The rigid beam is supported at its ends by two A-36 steel...

The rigid beam is supported at its ends by two A-36 steel tie rods. If the allowable stress for the steel is $\sigma_{\text {allow }}=16.2 \mathrm{ksi}$, the load $w=3 \mathrm{kip} / \mathrm{ft},$ and $x=4 \mathrm{ft},$ determine the smallest diameter of each rod so that the beam remains in the horizontal position when it is loaded.

Key Concepts

Static Equilibrium

This concept involves ensuring that the net force and net moment on a structure are both zero. In structural systems like beams supported by tie rods, the calculations for forces and moments help determine the reaction forces in each support. When applying a load, the structure must satisfy these equilibrium conditions to remain at rest and to prevent rotation or translation.

Stress Analysis

Stress analysis is the process of determining the internal forces in a material as a result of external loads. In the context of the tie rods, it involves calculating the tensile stress produced by the applied load and ensuring that this stress does not exceed the allowable limit for the material. This ensures the structural integrity and safety of the rod under load.

Cross-Sectional Area Geometry

The cross-sectional area of a rod is directly related to its ability to withstand stress, as stress is defined as force per unit area. For circular rods, the relationship is given by the formula A = (?/4)*d², where d is the diameter. Determining the minimal diameter that satisfies the allowable stress criteria is essential in designing safe and efficient structural members.

Structural Loading and Moment Calculations

This concept involves understanding how distributed loads create bending moments in a beam. Moment calculations are used to determine the forces transmitted to the tie rods, especially when the beam must remain horizontal under load. It integrates the effects of the load's magnitude and its position relative to the supports, ensuring that the design accounts for all possible induced stresses.

Transcript

00:01 In this question it is told that a body is rest on a frictionless horizontal surface and a constant horizontal force f is acting on the positive direction.

00:10 Now we have to find the value of work done on the body by the applied force from between time t is equal to 0 to 2 second.

00:19 So we have given that there is a constant acceleration.

00:29 Now we know that from equation of motion that is x is written as.

00:38 V0 t plus of half a t square here initial position x not is 0 so we can say that now from the graph for third point we have given that the distance is 0 .2 meter is equal to this is v0 into time is 1 second plus half of acceleration into this is 1 square from here we get this is 0 .2 is equal to v0 plus half a let this is our equation first and for fifth point we can say that from graph that is given to us, the distance is 0 .8 meters and this will equal to v0 into time is 2 second plus half a into this is 2 square.

01:43 So we can say this is 0 .8 is equal to v0 plus 2a let this is our equation 2.

01:49 Now from equation 1 and 2 we get the value of v0 and this will come out to be 0 and acceleration is 0 .4 meter per second square this is the value of acceleration now we can say that final velocity is given by that is v is equal to v0 plus of 80 now at t is equal to 2 second velocity is given by that is this is 0 plus acceleration is this is 0 .4 so 0 .4 into 2 so this will come out to be 0 .8 meter per second now we have to find the work done so we can say that from work energy theorem that work done is given by change in kinetic energy.

02:46 This is work energy theorem...

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