The beam weighs 280 lb/ft. Determine the internal normal force, shear force, and moment at point

The beam weighs 280 lb/ft. Determine the internal normal...

Key Concepts

Free-Body Diagram (FBD)

A free-body diagram is a simplified representation of a structural element isolated from its surroundings, showing all the external forces and moments acting on it. Creating an FBD is a crucial step in mechanical analysis, as it allows engineers to visualize and resolve the effects of loads, including distributed loads, point loads, and reactions, thereby serving as the basis for applying equilibrium equations.

Distributed Load

A distributed load refers to a load spread continuously along the length of a structural element, typically described in units like pounds per foot. This concept is essential in structural analysis since distributed loads cause both shear forces and bending moments in beams. Understanding how to represent and integrate distributed loads is fundamental to determining the internal force distribution along the member.

Static Equilibrium

Static equilibrium is a state where the sum of forces and the sum of moments acting on a body are zero. This principle is foundational in structural analysis and is used to solve for unknown internal forces and reactions in beams and other structures. By applying the conditions of equilibrium, one can determine how external loads are balanced by internal normal forces, shear forces, and bending moments.

Internal Shear Force

Internal shear force is the transverse force within a beam that arises due to applied loads, particularly distributed or point loads. It represents the force transfer across a section of the beam and is found by taking the summation of forces acting parallel to the beam's cross-section. This concept is key for ensuring that the beam design can handle induced stresses without failure.

Internal Bending Moment

The internal bending moment is the moment induced in a beam due to external loads, causing the beam to bend. It results from the distribution of forces along the beam and is essential for predicting deflections and stresses in the structure. Calculating bending moments involves integrating the effects of the applied load, often using the relationship between shear force and moment.

Internal Normal Force

Internal normal force refers to the axial force acting along the length of a structural element, either causing tension or compression. In many beam problems that deal solely with vertical distributed loads, the normal force is zero; however, in cases involving axial loads or inclined members, it must be taken into account. Recognizing when and how to calculate this force is crucial for a complete structural analysis.

Transcript

00:02 So in this problem we have a beam that is inclined and it is, let's see here, it is supported at the lower end down here by a pin joint and at upper here by essentially a roller.

00:22 So we only have a horizontal force up here and we have both horizontal and vertical forces down here.

00:29 I see there are basically it's under its the external load is essentially its own weight.

00:40 So this is a problem where we have to be a little bit careful about things.

00:44 Again, this makes a three, four, five triangle here, so we can get the angle from that.

00:53 But we have to be a little careful about this distributed load here.

00:57 Obviously we can lump it at the center when we look at the entire structure, but when we just look at a piece of it, then we have to be careful and move it.

01:09 Just consider the contributed load over that piece.

01:14 So for the entire structure, we have p2 plus p4 is zero, and p1 minus p3 is zero.

01:22 And then this distance here is three feet.

01:26 So we have minus three times p3.

01:29 Minus 8 because this distance is 8 times p4 and that has to be 0 so that would give us that gives us three equations and we could find p1 p2 and p4 from those i just solved everything altogether so then i took a cut here which is at point c in the in the figure and so we take a look at this little chunk and so now we have v our shear force and our normal force are inclined with respect to the x and the y axes.

02:13 So we need to take their components.

02:16 So the component of v in the x direction is 4 fifths v and it's negative.

02:24 Then we have plus p2.

02:26 And then the component of n in the x direction is three -fifths n.

02:33 So again the magnitude is n and v...

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