# MECHANICS OF SOLIDS COMPRESSION MEMBERS TUTORIAL 1 STRUTS. On completion of this tutorial you should be able to do the following.

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1 MECHANICS OF SOLIDS COMPRESSION MEMBERS TUTORIAL 1 STRUTS You should judge your progress by completing the self assessment exercises. On completion of this tutorial you should be able to do the following. Define a strut. Extend bending theory to struts. Solve the loads at which struts collapse. Solve the sideways deflection of struts. It is assumed that students doing this tutorial already familiar with the concepts of second moments of area, bending stress, bending moments and deflection of beams. D.J.DUNN 1

2 1. INTRODUCTION COMPRESSION MEMBERS Compression members are divided into three types. 1.1 SHORT In this case the member is called a column and will fail when the ultimate compressive stress of the material is exceeded. Failure will depend upon the type of material. Columns are usually made of brittle material (concrete, stone or cast iron) and failure will be of the brittle type which is basically a crumbling of the material. Columns were studied in Mechanical Science A. 1. LONG Long members are called struts. These bend and bow out when compressed beyond a certain limit. Failure occurs by bending when the maximum tensile stress is exceeded. 1.3 INTERMEDIATE These are short enough to produce failure by bending but also short enough to produce compression failure, especially if the material is brittle This tutorial covers struts. Intermediate and short compression members are covered in the next tutorial. D.J.DUNN

3 . STRUTS.1 DEFINITIONS A strut is a long thin compression member. It may collapse under a compressive load by buckling and bowing out as shown in fig.1. The diagram shows the member with its length horizontal but it is just as likely to be vertical. It is drawn this way so that the x-y coordinates are in the normal position at the left end. x measures the distance along the length and y is the deflection. Figure 1. SLENDERNESS RATIO L A strut is usually defined by its slenderness ratio. This is defined as S.R. k L is the effective length and k is the radius of gyration for the cross sectional area..3 RADIUS OF GYRATION k The radius of gyration is defined as k I is the nd moment of area and A is the cross sectional area. I A WORKED EXAMPLE No.1 Derive formulae for the radius of gyration of a circle diameter D and a rectangle width B and depth D. Circle 4 4 πd πd 4D I A k D D 4 Rectangle BD I 1 3 A BD k 3 BD 1BD D 1 D.J.DUNN 3

4 WORKED EXAMPLE No. Calculate the slenderness ratio of a strut made from a hollow tube 0 mm outside diameter and 16 mm inside diameter and 1. metres long. For a hollow tube the second moment of area is πd d π I 4637mm πd d π0 16 A mm 4 4 I 4637 k 6.4 mm A L 100 mm S.R k 6.4 mm SELF ASSESSMENT EXERCISE No.1 1. Find the radius of gyration and the slenderness ratio of a strut made from 5 m length of hollow tube 50 mm outer diameter and 40 mm inner diameter. (Ans 16 mm and 31.3) D.J.DUNN 4

5 . THEORY.1 CRITICAL LOAD The force applied to a strut is in an axial direction (the x direction) and not transversely (y direction) as it is for beams. Consider a long thin strut resting against a solid surface at one end and with a screw device at the other as shown. The distance from the end is x and the deflection is y. When the screw is tightened, the strut is forced to deflect sideways. The more the screw is turned the more the strut deflects. If the strut bends as shown, there must be a bending moment and a bending stress in the material. The applied bending moment is Fy. The force applied by the screw and the distance y will increase as the bending moment increases. At some point it will be found that the screw can be turned with no further increase in the force. This can be explained because the increase in deflection alone is sufficient to produce the required increase in the bending moment. The strut will go on bending until it fails (usually by exceeding the yield stress in the material and leaving it permanently bent). When this point is reached the strut has failed and the critical force F c has been reached. Figure Now consider a vertical strut with weights causing the compression. If the weights are less than the critical force F c the strut is unlikely to deflect very much as it is no longer forced to do so. However when the critical value is reached, the strut collapses suddenly and fails as there is nothing to stop it. This might be explained as follows. If the load is critical, the strut will start to deflect. As the distance y increases so will the bending moment. This in turn makes it deflects even further. This is a run away or unstable condition and the strut keeps on bending and fails. A strut is an unstable structure as collapse is sudden and without warning. Figure 3 D.J.DUNN 5

