3 axial loading 2015

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CHAPTER 3: AXIAL FORCES 3.1 Concept, Internal Forces 3.2 Normal Strain and Normal Stress 3.3 Stress – Strain diagram 3.4 Hooke’s law: Modulus of Elasticity 3.5 Deformation under axial loading 3.6 Safety Factor, Allowable Stress 3.7 Statically Indeterminate Systems 3.1 CONCEPT, INTERNAL FORCES • Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses induced under loading Statics analyses alone are not sufficient • Considering structures as deformable allows determination of member forces and reactions which are statically indeterminate • Determination of the stress distribution within a member also requires consideration of deformations in the member • Chapter is concerned with deformation of a structural member under axial loading Later chapters will deal with torsional and pure bending loads 3.1 CONCEPT, INTERNAL FORCES A structural member under axial loading means: Internal forces: only axial force N (along longitudinal axis of the bar) Method of Section: the same as discussed in previous chapter Imaginary cut is made through the body in the region where the internal loading is to be determined The two parts are separated and a free body diagram of one of the parts is drawn Only then applying equilibrium would enable us to relate the resultant internal force and moment to the external forces 3.1 CONCEPT, INTERNAL FORCES WHAT DOES IT LOOK LIKE? 3.1 CONCEPT, INTERNAL FORCES WHAT DOES IT LOOK LIKE? 3.1 CONCEPT, INTERNAL FORCES HOW CAN WE KNOW THAT IT IS A PURELY AXIAL LOADING PROBLEM? (1) Structure member is subjected to axial (longitudinal direction) forces only For example: OR (2) Structure members are connected together by hinges (so-called nodes) and there is no load on the member, just have loads at nodes For example: 3.1 CONCEPT, INTERNAL FORCES EXAMPLES: DERTEMINE INTERNAL FORCES OF STRUCTURES BELOWS? (1) (2) 3.2 NORMAL STRAIN AND NORMAL STRESS a force P produces a deformation  In engineering, we usually transform this force into stress and the deformation into strain and we define these as follows: P    stress A   L  normal strain     2P P  2A A  L P A 2    2L L  3.2 NORMAL STRAIN AND NORMAL STRESS Strain has no unit’s since it is a ratio of length to length Most engineering materials not stretch very mush before they become damages, so strain values are very small figures It is quite normal to change small numbers in to the exponent for 10-6( micro strain) Sign convention of normal stresses: x > : tensile stress x < : compression stress 3.2 NORMAL STRAIN AND NORMAL STRESS Normal stress: at a section far from the bar ends, the normal stress can be regarded uniform, or considered as average value Stress has units of force per unite area (kN/m2; N/m2…) Note: Most of engineering fields used kPa, MPa, GPa 3.5 DEFORMATION UNDER AXIAL LOAD Example 3.01 SOLUTION: • Divide the rod into components at the load application points E  29 10 6 psi D  1.07 in d  0.618 in Determine the deformation of the steel rod shown under the given loads • Apply a free-body analysis on each component to determine the internal force • Evaluate the total of the component deflections 3.5 DEFORMATION UNDER AXIAL LOAD SOLUTION: • Divide the rod into three components: • Apply free-body analysis to each component to determine internal forces, P1  60  103 lb P2  15  103 lb P3  30  103 lb • Evaluate total deflection, Pi Li  P1L1 P2 L2 P3 L3       E  A1 A2 A3  i Ai Ei          60 103 12  15 103 12 30 103 16      6 0.3 29 10   L1  L2  12 in L3  16 in A1  A2  0.9 in A3  0.3 in  75.9 103 in   75.9 103 in 3.5 DEFORMATION UNDER AXIAL LOAD Example 3.02 SOLUTION: The rigid bar BDE is supported by two links AB and CD • 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 Link AB is made of aluminum (E = 70 deflection at E given the deflections GPa) and has a cross-sectional area of 500 at B and D 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 3.5 DEFORMATION UNDER AXIAL LOAD SOLUTION: Free body: Bar BDE Displacement of B: B  PL AE   60 103 N 0.3 m   500 10-6 m2 70 109 Pa   514 10 m MB  0  30 kN  0.6 m   FCD  0.2 m FCD  90 kN tension  B  0.514 mm  Displacement of D: D  PL AE  30 kN  0.4 m   FAB  0.2 m  90 103 N 0.4 m   600 10-6 m2 200 109 Pa  FAB  60 kN compression  300 