Ace004 Strength of Materials

INSTITUTE OF AERONAUTICAL ENGINEERING (AUTONOMOUS) Dundigal – 500 043, Hyderabad Regulation: R16 (AUTONOMOUS) Course code: ACE004 STRENGTH OF MATERIALS – II B.TECH IV SEMESTER Prepared By SURAJ BARAIK Assistant Professor 1 Unit I Deflection in Beams 2 Deflection in Beams • Topics Covered Review of shear force and bending moment diagram Bending stresses in beams Shear stresses in beams Deflection in beams 3 Beam Deflection Recall: THE ENGINEERING BEAM THEORY M E y I R Moment-Curvature Equation v (Deflection) y NA x A B A’ B’ If deformation is small (i.e. slope is “flat”): 4 1 d R dx R y and (slope is “flat”) x B’ y 1 d2y A’ R dx 2 Alternatively: from Newton’s Curvature Equation y d2y dy2 R if 1 1 dx 2 dx 3 y f (x) R 2 dy 2 2 1 1 d y dx x R dx 2 5 From the Engineering Beam Theory: M E 1 M d2y I R R EI dx2 d2y EI M dx 2 Flexural Bending Stiffness Moment Curvature 6 Relationship A C B Deflection = y dy Slope = dx d2y A C B Bending moment = EI 2 y dx d3y Shearing force = EI dx3 d4y Rate of loading = EI dx4 7 Methods to find slope and deflection Double integration method Moment area method Macaulay’s method 8 Double integration method d2y 1 M Curvature Since, dx 2 EI dy 1 M dx C Slope dx EI 1 1 y M dx dx C1 dx C 2 Deflection EI Where C1 and C2 are found using the boundary conditions. Curvature Slope Deflection y R dy dx 9 Double integration method Simple supported W Slope Deflection L/2 L/2 dy A C B Slope = Deflection = y dx c 3 yc WL WL2 48EI A B 16EI L Uniform distributed load Slope Deflection x w/Unit length A C dy B Slope = Deflection = y dx c 3 yc 5 WL WL2 384 EI A B 24EI L 10 Macaulay’s method The procedure of finding slope and deflection for simply supported beam with an eccentric load is very laborious. Macaulay’s method helps to simplify the calculations to find the deflection of beams subjected to point loads. 11 Moment-Area Theorems • Consider a beam subjected to arbitrary loading, d d2y M dx dx2 EI D xD M d dx EI C xC xD M dx D C EI xC dx CD Rd dx R d • First Moment-Area Theorem: area under BM diagram between C and D. 9 - 11 12 M o me n t - Area Theorems • Tangents to the elastic curve at P and P’ intercept a segment of length dt on the vertical through C. M dt xd x dx EI x x D M 1 D 1 t x dx xMdx Ax C D EI EI EI xC xC A= total area of BM diagram between C & D x = Distance of CG of BM diagram from C • Second Moment-Area Theorem: The tangential deviation of C with respect to D is equal to the first moment with respect to a vertical axis through C of the area under the BM diagram between C and D. 9 - 12 13 Moment Area Method 14 An Exercise- Moment of Inertia – Comparison 1 Load Maximum distance of 4 inch to the centroid I2 2 x 8 beam Load 2 I1 Maximum distance of 1 inch to 2 x 8 beam the centroid I2 > I1 , orientation 2 deflectsless 15 UNIT II DEFLECTION BY ENERGY METHODS 16 Elastic Deflection Castigliano’sMethod Complementary Energy U’ Incremental: Deflection: Castiglino’s Theorem: ∆ U Q When a body is elastically deflected by any combination of loads, the deflection at any point and in any direction is equal to the partial derivative of strain energy (computed with all loads acting) with respect to a load located at that point and acting in that direction 17 Elastic Deflection Castigliano’s M e t h o d Energy and Deflection Equations Example: Axial Tension Stored Elastic Energy: Case 1 from Table 5.1: gives: For varying E and A: 18 Elastic Deflection Castigliano’s Method Energy and Deflection Equations (1) Obtain expression for all components of energy Table 5.3 (2) Take partial derivative to obtain deflection Castiglino’s Theorem: ∆ U Q 19 Elastic Deflection: Castigliano’s Method first compute Energy, then Partial Derivative to get deflection Here 2 types of loading: Bending and Shear magnitude @ x: Table 5.3 1. Energy: here it has two components: (23=8)*3*4 = 96 2. Partial Derivatives for deflection: 20 Elastic Deflection: Castigliano’s Method Table 5.3 TWO METHODS Differentiate after Integral Differentiate under Integral 21 Elastic Deflection: Castigliano’s Method m m Transverse shear contributes only <5% to deflection 22 Elastic Deflection: Castigliano’s Method •90° bend cantilever beam •shear neglected Use of “Dummy Load” Q=0 •Shear neglected => only 4 energy components: 1) BENDING portion a_b: Mab=Py (Tension and Compression mostly 2) BENDING portion b_c: Mbc=Qx +Ph negligible if torsion and bending 3) TENSION portion a_b: Q are present) 4) COMPRESSION portion b_c: P : 23 Elastic Deflection: Castigliano’s Method 2 •Eccentrically Load Column 500kg x 9.8m/s •No Buckling =4900 N Redundant Support Guy wire •Now Deflection known (=0) •Find necessary Tension Force F •Hence partial derivative of total elastic energy with respect to F must be zero •Omit zero derivatives - all energy terms above a - compression term below a Nm •Only bending term is left: M= (4900(Nm N)(1.2m))2 -FyNm = 5880 - Fy (Nm)2 m Nm3 m3 Nm3 m3 finite value 24 Energy Method External Work When a force F undergoes a displacement dx in the same direction as the force, the work done is Ue P/2 dUe F dx Ifthe total displacement is x the work become x Ue F dx 0 25 The work of a moment is defined by the product of the magnitude of the moment M and the angle d then if the total angle of rotation is the work become: dUe M d e U M/2 Ue M d 0 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 Maxwell–Betti Reciprocal theorem Consider a simply supported beam of span L as shown. Let this beam be loaded by two systems of forces P1and P2 separately as shown in the figure. Let u21 be the deflection below the load point P2 when only load P1is acting. Similarly let u12 be the deflection below load P1 , when only load P2 is acting on the beam. 43 44 The reciprocal theorem states that the work done by forces acting through displacement of the second system is the same as the work done by the second system of forces acting through the displacements of the first system. Hence, according to reciprocal theorem, P1 u12 P2 u21 Now, u12 and u21 can be calculated using Castiglinao’s first theorem. Substituting the values of u12 and u21 in equation we get, 5P L3 5PL3 P 2 P 1 48EI 48EI 45 Hence it is proved. This is also valid even when the first system of forces is P1, P2 ,...., Pn and the second system of forces is given by Q1, Q2 ,...., Qn . Let u1 , u2 ,...., un be the displacements caused by the forces P1, P2 ,...., Pn only and 1, 2 ,...., n be the displacements due to system of forces Q1, Q2 ,...., Qn only acting on the beam as shown in Fig 46 UNIT III STRESSES IN CYLINDERS AND SPHERICAL SHELLS 47 THIN CYLINDER 48 Introduction Cylindrical and spherical vessels are used in the engineering field to store and transport fluids. Such vessels are tanks ,boilers , compressed air receivers , pipe lines etc. these vessels, when empty, are subjected to atmospheric pressure internally as well as externally and the resultant pressure on the walls of the shell is nil. but whenever a vessel is subjected to an intenal pressure (due to air , water , steam etc.) its walls are subjected to tensile stresses. 49 Thin cylindrical shell. When t/d <= d/10 to d/15, it is called thin cylindrical shell. t = thickness of the shell d =internal diameter of shell. in thin cylindrical shells hoops stress and longitudinal stresses are constant over the Thickness and radial stresses are negligible. When t/d > d/10 to d/15, it is called THICK cylindrical shell. 50 Stresses in thin cylindrical shells : whenever, a thin cylindrical shell is subjected to an internal pressure (p). Its Walls are subjected to two types of tensile stresses. (a) Hoop stress (circumferential stress) (b) Longitudinal stress. 51 Hoop stress (circumferential stress) : 흈풄 Consider a thin cylindrical shell subjected to an internal Pressure as shown in fig. 휎푐 = circumferential stress in the shell material. p =internal pressure d =internal diameter of shell t =thickness of the shell. Total pressure, p = Pressure * Area =p.d.l Resisting area = A = 2.t.l 휎푐 = P/A = p.d.l/2.t.l 흈풄 = p.d/2t. 52 (b) Longitudinal stress (휎푙 ): Total pressure p = Pressure * pArea = p휋 4 푑2 Resisting area, A = 휋푑푡 휋푑 2 휎푙 = P/A = 4 휋푑 p 푡 흈풍 = pd/4t 53 Change in dimensions of a thin cylindrical shell due to internal pressure: Let, 훿푑=change in dia. Of shell 휎푙=change in length of shell Circumferential strain, 훿푑 휎 푝푑 푝 휀 = 푐 − 1 푑 퐸 푙 푚퐸 − 2푡퐸푑 = 휎 = 4푡푚퐸 ퟏ 휺ퟏ = 풑풅/ퟐ풕푬(ퟏ − ) ퟐ풎 Longitudinal strain, 훿푙 휎 푝푑 푝 휀 = 푙 − 푐 = 2 퐸 푚퐸 4 2푑푡푚 휎 − = 푡퐸 퐸 푙 풑풅 ퟏ ퟏ 휺 = ퟐ ퟐ풕푬 ퟐ − 54 풎 Change in volume of a thin cylindrical shell due to internal pressure: 훿푣 푉 + 훿푣 = 휋 = [ 푑 + 훿푑2 ∗ (−푙 +푉훿푙)] − ∗ 휋 4 2 Volume of 4 푑 ∗ 푙 shell, 휋 = [ 휋 (푑2 + 2훿푑 ∗ 푑 + 훿푑 ) ∗ (푙 + 훿푙)]2 4 2 − 4 ∗ 푑 휋 2 ∗ 푙 휋 V = ∗ 푑 = 푑2푙 + 2훿푑 ∗ 푑 ∗ 푙 + 훿푑2 ∗ 푙 + 푑2 ∗ 훿푙 + 2훿푑 ∗ 푑 ∗ 훿푙 − 푑2 휋 4 4 ∗푙 + 훿푑2 ∗훿푙 ∗ 푙 = 푑2훿푙 + 2훿푑 ∗ 푑 휋 4 4 ∗ 푙 Final volume, 훿푣= 2 휋 휋 푑 훿푙 + 2훿푑 ∗ 푑 ∗ 푙 / ∗ 휋 2 푉 2 4 V+훿푣 = 푑 + 푙 + 훿푙 푑 2∗ 푙 4 = + 훿푑 ∗ 훿푙 4 훿푙푑 푑 =휀2 + 2휀1 훿푑 훿 (휀1= , 휀 = 푙 푑 Change in 휹풗 =V(흐ퟐ + ퟐ )2 volume, 휺ퟏ) 푙 55 Thin spherical shells Consider a thin spherical shell subjected to internal pressure p as shown in fig.

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