Determining Equivalent Force for Distributed Loads in Engineering Analysis, Lecture notes of Engineering

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ENGR-1100 Introduction to

Engineering Analysis

Lecture 18

REDUCTION OF A SIMPLE DISTRIBUTED LOADING

In-Class Activities:

• