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Bài giảng SBVL1 Axial Load được rút gọn với những công thức và ví dụ chính có thể xem để hiểu và ôn lưu ý đây là bài giảng bằng tiếng anh cho các khóa CLC tuy nhiên ai đọc cũng có thể hiểu mong mọi người thành công
Axial Load Stress-Strain Normal stress Normal strain Hooke’s Law The two aluminum rods AB and AC have diameters of 10 mm and mm, respectively Determine the largest vertical force P that can be supported The allowable tensile stress for the aluminum is σallow = 150 MPa Elastic Deformation How to determine δ = displacement of one point on the bar relative to the other point N ( x) Hooke’s Law A ( x) σ =E ( x) ϵ dδδ ϵ= dδx L σ= For the entire length L of the bar, δ =∫ P = constant, AE = constant General loads, AE = constant on Li n Li SN δ =∑ i=1 ( AE )i Li S N is area of N on Li Sign Convention dδδ= N ( x) dδx E ( x ) A( x ) N ( x) dδx E ( x ) A( x ) Segments AB and CD of the assembly are solid circular rods, and segment BC is a tube If the assembly is made of 6061-T6 aluminum (E=68.9 GPa), determine the displacement of end D with respect to end A External Work and Strain Energy Work of a Force Axial Load The work done by dFz Strain Energy If no energy is lost: the external work done by the loads → converted converted into internal work called strain energy which is caused by the action of either normal or shear stress Conservation of Energy Strain energy of normal stress The strain energy If A=100 mm2, E = 200 Gpa, determine the horizontal displacement at point B N 1= P / √ ∂ N / ∂ P=1/ √ L 1=1 m N L Pδ=∑ 2 AE A1 = A Castigliano’s Theorem n δ j =∑ N i ( i =1 ∂ N i Li ) ∂ P ( AE)i N 3= P N =−2 P / √ ∂ N /∂ P=−2 / √ ∂ N /∂ P=1 L =2 m L 3= √ m A2 = A A3 = A E =E E 2= E E 1= E Method 2: Castigliano’s Theorem Method 1: Conservation of Energy P 2P δ= ( × ×1 + × ×2 + P×1×√ 3) 1 2 AE 3 √ √ √ √3 Pδ= (L N + L N + L N ) 2 AE 1 2 3 (3+ √ 3) P (3+ √ 3) P δ=4.732 mm→ δ= δ=4.732 mm→ δ= AE AE Statically indeterminate axially loaded member Principle of superposition Compatibility conditions Force Method for statically indeterminate problem 11 X1 p 0 Ni fi 0, P Ni fi X , P Ni1 fi 1, N1 N i21 11 Li L4 EA EA i 1 i i=1 X1= N4 i=2 N2 N3 i=3 p i 1 X N i N i1 L EA i i p 11 Ni fi X1 , P The equations of equilibrium are not sufficient to determine all the reactions on a member → converted statically indeterminate problem Compatibility conditions Determine the force developed in each bar Bars AB and EF each have a crosssectional area of 50 mm2 , and bar CD has a cross-sectional area of 30 mm2 Review problems The assembly shown in Fig.a consists of an aluminum tube AB having a cross-sectional area of 400 mm2 A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube If a tensile load of 80 kN is applied to the rod, determine the displacement of the end C of the rod Take Est = 200 Gpa, Eal = 70 GPa Rigid beam AB rests on the two short posts shown in Fig a AC is made of steel and has a diameter of 20 mm, and BD is made of aluminum and has a diameter of 40 mm Determine the displacement of point F on AB if a vertical load of 90 kN is applied over this point Take Est = 200 GPa, Eal = 70 GPa The rigid bar is supported by the pinconnected rod CB that has a crosssectional area of 14 mm2 Determine the vertical deflection of the bar at D when the distributed load is applied Review problems Determine the support reactions at the rigid supports A and C The material has a modulus of elasticity of E Determine the vertical displacement at C ... External Work and Strain Energy Work of a Force Axial Load The work done by dFz Strain Energy If no energy is lost: the external work done by the loads → converted converted into internal work... AE 1 2 3 (3+ √ 3) P (3+ √ 3) P δ=4.732 mm→ δ= δ=4.732 mm→ δ= AE AE Statically indeterminate axially loaded member Principle of superposition Compatibility conditions Force Method for statically... ϵ dδδ ϵ= dδx L σ= For the entire length L of the bar, δ =∫ P = constant, AE = constant General loads, AE = constant on Li n Li SN δ =∑ i=1 ( AE )i Li S N is area of N on Li Sign Convention dδδ=