STRENGTH OF MATERIALS FOR TECHNICIANS

STRENGTH OF MATERIALS FOR TECHNICIANS

This page intentionally left blank

Strength of Materials for Technicians J C DROTSKY BSC, NDT, NTTD Head of Department Strength of Materials Vaal Triangle Technikon Butterworths Durban/Pretoria

BUTTERWORTH PUBLISHERS (PTY) LTD 1984 ISBN: 0 409 11082 5 THE BUTTERWORTH GROUP South Africa BUTTERWORTH PUBLISHERS (PTY) LTD 8 Walter Place, Waterval Park, Mayville, Durban 4091 England BUTTERWORTH & CO. (PUBLISHERS) LTD 88 Kingsway, London WC2B 6AB Australia BUTTERWORTHS (PTY) LTD PO Box 345, North Ryde, NSW 2113 Canada BUTTERWORTH & CO. (CANADA) LTD 2265 Midland Avenue, Scarborough, Ontario MIP 4SI New Zealand BUTTERWORTHS OF NEW ZEALAND LTD 33-35 Cumberland Place, Wellington USA BUTTERWORTHS (PUBLISHERS) INC. 10 Tower Office Park, Woburn, Massachusetts 01801 House editors: Sarie Moolman, Gisela Hasse Set and designed by Dieter Zimmermann (Pty) Ltd, Johannesburg Cover design by Jenny Exton Printed by Sigma Press (Pty) Ltd, Pretoria

Contents Preface. ix 1 Simple trusses 1 1.1 Direct force 1 1.2 Moment of a force 2 1.3 Conditions for static equilibrium 2 1.4 Simple plane trusses 2 1.5 Assumptions 4 1.6 Free-body diagrams 4 1.7 Method of sections 4 1.8 Simple space trusses 10 1.9 Graphical solution 10 1.10 Analytical solution 13 2 Simple stress and strain 28 2.1 Direct stress and strain 28 2.2 Tension test 29 2.3 Factor of safety 33 2.4 Axially loaded members 33 2.5 Shear stress and shear strain 39 2.6 Modulus of rigidity 40 2.7 Single shear and double shear 40 3 Thin-walled pressure vessels and thin rotating cylinders 51 3.1 Thin-walled cylinder subjected to an internal pressure 51 3.2 Assumptions 51 3.3 Thin-walled sphere subjected to an internal pressure 53 3.4 Rotating thin cylinders 56 4 Torsion of circular shafts 60 4.1 Assumptions 60 4.2 Torsional rigidity (stiffness) and strength 62 4.3 Polar section modulus 62 4.4 Mean torque 62 4.5 Comparison of hollow and solid shafts 65 4.6 Compound shafts 66 4.7 Rigid-flange couplings 71 5 Close-coiled helical springs 80 5.1 Assumptions 80

5.2 Maximum shear stress 80 5.3 Deflection and strain energy 81 5.4 Stiffness of a spring 83 5.5 Compound springs 85 6 Shear force and bending moment 94 6.1 Beams 94 6.2 Types of load 95 6.3 Shear force and bending moment 95 6.4 Relationship between load, shear force and bending moment 99 6.5 Conclusions 100 6.6 Bending moment diagram by summation 101 6.7 Bending moment diagram by definition 107 6.8 Built-in beams 114 6.9 Curved beams 117 7 Temperature stresses 129 7.1 Introduction 129 7.2 Fixed ends 129 7.3 Materials in series 130 7.4 Materials in parallel 133 8 Strain energy due to direct stresses 141 8.1 Introduction 141 8.2 Gradually applied loads (dead loads) 141 8.3 Impact loads 142 8.4 Suddenly applied loads (live loads) 143 8.5 Rods in series and parallel 146 9 Second moment of area 153 9.1 Centroid of an area 153 9.2 Second moment of area (moment of inertia) 155 9.3 Theorem of parallel axes 156 9.4 Theorem of perpendicular axes 156 10 Bending stresses 164 10.1 Introduction 164 10.2 Assumptions 164 10.3 The beam formula 164 10.4 The (elastic) section modulus 167 10.5 Selection of a suitable section for a beam 167 10.6 Eccentric load on a short column 172 10.7 Reinforced concrete beams (simple tension reinforcement only) 177

