MECHANICS OF MATERIALS Third Edition Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University CHAPTER 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 Stress and Strain Axial Loading PENDIDIKAN TEKNIK MESIN JPTK FKIP UNS TEGANGAN DAN REGANGAN KERJA MESIN Tegangan Setiap material adalah elastis pada keadaan alaminya. Karena itu jika gaya luar bekerja pada benda, maka benda tersebut akan mengalami deformasi. Ketika benda tersebut mengalami deformasi, molekulnya akan membentuk tahanan terhadap deformasi. Tahanan ini per satuan luas dikenal dengan istilah tegangan. Secara matematik tegangan didenisikan sebagai gaya per satuan luas, atau: stress = =APoRegangan Deformasi per satuan panjang disebut dengan regangan. Secara matematis ditulis: strain normal = =LL oc 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 9 Normal Strain strain normalstress= == =LAPocoLAPAPoco== =22L LAPo oco= ==22 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 10 Stress-Strain Test 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 11 Stress-Strain Diagram: Ductile Materials 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 12 Stress-Strain Diagram: Brittle Materials 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf Contoh Soal Sebuah motor satu silinder dengan ukuran garis tengah silinder 70 mm. Tekanan pembakaran bahan bakar maksimum yang dihasilkan dalam ruang pembakaran sebesar 48 kg/cm2. Kepala silinder yang sekaligus sebagai penutup silinder dipasang pada blok silinder menggunakan 4 buah baut. Hitunglah: a). Gaya maksimum yang bekerja mendorong piston. b). Gaya yang bekerja pada setiap baut. c). Besarnya diameter baut yang harus digunakan, jika baut terbuat dari bahan St. 60 dengan angka keamanan 2. 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 14 Hookes Law: Modulus of Elasticity Below the yield stress Elasticity of Modulus or Modulus Youngs ==E Ec o Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of Elasticity) is not. 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 15 Elastic vs. Plastic Behavior If the strain disappears when the stress is removed, the material is said to behave elastically. When the strain does not return to zero after the stress is removed, the material is said to behave plastically. The largest stress for which this occurs is called the elastic limit. 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 16 Fatigue Fatigue properties are shown on S-N diagrams. When the stress is reduced below the endurance limit, fatigue failures do not occur for any number of cycles. A member may fail due to fatigue at stress levels significantly below the ultimate strength if subjected to many loading cycles. 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 17 Deformations Under Axial Loading AEPEE = = = oc c o From Hookes Law: From the definition of strain: Loc = Equating and solving for the deformation, AEPL= o With variations in loading, cross-section or material properties, =i i i i iE A L Po 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 18 Example 2.01 Determine the deformation of the steel rod shown under the given loads. in. 618 . 0 in. 07 . 1psi 10 296= = = d DESOLUTION: Divide the rod into components at the load application points. Apply a free-body analysis on each component to determine the internal force Evaluate the total of the component deflections. 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 19 SOLUTION: Divide the rod into three components: 22 12 1in 9 . 0in. 12= == =A AL L233in 3 . 0in. 16==AL Apply free-body analysis to each component to determine internal forces, lb 10 30lb 10 15lb 10 60333231 = = =PPP Evaluate total deflection, ( ) ( ) ( )in. 10 9 . 753 . 016 10 309 . 012 10 159 . 012 10 6010 291133 3 3633 322 211 1 =(((

+ +=||.|

