some crosssectional shapes are shown


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1 Chapter Axial Loaded Members.1 Introduction Axial loaded member : structural components subjected only to tension or compression, such as trusses, connecting rods, columns, etc. change in length for prismatic bars, nonuniform bars are determined, it will be used to solve the statically indeterminate structures, change in length by thermal effect is also considered stresses on inclined sections will be calculated several additional topics of importance in mechanics of materials will be introduced, such as strain energy, impact loading, fatigue, stress concentrations, and nonlinear behavior, etc.. Changes in Length of Axial Loaded Members consider a coil spring with natural length L subjected to an axial load P if the material of the spring is linear elastic, then P = k or = f P k : stiffness (spring constant) f : flexibility (compliance) with k f = 1 some crosssectional shapes are shown 1
2 prismatic bar : a member having straight longitudinal axis and constant cross section consider a prismatic bar with crosssectional area A and length L subjected to an axial load then = P / A P and = / L material is elastic = E L P L = L = CC = CC E E A E A : axial rigidity of the bar compare withp = k we have k = E A CC or f = L CC L E A Cable : used to transmit large tensile forces the crosssection area of a cable is equal to the total crosssectional area of the individual wires, called effective area, it is less than the area of a circle having the same diameter also the modulus of elasticity (called the effective modulus) of a cable is less than the modulus of the material of which it is made
3 Example 1 a Lshape frame ABC with b = 10.5 in spring constant c = 6.4 in pitch of the threads k = 4. lb/in p = 1/16 in if W = lb, how many revolutions of the nut are required to bring the pointer back to the mark? (deformation of ABC are negligible) M B = 0 => F = W b / c the elongation Then of the spring is = F / k = W b / c k = n p n = W b CCC = ( lb) (10.5 in) CCCCCCCCCCC = 1.5 revolutions c k p (6.4 in) (4. lb/in) (1/16 in) Example  the contraption shown in figure AB = 450 mm BC = 5 mm BD = 480 mm CE = 600 mm A BD = 1,00 mm 3 A CE = 50 mm 3 E = 05 GPa A = 1 mm P max =? ABC is rigid take the free body ABC, M B = 0 and F y = 0, we have 3
4 F CE = P F BD = 3 P the shortening of BD is F BD L BD (3 P) (480 mm) BD = CCCC = CCCCCCCCC = P x 106 (P : N) E A BD (05 GPa) (100 mm ) and the lengthening of CE is F CE L CE ( P) (600 mm) CE = CCCC = CCCCCCCCC = 11.6 P x 106 (P : N) E A CE (05 GPa) (50 mm ) a displacement diagram showing the beam is deformed from ABC to A'B'C' using similar triangles, we can find the relationships between displacements A'A" CC = B'B" CC A + CE or CCCC = BD + CE CCCC A"C' B"C' or A P x 106 CCCCCCCC = P x P x 106 CCCCCCCCCCCCC substitute for A = 1 mm and solve the equation for P P = P max = 3,00 N = 3. kn also the rotation of the beam can be calculated A'A" A + CE ( ) mm tan = CCC = CCCC = CCCCCCC = A"C' 675 mm 675 mm = 0.11 o 4
5 .3 Changes in Length Under Nonuniform Conditions consider a prismatic bar is loaded by one or more axial loads, use the free body diagrams, the axial forces in each segment can be calculated N 1 =  P B + P C + P D N = P C + P D N 3 = P D the changes in length of each segment are N 1 L 1 N L N 3 L 3 1 = CC = CC 3 = CC EA EA EA and the change in length of the entire bar is = the same method can be used when the bar consists of several prismatic segments, each having different axial forces, different dimensions, and different materials, the change in length may be obtained = n N i L i CCC i=1 E i A i when either the axial force N or the crosssectional area A vary continuously along the bar, the above equation no longer suitable 5