11 Shear stress in beams 188 11.1 Introduction 188 11.2 The shear stress equation 188 11.3 Assumptions 189 11.4 Built-up sections 193 12 Struts 199 12.1 Introduction 199 12.2 Euler formula 199 12.3 Slenderness ratio 204 12.4 Validity limit for Euler formula 204 12.5 Rankine-Gordon (Rankine) formula 204 12.6 Comparison between the Euler and Rankine-Gordon formulae 204 13 Catenaries 211 13.1 Common (uniform) catenary 211 13.2 Saddles and frictionless pulleys 215 13.3 The parabolic catenary 217 14 Testing of materials 225 14.1 Tension test 225 14.2 Compression test 227 14.3 Shear tests 227 14.4 Cold bend test 229 14.5 Hardness tests 230 14.6 Impact tests 231 14.7 Stress concentration and fatigue 233 Appendix A 237 Appendix B 238 Appendix C 246 Index 258

This page intentionally left blank

Preface This textbook covers the syllabi of the first- and second-year courses in Strength of Materials for the National Diploma in Engineering. It should also prove to be of value to students studying for the National Technical Certificate and first-year degree courses in Engineering. The theory of each chapter is developed in an understandable and logical manner to enable the student to follow the reasoning without difficulty. All assumptions are clearly stated to emphasise the limitations of formulae. Basic concepts and principles, such as the conditions for static equilibrium, are emphasised to minimise the number of formulae students need to memorise. Free-body diagrams are also used to simplify the analysis of forces in members of structures. Theoretical explanations are followed by worked examples in which the application of the different principles is illustrated. Furthermore, each chapter ends with a number of examples arranged in order of difficulty, with answers provided. The material is presented in such a way that the book can not only be used as a reference source in work situations, but also prepares students for follow-up courses in the National Higher Diploma in Engineering and the National Diploma in Technology. An abridged version of the structural steel tables is also included. Computers are currently used extensively in the engineering field and students should not only be encouraged to write their own programs, but should be confronted with more sophisticated programs. As the majority of students either own or have access to a microcomputer, a sample listing of a BASIC program which analyses plane trusses is included in this book. The author has found this type of program invaluable as it leads to a better understanding of the topic since a student can observe the immediate effect of changes in the original data without laborious recalculations. I wish to acknowledge the helpful suggestions made by my colleagues, especially Mr A. E. F. Goosen from the Department of Civil Engineering. I would also like to thank my wife Louisa and children Vicky, Tanya and Nico for their patience and encouragement. The Author

This page intentionally left blank

1 Simple trusses 1.1 Direct force There are two types of direct forces, namely tensile and compressive. Externally applied direct forces, P, are illustrated in figure 1.1. -d. Tension "D ^ Compression Figure 1.1 Internal forces are induced in the body to counteract the externally applied forces. The calculation of these internal forces are the main object of this chapter. The corresponding internal forces, F, are illustrated in figure 1.2. A member in tension is called a tie and a member in compression a strut. Figure 1.2 1.2 Moment of a force The moment M A of a force P about a point A is equal to the force multiplied by the shortest distance between the point A and the line of action of the force..'. M A = Px newton metres (N.m) 1

Figure 1.3 1.3 Conditions for static equilibrium If the forces acting on a body are in equilibrium, then - (a) the sum of all the forces in any direction must be zero (ΣΡ = 0); and (b) the sum of all the moments of the forces about any point must be zero (ΣΜ = 0). These two conditions must both be satisfied and are very important as they will be sufficient to solve a considerable number of problems that will be investigated in this field. Such problems are termed statically determinate. All the problems in this chapter are statically determinate. There are problems which cannot be solved by the equations of equilibrium only. To illustrate this, consider the propped cantilever in figure 1.4. 15 kn 13 kn I kn Figure 1.4 If the force T is unknown, the reaction R and the moment M in the wall will consequently also be unknown. Thus there are three unknowns and only two equations. This type of problem is termed statically indeterminate. 1.4 Simple plane trusses A simple plane truss is a two-dimensional assemblage of members, each member being joined at its ends to the foundation or to other members by frictionless pin joints. In order to form a stable structure (fig. 1.5(a)) a sufficient number of members have to be used, arranged in a suitable manner. The simplest stable arrangement is a triangle and therefore a truss consists of a number of triangles. A pinjointed structure consisting of four members assembled in a rectangle (fig. 1.5(b)) 2

