Outline: Frames Machines Trusses Properties and Types Zero Force Members Method of Joints Method of Sections Space Trusses 1
structures are made up of several connected parts we consider forces holding those parts together (in addition to any external forces that may be present) for the whole structure, these are internal forces, but we can consider each of the parts as a RB and can treat forces from connecting parts as external forces corresponding points on different parts have equal but opposite rxns [Newton s 3 rd Law] 2
in this chapter, we will look at: Frames: they support loads, are usually static and completely constrained, but they have at least one multi-force member (as in Figure on prior slide) Machines: they transmit and modify forces - they have moving parts and at least one multi-force member Trusses: they support loads, are static and fully constrained, and are made of straight members connected at joints - all parts are 2 force elements 3
For all types of structures: If the whole structure is in equilibrium then any part of the structure must also be in equilibrium Can separate a structure into individual parts (all in equilibrium) Can cut a structure and examine a piece of it (also in equilibrium) 4
A system designed to support loads (forces and moments) Made up of multiple members At least one member is acted upon by 3+ forces (not two-force elements) Can be internally stable or internally unstable Stable no matter what the external support reactions are the frame maintains its shape Unstable if external support reactions are removed the members of the frame move relative to each other. 5
frames are inter-connected rigid bodies start by analyzing the whole system to get the reactions at the supports (if possible frames may be statically indeterminate) then isolate members or parts, for each one separately draw a FBD and apply equations of equilibrium Total number of unknown force components must be less than or equal to the number of independent equations (3 per piece the 3 equations for the whole frame are NOT independent of the 3 equations for each piece) 6
In general: start with two force element and show their member forces guess at force senses, verify later by looking at signs then look at multi-force members remember that F x and F y forces are same magnitude but opposite sense on each side of a connection Write equilibrium equations for multi-force member FBDs and back-substitute answers to solve all of your FBDs 7
Example analysis: note that pins are assumed to be part of an RB if an external force is applied at a pin, make sure the external force is only shown being applied to the member with the pin in it (if the pin connects two members you can pick which RB the external force is applied to) 8
Determine the components of the forces acting on each member of the frame shown. 9
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Like frames (support loads, made of up multiple members at least one supports 3+ loads) BUT, primary purpose is to transmit and modify forces they transform input forces (I/P) into output forces (O/P) Generally we want to find the magnitude of the O/P knowing the I/P or vice versa (ratio of output:input is mechanical advantage ) Often, machines won t be fully constrained because they re designed to move but they are in equilibrium (assume they are at rest) 12
We apply force P on the pliers in order for the pliers to overcome force Q to cut a wire pliers are not rigid, so break them into two pieces 13
Use the same method as for frames Draw FBD for each member Put opposite forces at connections Include input forces (act on the machine) and the REACTIONS TO the output forces (output forces act on something else show what the something else does to the machine) Often there are no external support reactions 14
What forces are exerted on the bolt at E as a result of the 150 N forces on the pliers? 15
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Typical Truss Note: floor beams connect at joints Can analyse each side as a planar (2D) system 18
trusses are major engineering structures (bridges & buildings) they consist of straight (thin) members connected at joints the members are only connected at the ends (not in the middle) there s no AB, just AD and DB Joints are represented as pin connections (idealizations) All loads act at the joints (including member weights - half at each end) 19
a rigid truss is a 3 member truss (it won t collapse, but 4 members will collapse) we can construct a larger truss by adding pairs of members to existing joints, connected at a new joint 20
we can repeat this procedure indefinitely every time we get +2 members and +1 joints a truss constructed this way is called a simple truss the total number of members in it is: m = 2n -3 (where n = the # of joints) you don t always have to have triangles in trusses 21
We assume each end of a member has one force acting on it, along the member, with no couple (every member is a two-force element) therefore, a truss can be regarded as a group of pins and 2F members where the members are either in tension or under compression 22
Member in tension Member in compression comes from a member in compression comes from a member in tension For two members in tension connected with a pin: equal & opposite 23
Consider 4 (2 element) members connected with a pin such that they lie in 2 intersecting straight lines = 2 pairs of directly opposite forces where the magnitudes of the forces in opposing members are equal 24