6 consider a bar with varying crosssectional area and varying axial force for the element dx, the elongation is d = N(x) dx CCCC E A(x) the elongation of the entire bar is obtained by integrating L L N(x) dx = d = CCCC 0 0 E A(x) in the above equation, = P/A is used, for the angle of the sides is 0 o, the maximum error in normal stress is 3% as compared to the exact stress, for small, error is less, for large, more accurate methods may be needed Example 3 L 1 = 0 in A 1 = 0.5 in L = 34.8 in A = 0.15 in E = 9 x 10 6 psi a = 8 in b = 5 in P 1 = 100 lb P = 5600 lb calculate C at point C 6
7 taking moment about D for the free body BDE P 3 = P b / a = 5600 x 5 / 8 = 5000 lb on free body ABC R A = P 3  P 1 = = 900 lb then the elongation of ABC is n N i L i N 1 L 1 N L = CCC = CCC + CCC i=1 E i A i E A 1 E A = (900 lb) (0 in) CCCCCCCCCC + (100 lb) (34.8 in) CCCCCCCCCC (9 x 10 6 psi) (0.5 in ) (9 x 10 6 psi) (0.15 in ) = in in = in = C = in ( ) this displacement is downward Example 4 a tapered bar AB of solid circular cross section with length L is supported to a tensile load P, determine L A C = d A C d(x) CC = x C d(x) = d A x CC L B d B d A L A L A 7
8 the crosssectional area at distance x is [d(x)] d A x A(x) = CCCC = CCC 4 4 L A then the elongation of the bar is N(x) dx L B P dx (4L A ) 4 P L A L B dx = CCC = CCCCC = CCC C E A(x) L A E ( d A x ) E d A L A x 4 P L A 1 L B 4 P L A P L A L B  L A = CCC [ C] = CCC ( C  C ) = CCC CCC E d A x L A E d A L A L B E d A L A L B 4 P L L A 4 P L = CCC ( C ) = CCCC E d A L B E d A d B for a prismatic bar d A = d B = d = 4 P L CCC = P L CC E d E A.4 Statically Indeterminate Structures flexibility method (force method) [another method is stiffness method (displacement method)] consider an axial loaded member equation of equilibrium F y = 0 R A  P + R B = 0 one equation for two unknowns [statically indeterminate] both ends A and B are fixed, thus AB = AC + CB = 0 8
9 this is called equation of compatibility elongation of each part can be obtained AC = R A a CC BC = R B b  CC E A E A thus, we have R A a CC  R B b CC = 0 E A E A then R A = P b / L R B = P a / L and C = AC = R A a CC = P a b CCC E A L E A summarize of flexibility method : take the force as unknown quantity, and the elongation of each part in terms of these forces, use the equation of compatibility of displacement to solve the unknown force stiffness method to solve the same problem the axial forces R A and R B can be expressed in terms of C R A = E A CC C R B = E A CC C a b equation of equilibrium 9
10 R A + R B = P E A CC C + E A CC C = P => C = P a b CCC a b E A L and R A = P b / L R B = P a / L summarize of stiffness method : to select a suitable displacement as unknown quantity, and the unknown forces in terms of these displacement, use the equation of equilibrium to solve the displacement Example 5 a solid circular steel cylinder S is encased in a hollow circular copper subjected to a compressive force P C for steel : E s, A s for copper : E c, A c determine P s, P c, s, c, P s : force in steel, P c : force in copper force equilibrium P s + P c = P flexibility method for the copper tube c = P c L CC = P L CC  P s L CC E c A c E c A c E c A c for the steel cylinder s = P s L CC E s A s 10
11 s = c P s L CC = P L CC  P s L CC E s A s E c A c E c A c P s = E s A s CCCCCC P E s A s + E c A c P c = P  P s = E c A c CCCCCC P E s A s + E c A c s = P s C = P E s CCCCC c = P c C = P E c CCCCC A s E s A s + E c A c A s E s A s + E c A c the shortening of the assembly is = P s L CCC = P c L CCC = P L CCCCC E s A s E c A c E s A s + E c A c stiffness method : P s and P c in terms of displacement E s A s E c A c P s = CC P c = CC L L equation of equilibrium P s + P c = P E s A s CC + E c A c CC = P L L it is obtained = P L CCCCC same result as above E s A s + E c A c 11