is unstable because a small sideway force will result in the collapse of the struture. A structure with a redundant member is shown in figure 1.5(c). This type of structure is also termed over stiff. (a) Stable (b) Unstable (c) Redundant member Figure 1.5 The following equation is used to determine the state of a structure not attached to a foundation: m = 2j - 3 (1.1) Where m = number of members j = number of joints. If equation 1.1 is satisfied, the structure is statically determinate provided the members are correctly arranged. To illustrate why this statement is conditional, consider figure 1.6. In both cases m = 2j - 3, but (a) is stable and (b) unstable. Figure 1.6 If m < 2j 3 the structure is unstable and if m > 2j 3 there are redundant members in the structure. If the structure is attached to a foundation, it will be statically determinate if the members are correctly arranged and the following equation is satisfied: 2j = m + r (1.2) Where j = number of joints m = number of members r = number of reaction components in the x and y directions. 3

1.5 Assumptions The following assumptions are usually made in the analysis of trusses: (a) Buckling of members will not occur. (b) The forces in the members due to the mass of the structure are small compared to the applied loads and may be ignored. (c) All the members are connected by frictionless pin-joints. In practice the members are bolted, riveted or welded together, thus making the structure more rigid. This assumption simplifies the computations involved and gives conservative results. (d) All external forces, including the reactions at the supports, are applied at the joints and the structure is statically determinate. 1.6 Free-body diagrams A free-body diagram is a sketch of the body to be investigated, indicating all external forces and reactions. k Figure 1.7 Figure 1.7(b) shows a free-body diagram of the pylon and rope in figure 1.7(a). This type of diagram normally simplifies a problem. 1.7 Method of sections This is an analytical method for solving simple plane trusses. Procedure (a) Sketch a free-body diagram of the structure. (b) Determine the reactions at the supports, using the two conditions for static equilibrium, if they are required or necessary for further calculations. (c) Draw a section line through the structure, cutting those members in which the 4

forces are required. A maximum of three members should be cut by a section line. (d) The external forces on only one side of the section line must be considered. Indicate this direction on the section line. (e) Indicate the assumed force directions in the members that have been cut by the section line (on that side of the section line chosen in (d)). If the force direction is chosen correctly the answer will be positive. A negative answer will result if the direction was chosen incorrectly. (f) Take moments about any joint on the structure, considering only the external forces and forces in the members that have been cut as indicated in steps (d) and (e). The selected joint must have only one unknown force causing a moment about that joint. This force may then be calculated using the second condition for static equilibrium. If the answer is negative, the direction of the appropriate arrow must be changed before attempting any further calculations. This method is useful in cases where the forces in only a few members of a truss are required. It should be noted that step (b) may be omitted in cases where the reactions are not required for the calculations in step (f). Example 1.1 Determine - 1 whether the truss in figure 1.8 is statically determinate; 2 the magnitude of the forces in members BC, CE and ED; and 3 the magnitude of the reactions at A and F. Figure 1.8 Solution 1 Applying equation 1.2: 2x6 = 8 + 4, hence the truss is statically determinate. 2 A section line, such as xx (fig. 1.9(b)), devides the truss into two free bodies, one of which is shown in figure 1.9(a). The free bodies will be subjected to externally 5

applied loads and the member forces at the section line must be inserted to maintain equilibrium of forces. The unknown member forces may then be calculated applying the two conditions for static equilibrium. This generally results in simultaneous equations which then have to be solved. This can be avoided by taking moments about a joint, resulting in only one unknown force in the equation (i.e. the method of sections). Following the method outlined in paragraph 1.7: (a) Figure 1.9 (c) Consider section line xx (fig. 1.9(b)), and (d) that part of the structure to the right of xx. (e) The assumed directions of the forces in the members are as indicated in figure 1.9(b). (f) The length of CE, L CE = 2 tan 30 = -? IM C = 0.. 2 x 50 - - x F ED = 0.. F ED = 86,6 kn (strut) ΣΜ Ε = 0.. 2 x 50 -? x F BC cos 30 = 0.. F BC = 100 V3 kn (tie) ΣΜ Ό = 0.. 2xF C E -^X 100 cos 30-2 x 100 sin 30 = 0.'. F CE = 100 kn (strut) 6

