RODS: HERML SRESS ND SRESS CONCENRION Example 5 rod of length L, cross-sectional area, and modulus of elasticity E, has been placed inside a tube of the same length L, but of cross-sectional area and modulus of elasticity E. What is the deformation of the rod and tube when a force P is exerted on a rigid end plate as shown? What are the internal forces in the rod and the tube?
Example 5 (cont d) ube (, E ) Rod (, E ) L End plate P Example 5 (cont d) ube (, E ) P Rod (, E ) P FBD: P F / F R F / F F + Fx 0; P FR 0 F + F P R (9)
Example 5 (cont d) Clearly on equation is not sufficient to determine the two unknown internal forces FR and F. he problem is statically indeterminate. However, the geometry of the problem shows that the deformations δr and δ of the rod and tube must be equal, that is Example 5 (cont d) δ FR L E F R FR E δ F L E F E FR E E ube (, E ) Rod (, E ) P End plate L (0)
Example 5 (cont d) Substituting Eq. 0 into Eq. 9, therefore F F R R + F P FR E + E E FR + E P E + E P FR E ube (, E ) Rod (, E ) P End plate L P Example 5 (cont d) Substituting Eq. 0 into Eq. 9, therefore Or Rod Rod ( (, E, E) ) P E FR ns. E + E L L and from Eq.9, P E P E F P FR P E + E E + E ube ( (, E, E) ) P P End End plate ns.
Example 6 very stiff bar of negligible weight is suspended horizontally by two vertical rods as shown. One of the rods is of steel, and is ½-in in diameter and 4 ft long; the other is of brass and is 7/8-in in diameter and 8 ft long. If a vertical load of 6000 lb is applied to the bar, where must be placed in order that the bar will remain horizontal? ake E s 30 0 6 psi and E b 4 0 6. Example 6 (cont d) lso find the stresses in the brass and steel rods. 8 ft Brass 6000 lb Steel x 4 ft 0 ft
Example 6 (cont d) FBD F b F s 8 ft 96 in. F b 6000 lb x F s 4 ft 48 in. 0 ft Example 6 (cont d) wo independent equations of static equilibrium may be written for the free-body diagram. he possible equations are + + F b 6000 lb 0 ft x F s F 0; F + F 6000 0 () y s b M 0; F ( 0) 6000( x) 0 () b
Example 6 (cont d) Since no more independent equations of equilibrium can be written and there are three unknown quantities, the structure is statically indeterminate. One additional independent equation is needed. he problem requires that the bar remain horizontal. herefore, the rods must undergo equal elongations, that is δ (3) s δ b Example 6 (cont d) σl δ b δ s, but in generalδ E σ bl σ sl Eb Es δ b σ b(96) σ s (48) 6 6 4 0 30 0 6.85743 0 σ b.6 0 σ s 0 Fb Fs But σ b and σ s, therefore b s δ s δ b (4) (5) δ b
Example 6 (cont d) Substituting Eq.5 into Eq.4, thus Fb Fs 6.85743 0.6 0 0 b s he areas of brass and steel bar are ( 7 / 8) π b 4 π s 4 0.603 in ( / ) 0.9635 in (6) (7) (8) Example 6 (cont d) Substituting Eqs. 7 and 8 into Eq. 6, gives.40348 0 F 8.4874 0 0 (9) b F s Recalling Eqs. and, F b 0F + Fs 6000 0 b 6000x 0
Example 6 (cont d) he solution of the following system of simultaneous equations, gives F b, F s, and x: Fb + Fs 6000 0F 6000x 0 b (0).40348 0 F b 8.4874 0 F s 0 Example 6 (cont d) From the system of Equation 0: F b 500 lb F s 3500 lb x x 4.67 ft Hence, if the bar is to remain horizontal, the 6000-lb load should placed 4.67 ft from the steel rod as shown.
Example 6 (cont d) he stresses in brass and steel rods can be calculated from Eq.5 as follows: Fb σ b b Fs σ s x s 500 0.603 3500 0.9635 4,57.5 psi 4.6 ksi 7,85.3 psi 7.83 ksi hermal Stress emperature Strain Most materials when unstrained expand when heated and contract when cooled. he thermal strain due to one degree (0) change in temperature is given by α and is known is the coefficient of thermal expansion. he thermal strain due a temperature change of degrees is given by ε () α
hermal Stress otal Strain he sum of the normal strain caused by the loads and the thermal strain is called the total strain, and it is given by σ ε total εσ + ε + α () E hermal Stress Definition hermal stress is the stress that is induced in a structural member due a temperature change while the member is restrained (free movement restricted or prevented)
hermal Stresses ENES 0 ssakkaf temperature change results in a change in length or thermal strain. here is no stress associated with the thermal strain unless the elongation is restrained by the supports. reat the additional support as redundant and apply the principle of superposition. δ α( ) L α thermal expansion coef. δ δ + δ P 0 PL E α( ) L + 0 δ P PL E he thermal deformation and the deformation from the redundant support must be compatible. δ δ + δ P 0 P Eα( ) P σ E α( ) hermal Stress Example he bar B is securely fastened to rigid supports at the ends and is subjected to a temperature change. Since the ends of the bar are fixed, the total deformation of the bar must be zero δ total δ + δσ ε L + εσ L σ α L + L E (3)
hermal Stress Example (cont d) L B B δ P hermal Stress Example (cont d) If the temperature of the bar increases ( positive), then the induced stress must be negative, and the wall must push on the ends of the rod. If the temperature of the bar decreases ( negative), then the induced stress must be positive, and the wall must pull on the ends of the rod.