now consider and if 2 of 3 members are in a line and an applied force is in line with the 3 rd member, then the 3 rd member s force is equal and opposite to the applied force if there is no such applied force then the force in the 3 rd member = AC is zero such a member is called a zero force member (ZFM) and they are worth spotting early in problem solving, to simplify/speed up the analysis 25
if only 2 members connect at a joint and they have a common LOA, then their forces are equal and opposite in sense (a), but if they do not have a common LOA, then they will be zero force members (b) the trick is to spot these special situations 26
C connects 3 members, 2 in a line (no external F at C), F BC =0 same at K for JK Now J is the same along HL, so F IJ =0 also AC=CE, HJ=JL, and IK=KL since F C = F K = F J = 0 now look at I F HI =20 & F GI = F IK = F KL all of this would be very different with different external loads (the ZFMs would not necessarily still be ZFMs) you don t HAVE to catch these tricks, but it helps 27
dismember a simple truss each member is a 2F member and has its own FBD The pins also have FBDs Becomes the magnitude of the force at the end of each member is called the force in the member (though it s a scalar) truss analysis is finding the magnitudes and senses (compression or tension) of the forces in each member 28
since the truss is an RB in equilibrium, so each pin (is a particle) has 2 equilibrium equations: ΣF x = 0 and ΣF y = 0 if a truss has n pins, we have 2n equations to solve for up to 2n unknowns for a simple truss, m = 2n-3, so 2n = m+3 we can find the forces of all members + 3 rxns (e.g. R Ax, R Ay, R By ) using pins as FBDs the 3 equations we could use for the truss as a whole are not independent of these pin equations, but you can use them to find the rxns at the supports 29
To analyse a truss using the Method of Joints Start with the joints that only have 2 unknowns if all joints have 3+ unknowns, then analyze the whole structure first with the standard equations of equilibrium (ΣF x = 0,ΣF y = 0,ΣM A =0) to get R A & R B then do the M of J at each pin using force triangles (or polygons), or systems of equations For following truss assuming we know R A & R B : 30
Start with Joint A Know R A & dir n of F AC and F AD Find magnitude and sense of F AC and F AD Move to Joint D Already found F AD given P Know directions of F DC & F DB Find magnitude and sense of F DC and F DB 31
Move to Joint C Already found F CA and F CD Know direction of F CB Find magnitude and sense of F CB Joint B Already know all forces acting Can use it as a check 32
Determine the force in each member of the truss shown. 33
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MofS is more efficient if you don t need to solve for all of the internal forces, you may only be asked for forces for one or a few members MofS is based on the idea that if the whole truss is in static equilibrium then any section of the truss must also be in equilibrium Method: Cut the truss through the members that you are interested in solving for Draw the FBD for the section you are left with Ideally you have 3 unknowns (and 3 equations) You may have to repeat the process 35
For example - want F BD in Figure need effect of F BD on B or D with MofJ, we would solve for F BD at B or D with MofS, we pick a chunk of truss where F BD is an external force and we choose a chunk of truss with only 3 unknown external forces so it s solvable we now have 3 equations for 3 unknowns if we only need F BD, then we might get it with just one equation e.g. M E = 0 as usual, a (+) sign on your answer means that you guessed the right direction, while (-) means you didn t 36
Calculate the force in members FH, GH, and GI of the roof truss shown. 37
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several straight members joined at ends form a simple 3D/space truss similar to 2D trusses, add sets of 3 members to 3 existing joints and join them at one new joint (where all 4 joints do not lie in a plane) to expand a simple space truss start with 6 members and 4 joints, and then expand with +3 members and +1 joint each time (m = 3j 6) 41
for a space truss to be statically determinate and completely constrained, use balls, rollers and ball and socket joints to get 6 unknown rxns (then use 6 equations of equilibrium for 3D) for analyses, assume you have ball & socket joints although they are often welds therefore, each member is a 2F member and at each joint you ll have F x = 0, F y = 0, F z = 0 in a truss with j joints, you get 3j equations and with m = 3j-6 you get 3j = m + 6 (can solve for all the Fs in the members + 6 rxns at supports) start solving joints with 3 unknowns 42
Determine the forces acting on the members of the space truss shown. 43
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simple trusses simply supported (1 pin and 1 roller) are completely constrained and determinate trusses for any other truss, if m = the # of members, j = the # of joints, r = the # of rxns at supports, then we have 2j independent equations of equilibrium if (m+r) < 2j we have fewer unknowns than equations and the system is partially constrained (internally unstable) if (m+r) > 2j we have more unknowns than equations and the system is statically determinate (redundancies exists) if (m+r) = 2j we have the same # of unknowns as equations BUT it may not be determinate since some equations may not be able to be satisfied ONLY if all rxns and forces in members can be determined will it be completely constrained and determinate (otherwise it is improperly constrained) 47