Member BC CE ED Force (kn) 100 100 86,6 Type Tie Strut Strut 3 From figure 1.9(b): ΣΜ Ρ = 0.'. 50 X 4-2R A sin 45 = 0.. R A = 141,42 kn Let Rpv be the vertical component of the reaction at F. ΣΜ Α = 0.. 50 X 6 - Rpv x 2 = 0 RFV = 150 kn ΣΜ Β = 0.'. 50 X 4 - R FH X 2 = 0 R FH = 100 kn.. The reaction at F = Vl50 2 + 100 2 = 180,28 kn in a direction tan -1 (^) = 56,3 to the horizontal. Notes 1 These reactions could also be obtained by first determining the forces in members AB, BF and EF and then the equilibrant force at each joint. 2 For this type of truss it is not necessary to calculate the reactions at the supports (step (b)) before determining the forces in the members. Example 1.2 Use the method of sections to calculate the magnitude of the forces in members BC, EC, ED, BE and AE of the plane truss shown in figure 1.10. 7

Solution (a) 10 sin 60 10 cos 60 = 5 kn F ED sin 15 Figure 1.11 (b) For this example at least one reaction must be known before the forces in the members can be calculated. Sum of the horizontal forces = 0 (ΣΡ Η = 0).'. R AH = 10 cos 60 = 5 kn Angle BCD = 135.. angle CDF = 180-135 = 45.. L CF = V3x-^=-^=L FD 'VT γτ" ΣΜ Ό = 0.'. L ro x 10 sin 60 + L CF x 5 + 5$ί+ V5) - R AV (λ/2 + 2 x ψ) = 0.. 2@χ 10 x V^+ ^"χ 5 + 5(\/3"+ V2) - R AV (V2 + 2 x &) = 0 V2 2 γ2 γ2 V2.. R AV = 7,745 kn ΣΜ Α = 0.. 5 x ^ V2 + 10 X ^X 2 $f + V2) - 5 X ^ - R D (V2 + 2x^) = 0 V> 'V2 'V2 7.. R D = 5,915 kn To verify these answers we use the first condition for static equilibrium: ΣΡ ν = 0.. 5 + 10 x VI = 7,745 + 5,915, which is correct. 2 To calculate the forces in members BC, EC and ED: (c) Consider section line xx (fig. 1.11), and (d) that part of the structure to the right of xx. (e) Assume the forces in the members to be in the direction as indicated in figure 1.11. 8

(f) ΣΜ = 0 ^ίx 5,915 -^XF E D cos 15 + -^F ED sin 15 = 0 V2 F ED = 8,365 kn (tie) ΣΜ Ε = 0 ^?x 10 sin 60 --^Tx 5 + ^ x F BC - (^+^) x 5,915 = 0 2 2 2 2 V2 F BC = 12,5 kn (strut) ΣΜ 0 = 0 ^lx 12,5 +V3~X F EC -^x 10 sin 60 -^fx 5 = 0 V2 VI V2 F EC = 0,82 kn (strut) To calculate the forces in members BE and AE: y 10 sin 60 kn (a) (c) Consider section line yy (fig. 1.12), and (d) that part of the structure to the left of yy. (e) Assumed directions of forces are as indicated in figure 1.12. (f) ΣΜ Β = 0 Λ ^Χ 7,745 + -^Χ F sin 15 - -^ X 5 - & x F AE cos 15 V5'"" ~ " V2" AE ~ V2 ' V2.'. F AE = 3,88 kn (tie) ΣΜ Α = 0 Λ ^ Χ 5 - ^ Χ 12,5 - V3 x F BE = 0 V2 VI.. F BE = -5,3 kn (tie) Member Force (kn) Type BC EC ED AE BE 12,5 0,82 8,365 3,88 5,3 Strut Strut Tie Tie Tie