\|+ + = =AL PAL PAL PE E A L Pi i i i io in. 10 9 . 753 = o 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 20 Sample Problem 2.1 The rigid bar BDE is supported by two links AB and CD. Link AB is made of aluminum (E = 70 GPa) and has a cross-sectional area of 500 mm2. Link CD is made of steel (E = 200 GPa) and has a cross-sectional area of (600 mm2). For the 30-kN force shown, determine the deflection a) of B, b) of D, and c) of E. SOLUTION: Apply a free-body analysis to the bar BDE to find the forces exerted by links AB and DC. Evaluate the deformation of links AB and DC or the displacements of B and D. Work out the geometry to find the deflection at E given the deflections at B and D. 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 21 Displacement of B: ( )( )( )( )m 10 514Pa 10 70 m 10 500m 3 . 0 N 10 6069 2 6 -3 = ==AEPLBo| = mm 514 . 0BoDisplacement of D: ( )( )( )( )m 10 300Pa 10 200 m 10 600m 4 . 0 N 10 9069 2 6 -3 = ==AEPLDo+ = mm 300 . 0DoFree body: Bar BDE ( )( )n compressio FFtension FFMABABCDCDB kN 60m 2 . 0 m 4 . 0 kN 30 00 M kN 90m 2 . 0 m 6 . 0 kN 30 00D = = = + = + = =SOLUTION: Sample Problem 2.1 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 22 Displacement of D: ( )mm 7 . 73mm 200mm 0.300mm 514 . 0===''x x xHDBHD DB B+ = mm 928 . 1Eo( )mm 928 . 1mm 7 . 73mm 7 . 73 400mm 300 . 0=+==''EEHDHED DE EooSample Problem 2.1 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 23 Static Indeterminacy Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate. 0 = + = R L o o o Deformations due to actual loads and redundant reactions are determined separately and then added or superposed. Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations. A structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium. 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 24 Example 2.04 Determine the reactions at A and B for the steel bar and loading shown, assuming a close fit at both supports before the loads are applied. Solve for the reaction at A due to applied loads and the reaction found at B. Require that the displacements due to the loads and due to the redundant reaction be compatible, i.e., require that their sum be zero. Solve for the displacement at B due to the redundant reaction at B. SOLUTION: Consider the reaction at B as redundant, release the bar from that support, and solve for the displacement at B due to the applied loads. 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 25 SOLUTION: Solve for the displacement at B due to the applied loads with the redundant constraint released, E E A L PL L L LA A A AP P P Pi i i i i9L4 3 2 12 64 32 62 13433 2 110 125 . 1m 150 . 0m 10 250 m 10 400N 10 900 N 10 600 0=== = = = = = = = = = = = o Solve for the displacement at B due to the redundant constraint, ( ) = == = = = = = i Bi i i iRBE RE A L PL LA AR P P32 12 622 612 110 95 . 1m 300 . 0m 10 250 m 10 400Example 2.04 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 26 Require that the displacements due to the loads and due to the redundant reaction be compatible, ( )kN 577 N 10 577010 95 . 1 10 125 . 1033 9= ==== + =BBR LRE REoo o o Find the reaction at A due to the loads and the reaction at B kN 323kN 577 kN 600 kN 300 0= + = =AA yRR FkN 577kN 323==BARRExample 2.04 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 27 Thermal Stresses A temperature change results in a change in length or thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by the supports. ( )coef. expansion thermal = = A =o o o oAEPLL T P T Treat the additional support as redundant and apply the principle of superposition. ( ) 00= + A= + =AEPLL TP Too o o The thermal deformation and the deformation from the redundant support must be compatible. ( )( ) T EAP T AE P P TA = = A = = + =o o oo o o 0 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 28 Poissons Ratio For a slender bar subjected to axial loading: 0 = = = z yxxE o ooc The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic (no directional dependence), 0 = = z y c c Poissons ratio is defined as xzxyccccv = = =strain axialstrain lateral 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf 2 - 29 Generalized Hookes Law For an element subjected to multi-axial loading, the normal strain components resulting from the stress components may be determined from the principle of superposition. This requires: 1) strain is linearly related to stress 2) deformations are small E E EE E EE E Ezyxzzyxyzyxxovovocvoovocvovooc+ = + = + = With these restrictions: 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer Johnston DeWolf Dilatation: Bulk Modulus Relative to the unstressed state, the change in volume is ( )( )( ) | | | |( ) e) unit volum per in volume (change dilatation2 11 1 1 1 1 1=+ +=+ + = + + + = + + + =z y xz y xz y x z y xEeo o ov c c c c c c c c c For element subjected to uniform hydrostatic pressure, ( )( )modulus bulk 2 1 32 1 3== = =vvEkkpEp e Subjected to uniform pressure, dilatation must be negative, therefore 210 <