6 . EULER'S THEORY FOR COLLAPSE No strut can be perfectly straight and a force applied as shown will make it bend slightly when an axial compression load is applied. The direction in which it moves is random so let s sketch it so that it bows upwards on the diagram. Note that the force is not a transverse force in the y direction but an axial force applied in the x direction. Consider any distance x from the end. The strut has deflected a distance y. The bending moment at this point is M = F y. This will be a maximum at the point where the deflection is greatest so let this maximum value of y be denoted y m. Figure 4 The applied bending moment is M = -Fy (minus because it hogs) If the strut does not collapse, the internal bending moment must balance the applied moment and this is given by bending theory as d y M EI Fy dx (minus because it hogs) d y EI Fy dx Rearrange d y Fy 0 dx EI This is a second order differential equation with a standard solution. y Acos(cx) Bsin(cx) F A and B are constant of integration and c represents the expression EI A and B are solved from boundary conditions. We know that for the case illustrated the deflection is zero at the ends and a maximum at the middle. x = 0 y = 0 and at x = L y = 0 First substitute y = 0 and x = 0 into the solution. 0 = Acos(0) + B(sin 0) = A(1) + B(0) hence A = 0 Next substitute y = 0 and x = L 0 = Acos(cL) + Bsin (cl) = 0 cos(cl) + B sin(cl) from which B sin (cl) = 0 If B is zero the solution is always zero and this clearly is not the case. It follows that sin (cl)= 0 and this occurs when cl = 0, 180o, 360o, and so on. In radians this is cl=,, 3 and so on. In general cl = n where n is an integer. D.J.DUNN 6

7 We may state that cl nπ L F EI F n nπ π EI L and thecorresponding deflection is y Bsin(cL) which may be evaluated. If the strut does not have a symmetrical cross section, it will buckle about the axis with least resistance (smallest value of I). For a rectangular section B must always be the larger of the two dimensions. Figure 5 This is the formula usually given in exams and the above derivation should be practised prior to the exam. n is called the mode and its meaning is very real. A node is any point where the strut does not deflect. If the strut is restrained at any point (e.g. guy ropes on a mast) that point will be a node. The diagram shows what happens when the restraints are placed at the middle (n = ) and at equal distances of 1/3 of the length (n = 3). n = 1 is the fundamental mode. Figure 6 This derivation is due to Euler and the value of F is called Euler's critical load. D.J.DUNN 7

8 There are three other modes of importance which are governed by the way the ends are constrained. Half Mode with n = 0.5 occurs when one end is held rigidly and the other is unrestrained. Figure 7 Another double mode with n = occurs when both ends are held rigidly. Figure 8 An unusual mode with n = 1.43 occurs when one end is held rigidly but the other end is pinned (allowed to rotate) but not allowed to move sideways. Figure 9 D.J.DUNN 8

9 WORKED EXAMPLE No.3 A strut is m long and has a rectangular cross section 30 mm x 0 mm. The bottom is built into a ground socket and the top is completely unrestrained. Given E = 00 GPa calculate the buckling load. SOLUTION F = n EI/L This case is as shown in fig. 7 with n = 0.5 I = BD 3 /1 = x 10-8 m 4 F = 0.5 x00 x 10 9 x x 10-8 / F = 470 N WORKED EXAMPLE No.4 Repeat the previous problem but with the strut is pinned at the top and bottom and not allowed to move sideways. SOLUTION F = n EI/L This case is as shown in fig. 6 with n = 1 I = BD 3 /1 = x 10-8 m 4 F = 1 x00 x 10 9 x x 10-8 / F = 9870 N. D.J.DUNN 9

10 SELF ASSESSMENT EXERCISE No. 1. A steel strut is 0.15 m diameter and 1 m long. It is built in rigidly at the bottom but completely unrestrained at the top. Calculate the buckling load taking E = 05 GPa. (Ans kn).. A steel strut has a solid circular cross section and is 8 m long. It is pinned at the top and bottom but unable to move laterally at the ends. The strut collapses under a load of 00 kn. Taking E = 05 GPa calculate the diameter of the strut. (Ans mm). 3. A shaft is made from alloy tubing 50 mm outer diameter and 30 mm inner diameter. The shaft is placed between bearings 3 m apart so that the ends are constrained to remain horizontal. The shaft also has to take a horizontal axial load. Taking E = 10 GPa determine the maximum axial load before buckling occurs. (Ans. 140 kn). 4. A strut is 0. m diameter and 15 m long. It is pinned at both ends. Calculate Euler's critical load. E = 05 GPa (Ans kn) D.J.DUNN 10

11 .3 VALIDITY LIMIT OF EULER'S THEORY Euler's theory is inaccurate when the slenderness ratio is small. If the strut is very thin, then the material will simply crush under the axial compression. The slenderness ratio limit depends upon the material but generally if the ratio is less than 10 for steel or less than 80 for aluminium and its alloys, the crushing becomes important and failure will occur at loads smaller than those predicted by Euler. 3. DEFLECTION OF STRUTS It is of interest to know the deflection of a strut at loads less than the buckling value. This can be very complicated work but we do not have to go into full details. When the strut buckles, it fails because it reaches the elastic limit of the material in compression. The strut is put into compression by the load and the direct compressive stress is D = F/A Bending stress is also induced in the strut which tends to be tensile on one side and compressive on the other. This is given by : B = M/I Figure 10 is the distance from the neutral axis to the extreme edge in compression (this is denoted y in beam stress problems but y is unsuited to this case). M is the bending moment Fy. The total compressive stress is hence = F/A + M/I At the point of collapse this is the elastic limit in compression c and the deflection is ym. F Fymδ σ c A I F I y m σ c A Fδ D.J.DUNN 11

12 WORKED EXAMPLE No.5 A strut is made is made from 16 mm diameter steel bar. The buckling load is 400 N. The elastic limit in compression is 30 MPa. Calculate the deflection just prior to collapse. SOLUTION I = D 4 /64 = 3.17 x 10-9 m4. A =D /4 = 01 x 10-6 m. c =30 x 10 6 N/m. = D/ = m y m 30 x x x 10 x m or 51.6 mm 400 x D.J.DUNN 1

13 SELF ASSESSMENT EXERCISE No. 3 1.a A uniform slender elastic column of length L is pin jointed at each end and subjected to an axial compression load P. Show that the Euler crippling load occurs when P = EI/L where I is the relevant second moment of area of the column and E is the modulus of elasticity of the material. State any assumptions made. b A straight steel rod 0.5 m long and 0.01 m diameters loaded axially until it buckles. Assuming that the ends of the rod are pin jointed, determine the Euler crippling load. Assume E = 06 GPa. (Ans kn).a The Euler buckling load P for a slender strut of length L and second moment of area I, pin jointed at each end, is given by P = EI/L E is the modulus of elasticity of the material. Using this expression without proof, obtain the formula for the Euler buckling load when the strut is i. fixed (built in) at each end. ii. fixed at one end and pin jointed at the other. b) Fig. 11 shows a vertical pole 6 m long, pinned at the lower end and supported by a wire at the upper end. The pole consists of a tube 50 mm outside diameter and 40 mm inside diameter and the wire has an effective diameter of 6 mm. What is the maximum load P that this system can withstand before failure occurs? For steel assume that the modulus of elasticity E is 06 GPa and for the wire assume that the ultimate stress is 480 MPa. Figure 11 (Answer, the rope breaks before the pole buckles so the maximum value of P is 9.59 kn) D.J.DUNN 13

14 3.a A uniform slender strut of length L which is clamped at one end and free at the other is subjected to an axial compression load P as shown in fig.1. Show that according to EULER'S theory, the strut will buckle when P= (/L)EI where I is the minimum second moment of area of the strut and E is the modulus of elasticity for the material. b A straight steel rod 9 mm diameter is rigidly built into a foundation, the free end protruding 0.5 m normal to the foundation. An axial load is applied to the free end of the rod which deflects as shown in fig.1. Determine the following. i. Euler s buckling load. (636 N) ii. The deflection of the free end of the rod when the total compressive stress reaches the elastic limit. (3.6 mm) For steel assume E = 00GPa and the stress at the elastic limit is 300MPa. Figure 1 D.J.DUNN 14

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