1.8 Simple space trusses Space trusses are three-dimensional structures. The simplest stable space truss consists of six members joined to form a tetrahedron as shown in figure 1.13. Figure 1.13 A simple space truss can be constructed by the successive addition of three members and a joint to the tetrahedron. Computers are normally employed to solve the forces in the members and the deflection of the joints because of the extent of the calculations involved. In the next paragraphs a graphical and analytical method will be used to solve simple space trusses consisting of a single tetrahedron. 1.9 Graphical solution Example 1.3 A wall crane (fig. 1.14), consists of a boom CD in the yz plane and two cables AD and BD, suspending a weight of 10 kn. Determine the magnitude of the forces in the boom and the cables by graphical solution. f 10 kn Figure 1.14 10

Solution The length of any member, such as AD, may be calculated as follows: ÄÖ = VXE 2 + EÖ 2 = Vl 2 + 3 2 AD = VAO 2 + OD 2 = V(l 2 + 3 2 ) + 5 2 = 5,92 m (a) The following distances are calculated in a similar way: ÄB~ = V5 2r T7 = 5,lm BC = V&TY 2 = 5,4 m BD~ = V2 2 + 2 2 + 5 2 = 5,74 m AC = V?T^ = 5m CD = VO 2 + 3 2 + 5 2 = 5,83 m P \\/ R D /k X \ B B E_ // E y/v A l D / v^n^^^ D" j c C M (a) X (b) Figure 1.15 (b) Draw to scale a view representing the true distances between A, B and C (fig. 1.15(a)). Draw triangles AD'B and AD"C to represent members AD, BD and AD, CDjisjf they were lying flat. From D' and D" draw lines perpendicular to AB and AC respectively to intersect at D. Complete the view. (c Elongate CD, representing the boom to intersect XrTin E and join points E and D'. CD is chosen since it lies in the same plane as the line of action of the applied force. (d) Draw the baseline xx parallel to CD and project points A, B, C and E to xx. The true length of the boom CD will be represented in the auxilliary view. Complete the auxiliary view (fig. 1.15(b)). 11

(e) Draw to scale the line DM to represent the force of 10 kn and complete the parallelogram MNDO with sides parallel to CD and ED. ND represents the magnitude of the force in the boom CD. (f) Draw the line EO on the original view. It must be equal in length to EO in the auxiliary view because both are shown in true length. Complete the parallelogram D'POR with sides parallel to AD' and D'B. D'P represents the magnitude of the force in rope BD and D'R represents the magnitude of the force in rope AD..'. Force in AD = 5,1 kn (tension) Force in BD = 7,5 kn (tension) Force in CD = 12,6 kn (compression) Example 1.4 Determine the forces in all the members of the tripod crane shown in figure 1.16. Figure 1.16 Solution The method described in the previous example will again be used. (a) ÄD = V4 2 + 0 2 + 7 2 = 8,06 m ÄB" = Vll 2 + 5 2 = 12,1 m BD" = V7 2 + 5 2 + 7 2 = 11,09 m CD = V8 2 + 3 2 + 7 2 = 11,05 m BC~ = Vl 2 + 8 2 = 8,1 m AC = Vl2 2 + 3 2 = 12,4 m Note: Any one of lines AD, BD or CD (fig. 1.17) could be elongated in this case to obtain a point E (step (c) in example 1.3). The longest of these lines are normally elongated. 12

D N Figure 1.17 FAD = 6,8 kn (compression) F BD = 1,8 kn (compression) F CD = 3 kn (compression) 1.10 Analytical solution There are various methods for solving simple space trusses. The method of sections could be employed but it is tedious because of the work involved to take moments about a point in a three-dimensional space. Example 1.5 Calculate the forces in all the members of the shear leg shown in figure 1.18(a). 25 kn Figure 1.18 (a) Figure 1.18(b) 13

