga, σ all = 1 Ma and using a factor of safet of.5 in computing Euler s critical load for buckling. Determine the size of cross section if the column is to safel support (a) 100 kn load and (b) 00 kn load. Solution (a) 100 kn load cr =.5 (100) = 50 kn, = m, E = 1 Ga. In Euler s Equation 5. and solve for I, we have cr I E (50x10 )( ) 7.794x10 9 (1x10 ) 6 m 4 18
For a square of side a, we have I = a 4 /1, we write 4 a 7.794x10 1 6 a = 98. mm 100 mm We check value of normal stress in the column: 100 0.1 10Ma Since σ is smaller than the allowable stress, a 100 x 100 mm cross section is acceptable. (b) For the 00 kn load Solving again Equation 5. for I, but making now cr =.5(00) = 500 kn, we have I = 15.588 x 10-6 m 4. 4 a 15.588x10 1 6 a = 116.95 mm The value of normal stress is 00 0.11695 14.6Ma Since this value is larger than the allowable stress, the dimension obtained is not acceptable and we must select the cross section on the basis of its resistance to compression. We write all 00 1 16.67x10 a = 16.67 x 10 - m m a = 19.1 mm 10 x 10 mm cross section is acceptable. 19
EXME m x 00 mm x m 50 mm Figure 5.4 column with 6 meters height is connected to the beam as in Figure 5.4. With the connection between beam and column is pinned, Determine the critical buckling load of the column. E = 5000 Ma Solution Determine moment of inertia, I bh 1 50 00 5.6x10 mm 4 1 I xx = 8 bh 1 00 50.9x10 mm 4 1 I = 8 Deremine critical buckling load cr e 8 (5000)(5.6x10 ) (1.0 x6000) 771kN cr e 8 (5000)(.9x10 ) (1.0 x000) 18kN Therefore the critical buckling load is 771 kn 10
5. imitation of Euler Theor Since cr is proportional to I, the column will buckle in the direction corresponding to the minimum value of I. column can either fail due to the material ielding, or because the column buckles, it is of interest to the engineer to determine when this point of transition occurs. Because of the large deflection caused b buckling, the least moment of inertia I can be expressed as I r where: is the cross sectional area and r is the radius of gration of the cross sectional area, i.e.. Note that the smallest radius of gration of the column, i.e. the least moment of inertia I should be taken in order to find the critical stress. Dividing the buckling equation b, gives: where: cr is the compressive stress in the column and must not exceed the ield stress of the material, i.e. cr <, / r is called the slenderness ratio, it is a measure of the column's flexibilit. cr cr E / r 5.4 Secant Formula for Column When a column with simpl supported is compressed b an eccentricit applied axial as in Figure 5.4, the maximum compressive stress in the column is: e k max sec 5. S 11
The first term on the right-hand side of this equation represents the effect of direct compression and the second term represents the effect of bending of the column. Recalling that the section modulus S = I/c, where c is the distance from the neutral axis to the extreme fiber on the concave side of the column and also introducing the notation r I / for the radius of gration, we can express equation 5. in the form e x e Figure 5.4: Column with eccentricall applied axial force max ec k 1 sec r Next replacing k b /, we obtain ec max 1 sec 5.4 r r E This equation is called the secant formula for an eccentricall loaded column. It gives the maximum stress in the column as a function of the average compressive stress /, the eccentricit ratio ec/r, and the slenderness ratio /r. 1
EXME steel column of 54 x 54 x 107 kg UC section (Fig. 5.5a) with pinned ends is 8 m long. It carries a centrall applied load 1 = 980 kn and an eccentricall applied load = 140 kn on axis - at a distance of 400 mm from axis 1-1 (Fig 5.5b). a) Using a secant formula, calculate the maximum compressive stress in the column; 1 1 400 mm e 1 (a) (b) (c) Solution a) The two loads 1 and acting as shown in Fig. 5.5(b) are staticall equivalent to a single load = 110 kn acting with an eccentricit e = 50 mm (see Fig. 5.5(c). Using the table of properties, we find 110x10 1660 8000 11 8N / mm 70. 8 r ec e (50)(1660) r S 11x10 0.50 Substituiting into equation 5.4 using E = 00 kn/mm, we get max 18.6 N/mm. 1
5.5 err Robertson Formula Several different formulae have been devised that give a more realistic estimate of buckling loads than the Euler equation. The formula usuall used for structural steelwork is the err-robertson formula. Formulation is based on the assumption that the strut is initiall bent with a maximum offset of c 0 Notes: Origin fixed at strut mid length. v is increase in deflection due to. err formula p 1 1 1 ield 1 E ield 1 E ield E Based on test on circular mild steel column, Robertson proposed that a value of l e 0.00 could be used for mild steel columns. If a value for initial out of k c0v straightness c 0 is known, this value should be used to calculate k 14