## Introduction

For as long as structures have been designed and built, civil engineers have had the responsibility of analyzing and understanding the effects of loads on structures. [1] Real life examples of loads are:

Dead, Live, Wind, Snow, Earthquake, Hydrostatic and Thermal loads.[1] These loads can be applied individually or in combination with others. For the purposes of this course, the focus will be mainly on applications of static loads on stationary members or structures.

Structures are affected by different loads or forces. These loads and forces cause reactions from the supports of the structure, which are necessary in order for the structure to remain in equilibrium. Some of the most common types of loads are:

A point load is caused by a force acting on a single stationary point of the structure. For example, heavy equipment which acts as a dead load. A dead load is a load of fixed magnitude caused by a force (the weight of an object, beam on girder, etc.) acting permanently at a specific point on a structure.[1]

To find a point load the following equation can be used:

$F=ma$ [2]

where F is the total point load, m is the mass of the object, and a is the acceleration. For dead loads a=g, where g is the gravitational constant 9.81 meters per square second.

A UDL is caused by a constant force acting over an interval of a specific length. For example, the self-weight of a beam.[1] This dead load has a constant force acting over the length of the beam or member.

To find a UDL the following equation can be used:

$F = wL$ [2]

where F is the total force, w is the force per distance, and L is the length.

A triangular distributed load typically occurs in buildings when a beam or girder supports a triangular tributary area, or dependent on climate, when snow drifts due to wind.[1] This causes snow to pile up near an obstruction which creates the triangular shape of loading with the maximum loading against the obstruction. To find the resultant force of the triangular load the following equation can be used:

${F}=\frac{wL}{2}$ [1]

where w is the distributed load at the high side, and L is the length of the distributed load.

The resultant force acts 1/3 of the distance from the high side of the triangular load, at the x location of the triangle's centroid. [1]

Snow loading is another simple example of loads on inclined members. Many houses have a gable (triangularly shaped) roof, and thus the slope of the roof is inclined to a certain degree. Since most snow loads are distributed parallel to the ground, the following equation must be used to re-distribute the parallel load onto the surface of the structure:

${F}=\frac{wLcos\theta}{D}$ [3]

The other portion of the load acts on the inclined member as an axial load and can be determined using the following equation:

${F}=\frac{wLsin\theta}{D}$ [3]

where w is the parallel load, L is the length of the parallel load, $\theta$ is the angle of incline, D is the length of the roof (or general member).

Point Moment

A point moment is caused by a rotational force acting on a structural member about a specific point. When a structural analysis is performed, the point moment can be moved around, depending on what reaction forces must be solved.

$M = Fd$ [2]

where M is the total point moment, F is the force causing the rotation, and d is the perpendicular distance from the centre of rotation, to the line of action of the force.

Typically, loads do not occur individually, but rather they are a combination of multiple loads acting simultaneously. [1] Therefore there can be a uniformly distributed load due to a live load acting on a structure at the same time as a point load due to a dead load and/or a triangular load due to a snow load. An engineer must take into consideration that these loads can act on the structure at the same time when designing a structure. [1] When faced with a member subjected to load combinations, the key to solving the support reactions is to find all of the individual x, y, and moment values caused at the supports by the individual forces, and then sum them, as in reality they occur simultaneously. The examples below demonstrate how to solve basic problems involving a member subjected to the aforementioned types of loading.

Figure 1. Simply Supported Beam with a Point Load. By: Kristyn Boehme

## Example - Point Load on a Member

In this example, a 1 kN point load is applied on a 5m long simply supported beam, two meters from the left support (C) and three meters from the right support (A). The following solution demonstrates how to solve for the support reactions due to the point load on the beam.

$\sum M_c=0: (1kN)(3m) + A_y(5M) = 0$

Solving this equation results in the following:

$A_y = \frac35 kN$

Then, by equilibrium, the value of Cy can be found:

$\uparrow \sum F_y = 0: \\ C_y = 1kN - \frac35kN$

$C_y = \frac25 kN$

Refer to Figure 1.

Figure 2. Point Moment on a Member. By: Kristyn Boehme

## Example - Point Moment on a Member

A point moment on a member works in much the same fashion as a point load; the only difference being that the sum of moments is taken at the end of the member where the point moment is located. In this following example, a 1 kN-m point load is located at point C of a member, and the support reactions are found using equilibrium. Note that the positive direction for moments is taken as clockwise.

$\sum M_c = 0:$

$1 kNm + 5m*A_y = 0$

$A_y =-\frac15 kN$

From here, the value of Cy can be found using equilibrium of forces in the y direction.

$\uparrow\sum F_y = 0:$

$-\frac15 + Cy = 0$

$C_y = \frac15kN$

Refer to Figure 2.

Figure 3. Simply Supported Beam with Uniformly Distributed Load. By: Kristyn Boehme
Figure 4. Simply Supported Beam with Triangular Uniformly Distributed Load. By: Kristyn Boehme

## Example - Distributed Load on a Member

In order to calculate the support reactions of a member undergoing a distributed loading, the force of this loading must be considered a point load. In order to do this, the magnitude of the force is multiplied by the length of the member that it covers. This resultant force is applied at the centroid of the shape of the distributed load; for example, a rectangular load will have the resultant force applied in the middle (i.e. if the force covers an entire member, the force is applied in the centre of the member). In the following example, a uniformly distributed load of 1 kN/m is applied to a member with a length of 5m.

$Point Load = 1 kN/m * 5m = 5 kN$

$\sum M_a = 0:$

$5 kN * 2.5m - C_y * 5m = 0$

$C_y = 2.5 kN$

Solving for the other support,

$\uparrow\sum F_y = 0:$

$2.5 kN + A_y - 5 kN = 0$

$A_y = 2.5 kN$

Refer to Figures 3 and 4.

Figure 5: Uniformly distributed load over an inclined beam, By: Kristyn Boehme

## Example - Distributed Load on an Inclined Member

A distributed load on an inclined member functions in much the same way as on a normally (horizontally) positioned member. The key difference is that the resultant force of the distributed load must be divided into x and y components using simple trigonometry in order to properly solve for the reactions. The equations for the x and y component divisions can be found above on this page.

Refer to Figure 5.

## References

1. Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning.
2. Hibbeler, R.C.(2009). Engineering Mechanics: Statics and Dynamics (12th ed.). Upper Saddle River, NJ: Prentice Hall
3. Prof. Erochko, J. CIVE 3202 Lecture 5 Presentation (Lecture, CIVE 3203 Structural Analysis, Ottawa, ON, September 25, 2013.)