MECHANICS O SOIDS COMPRESSION MEMBERS TUTORIA INTERMEDIATE AND SHORT COMPRESSION MEMBERS Yo shold jdge yor progress by completing the self assessment exercises. On completion of this ttorial yo shold be able to do the following. Define an intermediate compression member. Derive and se the Rankine Gordon formlae. Solve the loads at which members fail. Define a colmn. Calclate the stresses in a colmn with an offset load. Calclate the position of the netral axis in a colmn with an off set load. It is assmed that stdents doing this ttorial already familiar with the concepts of second moments of area, bending stress, bending moments and deflection of beams. Stdents shold stdy the ttorial on strts before starting this one. D.J.DUNN 1
INTRODUCTION Yo shold recall that compression members fall into three grops, long (strts) intermediate and short (colmns). irst we need to now that a material may fail de to exceeding the ltimate (maximm) compressive stress. This is often referred to as the crshing stress. If this was the only factor casing failre the load that prodces it wold be given by the formla = A = Ultimate compressive load = ltimate compressive stress A = cross sectional area. In reality, there is often bending associated with the failre and this is especially tre with intermediate members. 1. INTERMEDIATE COMPRESSION MEMBERS It is fond that steel strts with a slenderness ratio of 80 to 10 fail at smaller loads than predicted by Eler. These are intermediate compression members in which compression and bending have an effect on failre. 1.1 RANKINE - GORDON THEORY GORDON sggested that for sch members, an empirical formla be sed (based on experimental data). RANKINE modified Gordon's formla. The following shows the reasoning for this formla. π EI E = Eler's critical load. and this applies to strts. = Ultimate compressive load = A and this applies to colmns. = ltimate compressive stress. A = cross sectional area. Rankine sggested that an intermediate compression member fails de to both bckling and compression to more or less degrees. Based on experimental data, it is fond that a reasonable prediction of the load at failre is given by the reciprocal formla. 1 1 1 E This rearranges to R E R = Rankine s Critical load. R E D.J.DUNN
The formla indicates that for slender members E dominates and for short members dominates. URTHER DEVEOPMENT Consider the frther development of Eler's formla for strts. Slendernes s Ratio S.R. Rearrange to make I the sbject k I A (S.R.) A I π EI Eler' s critical load for strtsis E Sbstitte the above formla for I nto this. π EA π EA E k k If we divide by thecross sectional area we convert the force into a nominal stress called Eler' s critical A I (S.R.) stress A k E E π E π Ek E...(1) A k If we plot this stress against slenderness ratio we get the reslt shown on figre 1. The graph is called Eler's Hyperbola. Next we consider the frther development of the Rankine formla. Sbstitte the following into Rankine s formla. = A e = EI/ R e e k A EA n EAn k A D.J.DUNN 3
D.J.DUNN 4 E where...() 1 A E 1 A EA n A a n k a k A k EA n R R Althogh there is a theoretical vale for the constant a based on material properties, it is sal to determine it from experiment. The experimental vale varies slightly from the theoretical. Typical vales of a are Material MPa a Mild Steel 35 1/7500 Wroght Iron 47 1/9000 Cast Iron 557 1/1600 Timber 35 1/3000 GRAPHICA REPRESENTATION The diagram shows E (eqation ) and R (eqation 1) plotted against slenderness ratio. The ltimate compressive stress is marked on the stress axis. The reslt shows how R tends to as the member becomes short and tends to E as the member gets longer. The region of interest is arond the S.R. = 80 point where the correct stress lays between the other two vales. igre 1
WORKED EXAMPE No.1 ind the Rankine critical load for a strt with an I section as shown given = 35 MPa, E = 05 MPa and = 16 m. The strt is bilt in rigidly at each end. SOUTION igre The strt will bend abot the axis of minimm resistance and hence minimm I so we mst determine which is the minimm I. HORIZONTA N-A The section is symmetrical so sbtract I for the two ct ots from the vale for the oter rectangle. Remember that for a rectangle I = BD 3 /1 abot its centre line. or the oter rectangle I = 50 x 30 3 /1 = 68.67 x 10 6 mm 4 or one ct ot I = 10 x 300 3 /1 = 70 x 10 6 mm 4 or the section I = 68.67 x 10 6 x 70 x 10 6 = 14.66 x 10 6 mm 4 VERTICA N-A Treat this as three rectangles and add them together. Two ends I = x 10 x 50 3 /1 = 6.04 x 10 6 mm 4 Middle I = 300 x 10 3 /1 = 5000 mm 4 Add together I = 6.06 x 10 6 mm 4 The minimm I is abot the vertical axis so se I = 6.06 x 10 6 mm 4 A = (50 x 10) x + (300 x 10) = 8000 mm. k = (I/A) = (6x10 6 /8000) = 57 mm mode n = a = 0.00016 6 6 A 8000x10 x35x10 R a k 1 n 66.kN 0.00016 1 16 0.057 D.J.DUNN 5
