东南大学：《建筑力学 Architectural Mechanics》课程教学课件（英文讲稿）A21 Dynamic Loading

Dynamic Loadingmi@se.er.cn

Dynamic Loading mi@seu.edu.cn

Contents·Static vs.Dynamic Loading（静载与动载）·Structural Members under Constant Acceleration （匀加速构件)·Structural Members under Constant Rotation （匀速转动构件)·Horizontal Impact onAxial Members（水平放置杆件的水平冲击)·Horizontal Impact onBeams（竖直放置梁的水平冲击）·VerticalImpactonBeams（水平放置梁的竖直冲击）·Vertical Impact with an Initial Velocity（带有初速度的竖直冲击）2

• Static vs. Dynamic Loading（静载与动载） • Structural Members under Constant Acceleration（匀加 速构件） • Structural Members under Constant Rotation（匀速转 动构件） • Horizontal Impact on Axial Members（水平放置杆件 的水平冲击） • Horizontal Impact on Beams（竖直放置梁的水平冲击） • Vertical Impact on Beams（水平放置梁的竖直冲击） • Vertical Impact with an Initial Velocity（带有初速度的 竖直冲击） 2 Contents

Static vs. Dynamic Loading. Static Loading: external loadsare gradually increased to adefinite value and subsequentlyare held in constant: Dynamic Loading: the magnitude and/ordirection of external loads vary as a function oftime. This type of loads generate significantacceleration within structural members thatcannot be neglected2.65 in万式1.10B16in.12ft3

• Static Loading: external loads are gradually increased to a definite value and subsequently are held in constant. Static vs. Dynamic Loading • Dynamic Loading: the magnitude and/or direction of external loads vary as a function of time. This type of loads generate significant acceleration within structural members that cannot be neglected. 3

Structural Members under Constant Acceleration. Given: P, A and a: Find: the stress developed in the steel ropeNR: Solution by the method of statnamic:P1. Inertia force: F =一agP2. Axial force in rope: N = P+Pa/ggg3. Normal stress developed in rope:Napaa1+(1+6AAggN△Od4.Dynamic load factor:N△a8Sstst2

P a Structural Members under Constant Acceleration • Given: P, A and a. • Find: the stress developed in the steel rope. • Solution by the method of statnamic: 1. Inertia force: 2. Axial force in rope: 3. Normal stress developed in rope: 4. Dynamic load factor: 1 d d d d st st st N a k N g                P F a g  1 d P a N P a P g g           (1 ) 1 d d st N P a a A A g g              Nd P Pa g/ 4

Structural Members under Constant Rotation. Given: O, A, p and D: Find: the stress developed in the flange of the flying wheel. Solution by the method of statnamic:1. Inertia force acting on unit length:Dla = p.1. A.0?2D2. Circumferential force:PADOD?243. Circumferential stress:xFna = po'D?Hd4A5

Structural Members under Constant Rotation • Given: ω, A, ρ and D. • Find: the stress developed in the flange of the flying wheel. • Solution by the method of statnamic: x y  FN FN 1. Inertia force acting on unit length: 2. Circumferential force: 2 2 N 0 1 sin 2 2 4 d D AD F q d          4 2 2 D A FNd d     3. Circumferential stress: 2 1 2 d D q A       D ω 5

Horizontal Impact on Axial MembersArea = A. To determine the maximum stress Omax-Assumethatthekinetic energy is(a)Btransferred entirely to the structure,U=Imvm- Assume that the stress-strain(b)diagram obtained from a static testV=OBis also valid under impact loading:HorizontalImpact: Consider a: Energy conservation requires,rod which is hit at its end with a1D2body of mass m moving with amaxdxd2EvelocityVo.2EA. For the case of a uniform rod.: Rod deforms under impact. Stressesreach a maximum value max andp2LmvAE2UAEUthendisappear.LL2EA6

Horizontal Impact on Axial Members • Horizontal Impact: Consider a rod which is hit at its end with a body of mass m moving with a velocity v0. • Rod deforms under impact. Stresses reach a maximum value max and then disappear. • To determine the maximum stress max - Assume that the kinetic energy is transferred entirely to the structure, 1 2 U mv  2 0 - Assume that the stress-strain diagram obtained from a static test is also valid under impact loading. 2 2 max 2 2 P U dV dx E EA      • Energy conservation requires, • For the case of a uniform rod, 2 2 2 0 2 P L UAE mv AE U P EA L L     6

