几何尺寸与公差论坛------致力于产品几何量公差标准GD&T (GDT:ASME)|New GPS(ISO)研究/CAD设计/CAM加工/CMM测量  


返回   几何尺寸与公差论坛------致力于产品几何量公差标准GD&T (GDT:ASME)|New GPS(ISO)研究/CAD设计/CAM加工/CMM测量 » 三维空间:产品设计或CAX软件使用 » CAD设计 » 产品功能分析
用户名
密码
注册 帮助 会员 日历 银行 搜索 今日新帖 标记论坛为已读


回复
 
主题工具 搜索本主题 显示模式
旧 2009-09-09, 04:15 PM   #1
huangyhg
超级版主
 
huangyhg的头像
 
注册日期: 04-03
帖子: 18592
精华: 36
现金: 249466 标准币
资产: 1080358888 标准币
huangyhg 向着好的方向发展
默认 help with composite beam

help with composite beam
a little background info. i'm a recent ei and civil eng. graduate. i have been given a problem to do at my work, that i haven't done for a couple of years now, and would like some help starting.
i have a composite beam with 2 different i values. the left end is fixed and the right end is pinned. the beam is 10ft long with 1ft being one i value, and the other 9ft as the other i value. there is a plf uniform load only on the 9ft section of the beam.
where do i begin? which method would be best to use? any help would be great.
for a 10 ft. long beam, the composite action isn't going to get you much...many times, short beams are left non-composite.
i'm not sure if you are designing a new beam or analyzing an existing beam, so here are both of my suggestions.
for design, why not use a beam with a consistant cross section and properties? i can't imagine there would be any significant cost savings by designing the beam differently for a length of one foot. also what do you mean by "composite", is it a slab on beam with studs or is it a steel beam encased in concrete? i would suggest looking at an alternate design if it is a steel beam encased in concrete. it has been my experience that this method is practically obsolete.
from an analysis standpoint, if you want to determine the deflection, you will need to consider the moment of inertia for each section of the beam respectively. you will not need the moment of inertia when you determine the moment and shear on the beam.
good luck.
it sounds like the slab edge stops prior to the end of the beam - since most of the composite effect is in moment resistance, you could probably just assume a complete composite system, design it, then check the small bending moment at the 1 ft. point for the beam-alone condition.
i think your negative bending moment may reach out further than 1 ft, so i would not use the composite section in any negative moment region.
what do you mean by composite? typically when we say "composite" beam we mean a steel beam with a concrete slab acting compositely with the beam to increase section properties or some other case where two materials work together to act as one unit.
you will have to check the stresses in the beam where i1 is as well as i2. the moments of inertia won't matter for bending and shear but will come into play for deflection and stresses. also check your details for continuity. if the beam sizes differ much you might have a hard time developing moment where the change of section occurs.
this sounds like an odd arrangement. the only practical example i can think of where this would occur is a retaining wall with different stem properties. even then a one-foot length of beam is pretty small. why not use the larger beam for the whole thing if it's only ten feet long? it might be cheaper anyway with simpler detailing.
thanks for the suggestions, although they don't help me specifically.
this is a beam that is already designed and cannot be changed. when i said "composite beam" i probably should have used nonprismatic or maybe a beam with a stepped haunch.
so, i would still like to find out how to start solving for the moment and shear in the beam, using a method to solve indeterminate structures.
this sounds more like a homework problem than anything else. in the real world, i would simply use a 2-d frame analysis program, coding the beam as 2 separate
i can assure you it is not a homework problem. most of the design work that i do at my work (alpine engineered products) is with truss design and using sophisticated computer software. i haven’t worked out a problem by hand since being in college.
i would prefer to try and solve this using a non computer aided program and just paper and pencil.
what kind of connection are you assuming between the two parts?
if the left is fixed and the right end is pinned and the middle is a hinge, the beam is determinate. you only need force and moment equilibrium to solve the problem. from there you can get reactions. with the reactions and load diagrams you can draw the sfd and the bmd. make sure the bending moment is zero at the hinge location.
if the beam-to-beam connection is not a hinge you have an indeterminate problem. you can use the principle of virtual work to solve the problem, among other methods. the beam section of the aisc manual has beam diagrams that would help you in that case instead of working it out the long way.
johnhuebbe: i would probably recommend the beam finite element method (also sometimes called matrix structural analysis).
in your case, you have a three dof (degree of freedom) system, meaning the unconstrained structure stiffness matrix is 3 by 3. so you would invert a 3 by 3 to solve for your nodal displacements. this could be done using a computational tool (e.g., matlab, mathcad, etc.) or strictly by hand using gaussian elimination.
the applied nodal load vector (which is all zeros in your case) minus the fixed-end force vector (for your 3 by 3 unconstrained system of equations) would be r = < 0-0.5*w*l1, 0+(w*l1^2)/12, 0 >, assuming your nodal translations are defined positive upward, your nodal rotations are defined positive counterclockwise, your uniform load is directed vertically downward, and assuming the 12-inch segment is at your pinned end (which you didn't state), where w = uniformly-distributed load, and l1 = length of left-hand segment (108 inch). assuming both segments are made of the same material having tensile modulus of elasticity e, the assembled, unconstrained structure stiffness matrix would look like this:
k(1,1) = 12*e*[(i1/l1^3)+(i2/l2^3)]
k(1,2) = 6*e*[(-i1/l1^2)+(i2/l2^2)]
k(1,3) = 6*e*i2/l2^2
k(2,1) = k(1,2)
k(2,2) = 4*e*[(i1/l1)+(i2/l2)]
k(2,3) = 2*e*i2/l2
k(3,1) = k(1,3)
k(3,2) = k(2,3)
k(3,3) = 4*e*i2/l2
solve for the nodal displacements, d = inverse(k)*r. after you obtain d, multiple the beam element stiffness matrix for each of your two elements by d to obtain the element end-forces. are you familiar with this method?
__________________
借用达朗贝尔的名言:前进吧,你会得到信心!
[url="http://www.dimcax.com"]几何尺寸与公差标准[/url]
huangyhg离线中   回复时引用此帖
GDT自动化论坛(仅游客可见)
回复


主题工具 搜索本主题
搜索本主题:

高级搜索
显示模式

发帖规则
不可以发表新主题
不可以回复主题
不可以上传附件
不可以编辑您的帖子

vB 代码开启
[IMG]代码开启
HTML代码关闭

相似的主题
主题 主题发起者 论坛 回复 最后发表
effective length of beam subject to lateral torsional buckli huangyhg 产品功能分析 0 2009-09-08 09:40 PM
cross beam bolted connection huangyhg 产品功能分析 0 2009-09-08 03:55 PM
beam strengthening huangyhg 产品功能分析 0 2009-09-07 03:04 PM
adding studs to existing composite beam huangyhg 产品功能分析 0 2009-09-07 10:13 AM
12 x 6.5x 38 thick x 41 residential basement i-beam con huangyhg 产品功能分析 0 2009-09-06 10:30 PM


所有的时间均为北京时间。 现在的时间是 08:05 PM.


于2004年创办,几何尺寸与公差论坛"致力于产品几何量公差标准GD&T | GPS研究/CAD设计/CAM加工/CMM测量"。免责声明:论坛严禁发布色情反动言论及有关违反国家法律法规内容!情节严重者提供其IP,并配合相关部门进行严厉查处,若內容有涉及侵权,请立即联系我们QQ:44671734。注:此论坛须管理员验证方可发帖。
沪ICP备06057009号-2
更多