Structures

Part of the Statics course offered by the Division of Applied Mechanics, School of Engineering and the Engineering and Technology Portal

Lecture

Structural engineering relies heavily on the strengths of materials and their ability to withstand forces of tension or compression. When used in conjunction with each other, as in the case of a truss, individual load bearing members both share and transmit loads, enabling the structure to accomplish much more than any individual member could alone.

Plane Trusses

One way of distributing a force across a large distance is by building a Plane Truss, which takes advantage of the principle of Equilibrium to translate forces along a system of interconnecting members.

A Warren Truss

A Pratt Truss

File:Simpletruss.PNG

A Simple Triangle Truss

The simplest truss is a triangle made of three points: A, B and C, and three members: AB, BC and CA. The method of distributing forces amongst many members relies on the engineer's ability to place a tensile or compressive force at any particular location. To properly sum forces at a particular point, one must be able to sum forces into that point and also away from it. Thus, AB and AC are in compression while BC is in tension.

Method of Joints

File:TrussJoint.jpg

The principle of equilibrium is only effective when applied to a single point. Therefore, the Method of Joints is applied to a truss-joint to determine the forces in each member of the truss. A place where the forces and members meet becomes the joint to be examined, and the principle of equilibrium is applied, thus solving the question of where the forces are transmitted in that truss around that joint.

Space Trusses

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Newton's Second Law for a static system (Equilibrium):

<math> \sum \vec F \ = M * \vec a = 0 </math>

The sum of the forces is equal to the mass times the acceleration. The 2nd Law tells us that if the object or system is motionless, the acceleration is equal to zero. Therefore the sum of the vector forces must be equal to zero.

Example:

Consider a table with four legs and a 200 kg object resting motionless in the center of the table. What force acts upon the bottom of each table leg <math> \vec F_L </math>?

The force upward on the legs is found by balancing these forces in the equation given by Newton's Law. Gravity is acting downward on the 200 kg object and the 100 kg table. Therefore, we may substitute the acceleration due to gravity on Earth <math> \vec g </math> which is <math> 9.81 \frac{m}{s^2} </math>, for <math> \vec a </math>.

Substitute in known values...

<math> \sum \vec F_y = M * \vec a = 0 = - (200 kg * 9.81 \frac{m}{s^2}) - (100 kg * 9.81 \frac{m}{s^2}) + (F_L * 4) </math>

And simplifying...

<math> \vec F_L = \frac{200 kg * 9.81 \frac{m}{s^2} + 100 kg * 9.81 \frac{m}{s^2}}{4} </math>

The unit <math> \frac{kg.m}{s^2} </math> is equivalent to the unit of force called a Newton N. Thus when multiplying through, the kilograms cancel out, and we may solve for <math> \vec F_L </math>...

<math> \vec F_L = \frac{1962 N + 981 N}{4} = 735.75 N </math>

Rotational Equilibrium

Lesson: In Statics the Sum of the Torques (Moments) is Equal to Zero.

The sum of all rotational forces, or torques, denoted by the capital Greek letter tau (<math>\Tau </math>), is also zero. Commonly used units for torque are Foot•pounds (ft•lb) and Newton•meters (N•m).

Newton's Second Law (applied to torques):

<math>\sum \tau = \sum (\vec \omega * I) = (\vec \omega_1 * I_1) + (\vec \omega_2 * I_2) + (\vec \omega_3 * I_3) + ... + (\vec \omega_n * I_n)</math>

Torques may also be calculated as forces times distances:

<math>\sum \vec \tau = \sum (\vec F * d) = \vec F_1 * d_1 + \vec F_2 * d_2 + \vec F_3 * d_3 + ... + \vec F_n * d_n </math>

Example:

File:LeverExample.png

Consider a massless lever with two weights attached and a single massless support:

Weight of Object 1, <math>\vec F_1</math> = 10 lb
Distance of Object 1 from fulcrum = 10 ft
Distance of support from fulcrum = 7 ft
Weight of Object 2, <math>\vec F_2</math> = 80 lb
Distance of Object 2 from fulcrum = 4 ft

What forces act upon the massless lever?

First let's write the static torque equation for the system. Forces 1 and 2 are actually the weights of the two objects...

<math>\sum \tau = \sum (\vec F * d) = \vec F_1 * d_1 + \vec F_2 * d_2 + \vec F_S * d_S = 0 </math>

Substitute in known values...

<math>\sum \tau = \sum (\vec F * d) = (-10 lb)*(10 ft) + (-80 lb)*(4 ft) + \vec F_S * (7 ft) = 0 </math>

Simplify...

<math>\sum \tau = \sum (\vec F * d) = (-100 ft.lb.) + \vec F_S * (7 ft) + (-320 ft.lb.) = 0 </math>

And solve for <math>\vec F_S</math> ...

<math>\vec F_S * (7 ft) \ = (420 ft.lb.) </math> and therefore <math>\vec F_S = (420 ft.lb.) / (7 ft) \ = 60 lb.</math>

Assignments

Activities:

• Create an activity

Readings:

• Peruse the appropriate sections of Wikibooks:Statics
• Introduction to Structural Design, Virginia Tech.

Study guide:

1. Wikipedia article:Plane Truss
2. Wikipedia article:Tension
3. Wikipedia article:Compression
4. Wikipedia article:Method of Joints
5. Wikipedia article:Method of Sections