12 Example 6 a horizontal bar AB is pinned at end A and supported by two wires at points D and F a vertical load P acts at end B (a) ( all ) CD = 1 ( all ) EF = wire CD : E 1, d 1 ; wire EF : E, d P all =? (b) E 1 = 7 GPa (Al), d 1 = 4 mm, L 1 = 0.4 m E = 45 GPa (Mg), d = 3 mm, L = 0.3 m 1 = 00 MPa = 15 MPa P all =? take the bar AB as the free body M A = 0=> T 1 b + T (b)  P (3b) = 0 i.e. T 1 + T = 3 P assume the bar is rigid, the geometric relationship between elongations is = 1 1 = T 1 L 1 CC = f 1 T 1 = T L CC = f T E 1 A 1 E A f = L / E A is the flexibility of wires, then we have f T = f 1 T 1 thus the forces T 1 and T can be obtained T 1 = 3 f P CCCC T = 6 f 1 P CCCC 4 f 1 + f 4 f 1 + f 1
13 the stresses of the wires are T 1 3 P f 1 A 1 (4 f 1 + f ) 1 = C = CC ( CCC ) => P 1 = CCCCCC A 1 A 1 4 f 1 + f 3 f T 6 P f 1 A (4 f 1 + f ) = C = CC ( CCC ) => P = CCCCCC A A 4 f 1 + f 6 f P allow = minimum (P 1, P ) (b) numerical calculation A 1 = d 1 / 4 = 1.57 mm A = d / 4 = mm f 1 = L 1 E 1 / A 1 = 0.44 x 106 m/n f = L E / A = x 106 m/n with 1 = 00 MPa and = 15 MPa we can get P 1 =.41 kn and P = 1.6 kn then P allow = 1.6 kn at this load, Mg = 175 MPa, at that time Al = 00 (1.6/.41) = 105 MPa < 00 MPa.5 Thermal Effects, Misfits and Prestrains temperature change => dimension change => thermal stress and strain for most materials, thermal strain T is proportional to the temperature change T T = T : thermal expansion coefficient (1/ o C or 1/ o F) T : increase in temperature 13
14 thermal strain usually are reversible, expand when heard and contract when cooled no stress are produced for a free expansion body but for some special material do not behave in the customary manner, over certain temperature range, they expand when cooled and contract when heated (internal structure change), e.g. water : maximum density at 4 o C for a bar with length L, its elongation t due to temperature change T is t = t L = ( T) L this is the temperaturedisplacement relation no stress are produced in a statically determinate structure when one or more members undergo a uniform temperature change temperature change in a statically indeterminate structure will usually produce stress in members, called thermal stress for the statically indeterminate structure, free expansion or contraction is no longer possible thermal stress may also occurs when a member is heated in a nonuniform manner for structure is determinate or indeterminate 14
15 Example 7 a prismatic bar AB of length L the temperature is raised uniformly by T F y = 0 R A = R B = R displacement at A due to T : t = ( T) L ( ) R : R = R L / E A ( ) A = t  R = 0 ( T) L = R L CC E A R = E A ( T) and = R / A = E ( T) the stress is compressive when the temperature of the bar increases Example 8 a sleeve and the bolt of the same length L are made of different materials sleeve : A s, s bolt : A b, b s > b temperature raise T, s, b, =? 15
16 take a free body as remove the head of the bolt for temperature raise T 1 = s ( T) L = b ( T) L if s > b => 1 > the force existing in the sleeve and bolt, until the final elongation of the sleeve and bolt are the same, then 3 = P s L CC 4 = P b L CC E s A s E b A b equation of compatibility = 13 = + 4 s ( T) L  P s L CC = b ( T) L + P b L CC E s A s E b A b equation of equilibrium P b = P s it is obtained ( s  b ) ( T) E s A s E b A b P b = P s = CCCCCCCCCCC E s A s + E b A b the stresses in the sleeve and bolt are P s ( s  b ) ( T) E s E b A b s = C = CCCCCCCCC A s E s A s + E b A b P b ( s  b ) ( T) E s A s E b b = C = CCCCCCCCC A s E s A s + E b A b and the elongation of the sleeve and bolt is 16
17 = ( s E s A + b E b A b ) ( T) L CCCCCCCCCCC E s A s + E b A b partial check : if s = b =, then P b = P s = 0, and = ( T) L (O.K.) stiffness method : choose the final displacement as an unknown quantity E s A s P s = CC [ s ( T) L  ] L E b A b P b = CC [  b ( T) L] L P s = P b, it is obtained ( s E s A + b E b A b ) ( T) L = CCCCCCCCCCC same result E s A s + E b A b Misfits and Prestrains For the length of the bars slightly different due to manufacture if the structure is statically determinate, no prestrains and prestress if the structure is statically indeterminate, it is not free to adjust to misfits, prestrains and prestresses will be occurred if CD is slightly longer, CD is in compression and EF is in tension 17
