The truss shown consists of six members and is supported by a ball-and-socket at B, a short link

The truss shown consists of six members and is supported by...

Key Concepts

Force Resolution

Force resolution is the process of decomposing a force into its orthogonal components, typically horizontal and vertical. This technique is essential in structural analysis, especially when working with truss members, because it allows one to apply equilibrium equations effectively by addressing the contributions of these components individually.

Free-Body Diagrams

Free-body diagrams are simplified graphical representations that illustrate all forces acting on a structural element. They are critical in structural analysis because they help in organizing and visualizing the applied forces, support reactions, and internal member forces, thereby facilitating the setup of equilibrium equations.

Static Equilibrium

Static equilibrium is the condition where an object or a structure remains at rest because the sum of all forces and moments acting on it is zero. In structural analysis, this fundamental principle allows for setting up equations that ensure every external load, internal force, and reaction balance, which is essential in determining unknown forces in a structure.

Truss Analysis

Truss analysis focuses on determining the forces within the members of a truss, a structure composed of straight members connected at joints. In this analysis, assumptions—such as members carrying only axial forces and joints acting as perfect pins—simplify the process, making it possible to isolate and solve for unknown member forces.

Method of Joints

The method of joints is a systematic technique used in truss analysis where each joint is considered separately. By applying the conditions of static equilibrium to the forces in both the horizontal and vertical directions at a joint, the force in each connecting member can be calculated incrementally.

Transcript

00:02 Here says 6 .36 problem 6 .3 6 .3 here the space trust loading is shown the truss is symmetric above at san yacci so we can show only for half the trust and the other half will be same now the other thing here in d there is two sockets so because of that sockets there will be one along x and one along y so let's say if there is a first reaction dx because of this circuit and d y because of this socket and d y because of this socket and here in b actually this is a pinned socket so because of this support there will be direction all the reaction in all the directions in along x y z direction there will be bx and and and b y and and d z and here there is another one and the support c so because of this one there will be 4 cell of y -dange.

02:07 So this one, there will be cy, c -y.

02:20 Okay? yeah.

02:23 The measurements are given in meter.

02:28 So now from symmetry, we can write that bx will be equal to d -h dx and the b y will be equal to d y okay now let's apply equilibrium equation in 3d so equilibrium equation yeah summation along x equal to zero uses bx equal to dx the only thing is bx equal to dx so that means dx and d h will be equal to zero now summation along z equal to zero equal to b z equal to b z equal to summation at momentum is at c along the same direction and the joint c okay equal to c gives us the two meter this two meter into the 2 one night four total that is yeah 2 1 8 for newton minus 2 into 2 .8 meter of that is just because b .y equal to d y so that the result and will be 2 into that thing that's why it is too can equal to 0 so so from here we can write b y will be equal to 780 newton now let's move to that joint a, let's take a, join a, and join a c, and c and a, dene, and a c and a c and a, d.

05:57 Again let's find the distances distance a b will be equal to square root of minus 0 .8 meter square plus 4 .8 meters square plus 2 .1 meters square so actually this minus point it gets good and yeah so that is 5 .3 meter and the distance a c is 2 meter square plus 4 .8 meter square plus zero square so this gives us 5 .2 5 .2 meter now about ad that's my red same a d will be the minus point 8 .8 meters square plus 4 .8 meters square plus minus of 2 .1 meters square.

07:29 So square root of that will be 5 .3 meter.

07:39 Yeah.

07:44 Now let's find vector f of ab that will be the magnitude of f of a b into vector ab divided by ab so that means f a b divided by ab so that means f ab divided by ab we just found that is 5 .3 5 .3 meter times vector b is actually minus 0 .8 in i .companment that is x component next direction moment 4.

08:33 8 j by direction component plus 2 .1 along set direction in the moment yeah now about that's function fine here it's yeah fac c that reface equal to 9 inch of fac in do let's say by ac so fac c divided by ac we just found that is 5 .2 times 2 i plus 4 .8 j the next one will be yes yes f a d vector f a d equal to f a d 90 n2 a d times letter a d divided by magnitude of a that is f a d divided by 80 is 5 .3.

09:56 I found it.

09:57 We calculated that value.

09:59 Yeah.

10:01 0 .8 i.

10:05 Minus 4 .8 j minus 2 .1 k.

10:15 Yeah.

10:17 Let's draw the diagram for you.

10:21 That gives you a better understanding.

10:25 If this is a, there is here.

10:29 There is here.

10:45 And f -a -b and f -a and d and f -a -c okay and let's denote that i change a single at me this is y okay this is y and this is y and let's see this is red and and yeah, this is x, okay? but is that why and it's a perpendicular to each other? you know that.

11:56 Now for component at eye direction, let's find.

12:04 So the component at eye direction.

12:24 So to find the component at eye direction, you have to find summation along h equal to zero.

12:30 So that will be minus of 0 .8 divided by hibernases 5 .3 we found that times the fab in plus 2 by 5 .2 f.

13:04 F .c plus 0 .8 by 5 .3 f...

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