Determine the force in each member of the truss shown. State whether each member is in tension o

Determine the force in each member of the truss shown. State...

Key Concepts

Static Equilibrium

Static equilibrium is a fundamental concept in structural analysis where the sum of all forces and moments acting on a system is zero. This principle underpins the calculations in truss analysis, ensuring that the structure remains stable and that each member's force contributes appropriately to maintaining this balance.

Truss Analysis

Truss analysis involves evaluating statically determinate structures composed of straight members connected at joints, usually assumed to be pin-connected. This analysis is used to calculate the internal forces acting in each member when the structure is subject to loads, thereby ensuring the structural integrity under various loading conditions.

Method of Joints

The method of joints is a technique for analyzing the forces in each member of a truss by isolating a joint and applying conditions of equilibrium. By ensuring the sum of forces in both the horizontal and vertical directions equals zero at each joint, the forces in the connected members can be systematically determined.

Method of Sections

The method of sections involves cutting through a truss to isolate a section of the structure and applying equilibrium equations. This technique is particularly useful for determining the forces in specific members without having to analyze the entire truss, making the process more efficient for large structures.

Tension and Compression

Tension and compression are the two primary types of forces experienced by truss members. A member in tension is being pulled apart, while a member in compression is being pushed together. Identifying whether a member is in tension or compression is crucial for understanding its behavior under load and for designing safe, efficient structures.

Transcript

00:02 So this is this diagram is given in the question and it's given as the anchor b -a -f and the anchor b -g -f and the anchor d -g -h and the angle d -e -h is 30 degree.

00:21 And the distance from the point a to f and from the point f to g and from the point f to g and from the point g to h and from the point g to h and from the point g to h and from the point h to the point e all are equal and it's denoted as a small letter alphabetic a and there is two forces in the joint f and the joint h both are four kilo tonneuton it's vertically acting down so here we have a roller support at point e so in point point b, we have a roll of support.

01:17 So we know the specialty of roller support, produce it first.

01:22 So because of this type of support, we have one external unknown reaction, vertical force.

01:33 So we have vertical force.

01:39 And in point a, there is a pin support.

01:43 So in point a there is spin support.

01:57 So due to this support, you know there will be two forces, one vertical and a horizontal.

02:05 So there is vertical plus horizontal force.

02:18 Is a support here right here and a horizontal force so let's denote this force in the diagram itself let's take back to denote this force and in coin a there is a horizontal force let's name it as h a and and because this seapint support there will be a vertical force also let's name it as va and because of this roller support there will be a vertical force here let's call it as ve so you turn out that first send everything which you create that's yeah so first first we need to calculate the unknown reaction.

03:35 So for calculating unknown reaction, we need to write equation of equilibrium.

03:41 So equation of equilibrium.

04:00 So equation of equilibrium, let's write it.

04:03 First, solving its direction, summation of h equal to zero gives us the horizontal force ha minus there is no other force equal to 0 that uses ha equal to 0 now let's the summation of m .e equal to 0 that uses vertical adding at point aba into 4a minus 4 into 3a minus 4 into 3a minus 4 into a equal to 0 that is here from this diagram we can see here ve from if you take from ve if we consider from ve that is this is same right so if we consider from ve, a plus a plus a plus a plus a, there is 4a multiplied va will be the momentum here.

05:31 And the 4 into there is 3a, so 4 into 3a is this term, and about this 4, 4 into a is this term.

05:44 So let's reserve this and we will get the value of va equal to 4 kilo 4 kilo and summation let's resolve right now why direction summation of y we can write the abatea force at the point e matching at the point e ve plus this v a v a minus the forces given in the questions to four kilometers so these two forces are these two forces this four kilogram time is mentioned in as this one night so here i can write the is also for kilo nutal.

07:05 So we got the value of va and ve and h .a.

07:15 So let's now go to the joint a.

07:24 So consider joint a.

07:31 So let's write the equations of equilibrium for joint a.

07:36 So equations of equilibrium for joint a means summation of y will be 0 that gives us s, a, b, sine 30, sine angle of 30, minus 4 equal to 0.

08:04 That is, let's resolve the forces at joint here.

08:09 Here so you will get a better understanding so let's consider this is the point a so there will be a force in at an angle let's call it as sab and there be a horizontal force s a f and another vertical va we already got the value va that is for kilo let's write it here and the anchor between sab and saf that is that will be 30 degree and so let's go back to the survey so here that's way sab and sab and sab sine degree of 30 will be the reserving vector along yr 6 so the vector sab is resolved along y a axis and x axis sab into sign 30 will be along y a axis and that minus this va 4 kiloton will be zero so from here we can write sab will be equal to h kiloons 8 kilo matter and we know this sab is coming inward so this sab is coming inward so that will create compression let's denote it by letter c now to sew along at sizes summation of x equal to zero gives us s .a .f.

10:44 Minus h into cost 30 equal to 0.

11:01 So s .a .f.

11:05 This saf and this is reserved in sacc as sab into cost 30.

11:13 Scb is calculated.

11:15 We already calculated.

11:17 You got the value as 8 kilo -newton.

11:19 So the equation will be saf minus 8 kiloton into cost 30 equal to 0.

11:29 So you can find the value of saf from here that will be 6 .928 kiloton 6 .928 kiloton and you know, this saf is going out from the joint a, so that is considered as a tension.

12:08 Let's denote it as d...

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