18 if P is added, additional strains and stresses will be produced Bolts and Turnbuckles for a bolt, the distance traveled by the nut is = n p where p is the pitch of the threads for a doubleacting turnbuckle, the shorten is = n p Example 9 (a) determine the forces in tube and cables when the buckle with n turns (b) determine the shorten of the tube 1 = n p = P s L / E s A s 3 = P c L / E c A c eq. of compatibility 1  = 3 n p P s L CCC = P c L CCC (1) E s A s E c A c eq. of equilibrium P s = P c () (1) and () P s = n p E c A c E s A s CCCCCCCC P c = 4 n p E c A c E s A s CCCCCCCC L (E c A c + E s A s ) L (E c A c + E s A s ) Shorten of the tube is 18
19 3 = P c L CCC = 4 n p E s A s CCCCCCC E c A c E c A c + E s A s 6 Stresses on Inclined Sections consider a prismatic bar subjected to an axial load P the normal stress x = P / A acting on mn in 3D and D views are shown also the stress element in 3D and D views are presented (the dimensions of the element are assumed to be infinitesimally small) we now to investigate the stress on the inclined sections pq, the 3D and D views are shown the normal and shear forces on pq are calculated N = P cos V = P sin the crosssectional area of pq is A 1 = A / cos thus the normal and shear stresses on pq 19
20 are N P cos = C = CCCC = x cos A 1 A / cos V P sin =  C =  CCCC =  x sin cos A 1 A / cos sign convention : positive as shown in figure also using the trigonometric relations, we get x = C (1 + cos ) x =  C sin it is seen that the normal and shear stresses are changed with the angle as shown in figure maximum normal stress occurs at = 0 max = x maximum shear stress occurs at = 45 o max = x / the shear stress may be controlling stress if the material is much weaker then in tension, such as 0
21 short block of wood in compression mild steel in tension (Luder's bands) Example 10 a prismatic bar, A = 100 mm P = 90 kn = 5 o determine the stress state at pq section show the stresses on a stress element x = P  C = 90 kn  CCCC =  75 MPa A 100 mm = x cos = ( 75 MPa) ( cos 5 o ) = MPa =  x sin cos = 8.7 MPa to determine the complete stress state on face on face ab, = 5 o, the stresses are calculated ad, = 115 o, the stresses are = x cos =  75 cos 115 o = MPa shear stress is the same as on face ab, the complete stress state is shown in figure Example 11 a plastic bar with square cross section of side b is connected by a glued joint along plane pq P = 8000 lb = 40 o 1
22 all = 1100 psi all = 600 psi ( glude ) all = 750 psi ( glude ) all = 500 psi determine minimum width b A = P / x = 40 o =  =  50 o x = CCC x =  CCCCC cos sin cos (a) based on the allowable stresses in the glued joint = psi =  50 o ==> x = psi = psi =  50 o ==> x = psi (b) based on the allowable stresses in the plastic x = psi max = 600 psi occurs on the plane at 45 o = x / => x = psi (c) minimum width of the bar, choose x = psi, then A = 8000 lb CCCC = 7.88 in 1015 psi b min = A = 7.88 in =.81 in, select b = 3 in.7 Strain Energy the concept of strain energy principles are widely used for determining the response of machines and structures to both
23 static and dynamic loads consider a prismatic bar of length L subjected to tension force P, which is gradually increases from zero to maximum value P, the loaddeflection diagram is plotted after P 1 is applied, the corresponding elongation is, additional force dp 1 produce d 1, the work done by P 1 is d W = P 1 d 1 and the total work done is W = P 1 d 1 0 the work by the load is equal the area under the loaddeflection curve strain energy : energy absorbed by the bar during the load process thus the strain energy U is U U = W = P 1 d 1 0 referred to as internal work the unit of U and W is J (J = N.m) [SI], ftlb [USCS] during unloading, some or all the strain energy of the bar may be recovered 3