Solution ΣΜΛΒ = 0.. 25 x 4 = R C v x 3.. Rev = 33,33 kn H 33 λ/3 2 4- l 2.. R c = ^. = 33,33 X W0 = 35,13 kn sin β 3.. F CD = 35,13 kn.. 25 X 1 = (R AV + RBV)3.'. R AV = 4,167 kn since RAV ~~ RBV,. R A = iig = 4467 x^l +22 + 32 = 7,48 KN sin oc 3 F AD = F BD = 7,48 kn An easier method is to make use of the first condition for equilibrium of a static body. Even with this method the components of the forces in the three principal directions are required, involving quite a number of angles. This method may be simplified by employing tension coefficients. 1.10.1 Tension coefficients In figure 1.19, consider the line member AB in the x-y plane with the coordinates A(0,0) and B (Δχ,Δν) and assume that it is subjected to a tensile force F AB. Β(Δχ, Ay) Figure 1.19 Consider point A: Δχ The component of F A B in the x direction = F A B COS Θ = F A B X ^^ LAB Ay The component of F A B in the y direction = F A B sin Θ = F A B X - LAB Define the tension coefficient of AB as force in member AB _ FAB t AB = length of member AB L AB.". The component of F AB in the x direction = t AB x Δχ, and the component of F AB in the y direction = t AB x Ay. (1) (2) 14

Notes 1 With this method a third dimension can easily be accommodated. The component of F AB in the z direction = t AB x Δζ (= 0 in this case). 2 The use of angles to calculate the components of the forces has been totally eliminated. Sign convention 1 It will be assumed that tensile forces are positive and compressive forces nega- 2 The'assumed positive directions of the x, y and z axes must be indicated. Example 1.6 Solve example 1.3 using tension coefficients. f 10kN Figure 1.20 Solution There are three unknown forces, therefore we require three equations. If thehtast condition for static equilibrium (ΣΡ Χ = 0; ΣΡ, = 0; ΣΡ 2 = 0) is applied to joint D, three equations can be obtained from which the tension coeffccients can be calculated. Consider joint D: (1) Indicate the positive directions of the three principal axes at joint D. Remembering that the x component of F AD = length component of AD in the x direction multiplied by t A D = _ 3t AD / etc. 15

(2) Direction X y z (c) - (-5), (b) + 3 x (d), 2 x (e) - 5 x (a), ' tad = % substituting in (e) gives Equation -3t AD + 2t BD + 0 = 0-10 + lt AD + 2t BD - 3t CD = 0 5t A D 5t B D 5tcD = 0 tad + t BD + t CD = 0 4t AD + 5t BD = 10 23t AD = 20 tßd = / substituting in (d) gives. 50 t CD - -^ ' FAD = LAD X t AD = V3 2 + l 2 + 5 2 X ^ = 5,144 kn (tensile) 23 FßD = LßD X tßd : V2 2 + 2 2 + 5 2 x = 7,493 kn (tensile) (a) (ΣΡ Χ = 0) (b) (XP y = 0) (c) (ΣΡ Ζ = 0) (d) (e) FCD = L CD x tco = VO 2 + 3 2 + 5 2 x (-fj) = -12,676 kn (compressive) Example 1.7 Solve example 1.4 using tension coefficients. Figure 1.21 16

Solution (1) The positive directions of the principal axes are indicated at joint D in figure 1.21. (2) Consider joint D: Direction Equation x -4t AD + 7t BD + 8t CD = 0 (a) (ΣΡ Χ = 0) y 0 + 5t BD - 3t CD = 0 (b) (ΣΡ γ = 0) z -9-7t AD - 7t BD -7t CD = 0 (c) (ΣΡ Ζ = 0) (c) -s- (-7), t AD + t BD + tco = -f (d) 4 X (d) + (a), 0 + llt ro + 12t CD = -f (e) 4 X (b) + (e), 31t BD = -f ' *BD = -0,165 9, substituting in (e) gives t CD = 0,276 5 and substituting in (d) gives tad = -0,843 3 * FAD = LAD X t AD = VO 2 + 4 2 + 7 2 x (-0,843 3) = -6,800 kn (compressive) FBD = L BD X t BD = V7 2 + 5 2 + 7 2 X (-0,165 9) = -1,840 kn (compressive) FCD = L CD X t CD = V8 2 + 3 2 + 7 2 X (-0,276 5) = -3,054 kn (compressive) 1.10.2 Equivalent tension coefficient If the externally applied load is not in the direction of one of the principal axes, then the components of the force in the x, y and z directions must be calculated. To simplify these calculations an equivalent tension coefficient is introduced. A force of 7 kn is applied at point A in figure 1.22 and its line of action passes through point E. 7kN Λ * *A / \ y i i Figure 1.22 y E 17