SE ASSESSMENT EXERCISE No.1 1. A strt is 0. m diameter and 15 m long. It is pinned at both ends. Calclate Eler's critical load. Take E = 05 GPa (Answer 706.5 kn). A strt has a rectanglar section 0. m x 0.1 m. It is 8 m long. The bottom is bilt in and the top is free. Dring a test, it bckled at 164 kn. Given the ltimate compressive stress of the material is 345 MPa : Calclate the Rankine constant "a". (0.000133) A second strt made of the same material has a rectanglar section 0.3 m x 0.m. It is 6 m long and pinned at both ends. Using the constant a fond previosly, find the Rankine bckling load. (Answer 8.5 MN) D.J.DUNN 6
. COUMNS A colmn is a thick compression member. Strts fail de to bending bt colmns fail in compression. Colmns are sally made of brittle material which is strong in compression sch as cast iron, stone and concrete. These materials are weak in tension so it is important to ensre that bending does not prodce tensile stresses in them. If the compressive stress is too big, they fail by crmbling and cracking OSET OADS igre 3 Colmns often spport offset loads and these prodce bending stresses that combine with the compressive stress. This is illstrated in figre 4. igre 4 If a load is applied on the centre of the section, the stress in the colmn will be a direct compressive stress given by D = - /A Remember that compressive stresses are always negative. When the load is applied a distance 'x' from the centroid, a bending moment is indced in the colmn as shown. The bending moment is M = x where x is the off set distance. D.J.DUNN 7
rom the well known formla for bending stress we have B = My/I y is the distance from the centroid to the edge of the colmn. The stress prodced will be +ve (tensile) on one edge and -ve (compressive) on the other. On the compressive edge this will add to the direct compressive stress making it larger so that = B + D = -My/I - /A On the tensile edge the reslting stress is Sbstitte M = x = B + D = My/I - /A = = xy/i - /A WORKED EXAMPE No. A colmn is 0.5 m diameter and carries a load of 500 kn offset from the centroid by 0.1m. Calclate the extremes of stresses. SOUTION = 500 kn x = 0.1 m y = D/ = 0.5 m Tensile Edge = B + D = xy/i - /A = 500 000 x 0.1 x 0.5 /( x 0.54/64) - 500 000/( x 0.5/4) = 1.58 MPa (Tensile) Compressive Edge = B + D = -xy/i - /A = -500 000 x 0.1 x 0.5 /( x 0.54/64) - 500 000/( x 0.5/4) = -6.61MPa (compressive) D.J.DUNN 8
NEUTRA AXIS The netral axis is the axis of zero stress. In the above example, the stress varied from 1.58 MPa on one edge to -6.61 MPa on the other edge. Somewhere in between there mst a vale of y which makes the stress zero. This does not occr on the centroid bt is by definition the position of the netral axis. Ideally this axis shold not be on the section at all so that no tensile stress occrs in the colmn. The position of the netral axis can easily be fond by drawing a stress distribtion diagram and then either scaling off the position or calclate it from similar triangles. WORKED EXAMPE No.3 Determine the position of the netral axis for the colmn in example. SOUTION Drawing a graph of stress against position (y) along a diameter we get the figre shown (not drawn to scale). If it is drawn to scale the position of the netral axis may be scaled off. igre 5 Using similar triangles we arrive at the soltion as follows. A + B = 0.5 A = 0.5 - B A/1.58 = B/6.61 (0.5 - B)/1.58 = B/6.61 3.3105-6.61B = 1.58 B B = 0.406 m D.J.DUNN 9
SE ASSESSMENT EXERCISE No. 1. A colmn is 0.4 m diameter. It has a vertical load of 300 kn acting 0.05m from the centroid. Calclate the stresses on the extreme edges. (Answers 0 MPa and -4.77 MPa).. A colmn of diameter D has an offset load. Show that if the stress on one edge is zero, the offset position is D/8. 3. A colmn is made from cast iron tbe 0.4 m otside diameter with a wall 40 mm thick. The top is covered with a flat plate and a vertical load of 70 kn is applied to it. Calclate the maximm allowable offset position of the load if the material mst always remain in compression. (Answer 0.08 m) 4. A hollow cast iron pillar, 38 cm otside diameter and wall thickness 7.5 cm, carries a load of 75 kn along a line parallel to, bt displaced 3 cm from, the axis of the pillar. Determine the maximm and minimm stresses in the pillar. What is the maximm allowable eccentricity of the load relative to the axis of the pillar if the stresses are to be compressive at all points of the cross section? D.J.DUNN 10