24 if P is maintained below the linear elastic range U P = CC = W = P L CC U = P L CCC = E A CCC E A E A L also k = E A CC thus U = P CC = k CC L k the total energy of a bar consisting of several segments is n n N i L i U = U i = CCC i=1 i=1 E i A i the strain energy for a nonprismatic bar or a bar with varying axial force can be written as L P x dx U = CCC 0 E A x displacements caused by a single load U = W P = CC ==> = U CC P U = U AB + U BC strain energy density u is the total strain energy U per unit volume for linear elastic behavior U U P L 1 x E x u = C = CC = CCC CC = CC = CC = CC V A L E A A L E 4
25 modulus of resilience u r pl u r = CC E pl : proportional limit resilience represents the ability of the material to absorb and release energy within the elastic range modulus of toughness u t is the area under the stressstrain curve when fracture, u t represents the maximum energy density can be absorbed by the material strain energy (density) is always a positive quantity Example 1 3 round bars having same L but different shapes as shown when subjected to the same load calculate the energy stored in each bar U 1 = P L CCC E A n N i L i P (L/5) P (4L/5) P L U 1 U = CCC = CCCC + CCCC = CCC = CC i=1 E i A i E A E (4A) 5 E A 5 P n N i L i P (L/15) P (14L/15) 3 P L 3 U 1 U 3 = CCC = CCCC + CCCCC = CCC = CC i=1 E i A i E A E (4A) 0 E A 10 the third bar has the least energyabsorbing capacity, it takes only a small amount of work to bring the tensile stress to a high value when the loads are dynamic, the ability to absorb energy is important, the 5
26 presence of grooves is very damaging Example 13 determine the strain energy of a prismatic bar subjected to (a) its own weight (b) own weight plus a load P (a) consider an element N(x) = A (L  x) : weight density dx L [N(x)] dx L[ A(L  x)] dx A L 3 U = CCCC = CCCCCC = CCC 0 0 E A(x) E A 6 E it can be obtained from the energy density = N(x) CC = (L  x) A u = CC = (L  x) CCCC E E L L [ A(L  x)] dx A L 3 U = u dv = u (Adx) = CCCCCC = CCC 0 0 E A 6 E same result as above (b) own weight plus P N(x) = A (L  x) + P L [ A(L  x) + P] dx A L 3 P L P L U = CCCCCCCC = CCC + CCC + CCC 0 E A 6 E E E A note that the strain energy of a bar subjected to two loads is not equal to the sum of the strain energies produced by the individual loads 6
27 Example 14 determine the vertical displacement of the joint B, both bar have the same axial rigidity E A equation of equilibrium in vertical direction, it is obtained F = P CCCC cos the strain energy of the two bars is F L 1 P H U =.CCC = CCCCC L 1 = H / cos E A 4 E A cos 3 the work of force P is W = P B / B equating U and W and solving for B B = P H CCCCC E A cos 3 this is the energy method to find the displacement, we did not need to draw a displacement diagram at joint B Example 15 a cylinder and cylinder head are clamped by bolts as shown d = 0.5 in d r = in g = 1.5 in t = 0.5 in L = 13.5 in compare the energy absorbing of the three bolt configurations 7
28 (a) original bolt n N i L i P (g  t) P t U 1 = CCC = CCCC + CCC i=1 E i A i E A s E A r A s = d CC A r = d r CC 4 4 thus U 1 can be written as P (g  t) P t U 1 = CCCCC + CCC E d E d r (b) bolt with reduced shank diameter P g P g U = CCC = CCC E A r E d r the ratio of strain energy U / U 1 is U g d C = CCCCCC = CCCCCCCCCCCCC = 1.40 U 1 (g  t) d r + t d ( ) x 0.5 (c) long bolts P (L  t) P t U 3 = CCCCC + CCC E d E d r the ratio of strain energy U 3 / U 1 is U 3 CC = (L  t) d r + t d CCCCCCCC U 1 [(g  t) d r + t d ] = ( ) CCCCCCCCCCCCCCC = 4.18 [( ) ] thus, the long bolts increase the energyabsorbing capacity 8
29 when designing bolts, designers must also consider the maximum tensile stresses, maximum bearing stresses, stress concentration, and other matters.8 Impact Loading.9 Repeated Loading and Fatigue.10 Stress Concentrations.11 Nonlinear Behavior.1 Elastoplastic Analysis 9
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