hermal Stress Example (cont d) his means that if end B were not attached to the wall and the temperature drops, then end B would move to B, a distance δ ε α L (4) L as shown in the figure hermal Stress Example (cont d) herefore, for total deformation of the bar to be zero, the wall at B must apply a force P σ of sufficient magnitude to move end B through a distance δ P ε σ L (σ/e) L so that the length of the bar is a gain L, which is the distance between the walls.
hermal Stress Example (cont d) L B B δ P hermal Stress Since the walls do not move, then Or δ δ δ P P δ δ P + δ and hence, the total deformation of the bar is zero. 0
Example 7 0-m section of steel (E 00 Gpa and α.9 0-6 / 0 C) rail has a crosssectional area of 7500 mm. Both ends of the rail are tight against adjacent rails that can be assumed to be rigid. he rail is supported against lateral moment. For an increase in temperature of 50 0 C, determine a) he normal stress in the rail. b) he internal force on the cross section. Example 7 (cont d) (a) he change in length can be calculated as follows: δ ε L αl.9 0 (0)(50) 5.95 mm he stress required to resist a change in length of 5.95 mm is computed as 9 Eδ 00 0 (5.95 0 σ L 0 3 ) 9 MPa
Example 7 (cont d) (b) he internal force on a cross section of the rail is computed as follows: 6 F σ 9 0 (7500 0 893 kn ) 89.5 0 3 N General Notes on hermal Stress and hermal Deformation he results that was discussed earlier apply only in the case of a homogenous rod (or bar) of uniform cross section. ny other problem involving a restrained member undergoing a change in temperature must be analyzed on its own merits.
General Notes on hermal Stress and hermal Deformation However, the same general approach may be used, that is, we may consider separately the deformation due to temperature change and the deformation due to the redundant reaction and superimpose the solutions obtained. Example 8 Determine the values of the stress in portion C and CB of the steel bar shown when a the temperature of the bar is 50 0 F, knowing that a close fit exists at both of the rigid supports when the temperature is +70 0 F. Use E 9 0 6 psi and α 6.5 0-6 / 0 F for steel.
Example 8 (cont d) C B rea 0.6 in rea. in in in Example 8 (cont d) he reaction at the supports needs to be determined. Sine the problem is statically indeterminate, the bar can be detached from its support at B, and let it undergo the temperature change 0 ( 50) (75) 5 F
Example 8 (cont d) C B C B δ δ B L L C B R B Example 8 (cont d) he total deformation that correspond to this temperature change becomes δ α L 6.5 0 ( 5)(4) 0.095 in he input data is L L in 0.6 in F F R B. in E 9 0 6 psi
Example 8 (cont d) C B F L F L δ B + E E RB.0345 0 R 6 B 9 0 + 0.6. Expressing that the total deformation of the bar must be zero, hence δ δ + δ 0; 0.095 +.0345 0 R from which R B 8.85 0 3 R B 0 lb 8.85 kips Example 8 (cont d) herefore, F F 8.85 kips, and C B σ (stress in C) σ (stress in CB) F F 8.85 3.4 ksi 0.6 8.85 5.7 ksi.
Stress Concentrations he stresses near the points of application of concentrated loads can values much larger than the average value of the stress in a member. When a structural member contains a discontinuity, such as a hole or a sudden change in cross section, high localized stresses can also occur near the discontinuity (Figs and ). Stress Concentrations Fig.. Stress distribution near circular hole in flat bar under axial loading
Stress Concentrations Fig.. Stress distribution near fillets in flat bar under axial loading Stress Concentrations Hole Fig. 3 Discontinuities of cross section may result in high localized or concentrated stresses. σ K σ max ave
Stress Concentrations Fillet Fig. 4 Stress Concentrations o determine the maximum stress occurring near discontinuity in a given member subjected to a given axial load P, it is only required that the average stress σ ave P/ be computed in the critical section, and the result be multiplied by the appropriate value of the stress-concentration factor K. It is to be noted that this procedure is valid as long as σ max σ y
Stress Concentrations Example 9 Determine the largest axial load P that can be safely supported by a flat steel bar consisting of two portions, both 0 mm thick, and respectively 40 and 60 mm wide, connected by fillets of radius r 8 mm. ssume an allowable normal stress of 65 MPa. SOLUION: Determine the geometric ratios and find the stress concentration factor from Fig. 4. Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor. pply the definition of normal stress to find the allowable load. Stress Concentrations Example 9 D d 60mm.50 40mm K.8 r d 8mm 40mm 0.0 Determine the geometric ratios and find the stress concentration factor from Fig. 4. Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor. σ 65MPa σ max ave 90.7 MPa K.8 pply the definition of normal stress to find the allowable load. P σ ave ( 40mm)( 0mm)( 90.7 MPa) 3 36.3 0 N P 36.3kN