Sample ProblemSolution:Duetothechangeindiameter,thenormal stress distribution is nonuniformBFindthestaticloadPwhichproducesthe same strain energy as the impact.A.Evaluatethemaximum stressArea = 4Aresulting from the static load PHorizontalImpact:Body of massmwith velocity yo hits the end of thenonuniform rod BCD. Knowing thatthediameter of the portion BC is twicethe diameter of portion CD, determinethe maximum value of the normalstress in the rod

Sample Problem Horizontal Impact: Body of mass m with velocity v0 hits the end of the nonuniform rod BCD. Knowing that the diameter of the portion BC is twice the diameter of portion CD, determine the maximum value of the normal stress in the rod. Solution: • Due to the change in diameter, the normal stress distribution is nonuniform. • Find the static load P which produces the same strain energy as the impact. • Evaluate the maximum stress resulting from the static load P 7

FindthestaticloadPwhichproducesthe same strain energy as the impact.p2 (L/2)p2 (L/2)5 P?LULB2EA2E(4A)16 EA16UEA8mv.EAAV55LLArea= 4A: Evaluate the maximum stress resultingSolution:from the static load P:Due to the change in diameter,Pthenormal stress distributionisa.maxAnonuniform.16UEU=Imv?5.ALP?La8mv.Edy2E2EAAL58

Solution: • Due to the change in diameter, the normal stress distribution is nonuniform. 1 2 2 0 2 2 2 2 U mv P L dV E EA      • Find the static load P which produces the same strain energy as the impact.       2 2 2 2 0 2 2 5 2 2 4 16 16 8 5 5 P L P L P L U EA E A EA UEA mv EA P L L      • Evaluate the maximum stress resulting from the static load P max 2 0 16 5 8 5 P A UE AL mv E AL     8

Horizontal Impact on BeamsBSolution:Kinetic energy of the block.Um = mve /2·Find the static load P which producesthe same strain energy as the impact.OFor loaded beam as shown,m1 ?L31-ID2 48EI296EIU48mv,EIPLLALHorizontalImpact:A block of mass m.Maximumstressmoving with a velocity V. hits squarelyPLc3mv, EIM..Cmaxthe prismaticmember AB at its midpointomax141V(1/c)"LC. Determine (a) the equivalent static.Maximum deflectionload P, (b) the maximum stress Omax inmveLPL3the member, and (c)the maximumAdeflection △max at C.48EI48EI9

Horizontal Impact on Beams Horizontal Impact: A block of mass m moving with a velocity v0 hits squarely the prismatic member AB at its midpoint C. Determine (a) the equivalent static load P, (b) the maximum stress σmax in the member, and (c) the maximum deflection Δmax at C. 9 Solution: • Kinetic energy of the block. 2 0 2 U mv m  • Find the static load P which produces the same strain energy as the impact. For loaded beam as shown, 2 3 2 0 3 3 1 1 2 2 48 96 48 P L U P EI EIU mv EI P L L             • Maximum stress   2 max 0 max 2 3 4 M c mv EI PLc I I I c L     • Maximum deflection 3 2 3 0 max 48 48 PL mv L EI EI   

Vertical Impact on BeamsSolution:Work done by the impact loadhU=W(h+a)BFind the static load P which producesAthe same strain energy as the impact.Treat the beam as a spring:p2VerticalImpact:Ablock ofweightk32k22W is dropped from a height h ontothefreeend of thecantileverbeam: Energy conservation requires,Determinethemaximumvalueofthe1k =W(h+△a)stresses in the beam.2+A)=△,(h+Ad)2K3-2a-2△h=010

Vertical Impact on Beams Vertical Impact: A block of weight W is dropped from a height h onto the free end of the cantilever beam. Determine the maximum value of the stresses in the beam. 10 Solution: • Work done by the impact load. U W h     d  • Find the static load P which produces the same strain energy as the impact. • Treat the beam as a spring: 2 1 1 2 2 2 2 d d P U P k k      • Energy conservation requires,       2 2 2 1 2 1 2 2 2 0 d d d d st d d st d st k W h W h h k h                     