10 m  MD   D  0.300 mm  3.5 DEFORMATION UNDER AXIAL LOAD Displacement of D: BB BH  DD HD 0.514 mm 200 mm   x  0.300 mm x x  73.7 mm EE  HE  DD HD E 0.300 mm  400  73.7 mm 73.7 mm  E  1.928 mm  E  1.928 mm  3.5 DEFORMATION UNDER AXIAL LOAD Example 3.03 3.6 SAFETY FACTOR – ALLOWABLE STRESS ALLOWABLE LOAD / ALLOWABLE STRESS  Max load that a structural member/machine component will be allowed to carry under normal conditions of utilization is considerably smaller than the ultimate load  This smaller load = Allowable load / Working load / Design load  Only a fraction of ultimate load capacity of the member is utilised when allowable load is applied  The remaining portion of the load-carrying capacity of the member is kept in reserve to assure its safe performance  The ratio of the ultimate load/allowable load is used to define FACTOR OF SAFETY  FACTOR OF SAFETY = ULTIMATE LOAD/ALLOWABLE LOAD  FACTOR OF SAFETY = ULTIMATE STRESS/ALLOWABLE STRESS 3.6 SAFETY FACTOR – ALLOWABLE STRESS SELECTION OF F.S Variations that may occur in the properties of the member under considerations The number of loadings that may be expected during the life of the structure /machine Types of loadings that are planned for in the design, or that may occur in the future Types of failures that may occur Uncertainty due to the methods of analysis Deterioration that may occur in the future because of poor maintenance / because of unpreventable natural causes The importance of a given member to the integrity of the whole structure 3.6 SAFETY FACTOR – ALLOWABLE STRESS • Factor of Safety: n a ctua l strength Fa ctor of sa fety n  1 required strength • Allowable Stresses, Ductile Materials: Brittle Materials: allow ( or ) y a llow  n ult allow  n 3.6 SAFETY FACTOR – ALLOWABLE STRESS Saint-Venant’s Principle • Loads transmitted through rigid plates result in uniform distribution of stress and strain • Concentrated loads result in large stresses in the vicinity of the load application point • Stress and strain distributions become uniform at a relatively short distance from the load application points • Saint-Venant’s Principle: Stress distribution may be assumed independent of the mode of load application except in the immediate vicinity of load application points 3.7 STATICALLY INDETERMINACY • Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate • A structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium • Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations • Deformations due to actual loads and redundant reactions are determined separately and then added or superposed   L R  3.7 STATICALLY INDETERMINACY Example 3.04 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 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 • Solve for the displacement at B due to the redundant reaction 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 reaction at A due to applied loads and the reaction found at B 3.7 STATICALLY INDETERMINACY SOLUTION: • Solve for the displacement at B due to the applied loads with the redundant constraint released, P1  P2  P3  600 103 N P4  900 103 N A1  A2  400 10 m A3  A4  250 10 m L1  L2  L3  L4  0.150 m Pi Li 1.125 109 L    A E E i i i • Solve for the displacement at B due to the redundant constraint, P1  P2   RB A1  400 10 m L1  L2  0.300 m  A2  250 10 m  Pi Li 1.95 103 RB δR    A E E i i i 3.7 STATICALLY INDETERMINACY • Require that the displacements due to the loads and due to the redundant reaction be compatible,   L R    1.125 109 1.95 103 RB    0 E E RB  577 103 N  577 kN • Find the reaction at A due to the loads and the reaction at B  Fy   RA  300 kN  600 kN  577 kN RA  323 kN R A  323 kN RB  577 kN 3.7 STATICALLY INDETERMINACY • MORE EXAMPLES - PRACTICES ... 1 03 lb P3  30  1 03 lb • Evaluate total deflection, Pi Li  P1L1 P2 L2 P3 L3       E  A1 A2 A3  i Ai Ei          60 1 03 12  15 1 03 12 30 1 03 16      6 0 .3 29 10... used kPa, MPa, GPa 3. 2 NORMAL STRAIN AND NORMAL STRESS 3. 3 STRESS – STRAIN DIAGRAM 3. 3 STRESS – STRAIN DIAGRAM Ductile Materials 3. 3 STRESS – STRAIN DIAGRAM Brittle Materials 3. 3 STRESS – STRAIN... 6 0 .3 29 10   L1  L2  12 in L3  16 in A1  A2  0.9 in A3  0 .3 in  75.9 10 3 in   75.9 10 3 in 3. 5 DEFORMATION UNDER AXIAL LOAD Example 3. 02 SOLUTION: The rigid bar BDE is supported