Determine the force in member D E and in each of the members located to the left of D E for the

Determine the force in member D E and in each of the members...

Key Concepts

Truss Analysis

Truss analysis is a method used in structural engineering to determine the forces acting in each member of a truss, which is a framework composed of interconnected elements. This analysis is crucial for ensuring that the forces within each member do not exceed the material limits, thereby maintaining safety and structural integrity.

Static Equilibrium

Static equilibrium refers to the state in which a structure is at rest with all forces and moments balanced. In the context of truss analysis, applying the conditions of static equilibrium (i.e., the sum of forces and the sum of moments being zero) allows engineers to solve for the unknown internal forces in each member.

Method of Joints and Method of Sections

The method of joints and the method of sections are two primary techniques used for analyzing truss structures. The method of joints involves isolating a joint and applying equilibrium conditions to find the forces in the connecting members, while the method of sections involves cutting through the truss to analyze a segment. Both techniques are integral to assessing the load distribution throughout the structure.

Internal Forces in Structural Members

Internal forces in structural members, such as tension and compression, are the forces that develop in the elements of a truss due to external loads. Determining these forces is essential for ensuring that the members are properly designed to withstand the loads, preventing failure due to overstressing or buckling.

Tension and Compression

Tension and compression refer to the types of internal forces acting on truss members. Tension forces attempt to elongate a member, pulling it apart, while compression forces attempt to shorten a member, pushing it together. Correctly identifying whether a member is in tension or compression is vital for material selection and design, as different materials and cross-sectional shapes respond differently to these forces.

Transcript

00:01 So here in this problem we are given this trust and we have to calculate the force in the member d .e and each of the member located to the left side of the d .e.

00:10 That means we have to consider this side of the truss and we have to calculate the forces in this side of the member.

00:19 Now for that will be first calculating the angle alpha and this angle beta.

00:26 Now if you consider this triangle by using this triangle we can, we can't.

00:33 Can say that tangent alpha would be equals to 6 .72 divided by 10 .54.

00:43 So from here angle alpha comes out to be as nearly 32 .52 degree.

00:52 Next i will be calculating this angle alpha for that we can consider this triangle.

01:03 So from here tangent beta would be equal to this height is 6 .72 divide by this total height so this distance is 23 .04 so from here the angle beta comes out to be as nearly 16 .26 degrees now we have calculated angle alpha and beta now next will be calculating the reactions a x, a y and hyy so for that we'll be using the equilibrium condition the first one is sigma fx equals to zero and the forces towards this direction will be taking positive and towards this direction will be taking negative similarly in why direction forces towards in the vertical upper direction will be taking positive and in this direction will be taking negative and the moment in the interclockwise direction will be taking positive and the moment in the clockwise direction will be taking negative so now we'll be balancing the forces in the horizontal direction and we have only one force in the horizontal direction that is the reaction force your x so we have a equals to 0 now next we'll be calculating the value a for that we can do one thing we can take moment about point h so sigma mh is equals to 0 now the first one is this this is a y times the perpendicular distance this thing is a 4 times 5 .76 and since the force is making moment in the clockwise direction so we have to take a negative sign plus so this is 400 pound times the perpendicular distance to point h so this is four times 5 .76 and since the force is making the interclockwise direction moment so we have to take a positive sign similarly plus 800 times 3 multiplied with 5 .76 feet plus 800 multiplied with two times this is two times 5 .76 then plus 800 times times 5 .76 and it is equals to 0 now we're going to solve this and we'll get the value of a y as 1600 pound now next i will be considering the free boarded diagram of the joint a so this is a free board diagram of the joint a and the force in the member ac we have assumed to be in tensile in nature and in the member a b we have assumed the force it is acting like this that means the member would be tension but if the value comes out to be as negative after solving it so our assumed direction would be wrong and member would be in compression so let's find the values of forces so for that we can do we can take first a equilibrium condition that is sigma f x equals to zero we'll be balancing the force in the x direction so we have f ac cosine alpha plus f bc cosine beta is equal to 0 now we can put the values so if ac we have to find and cosine alpha we know it is 32 .52 so we can plug it here this is cosine 32 .52 plus fb this is ab this is fab so plus fab we have to calculate times cosine beta so beta we have 16 .26 and it is equals to zero so from here we'll get the relation between these forces fac and fab as fac equals to minus 1 .38 fab...

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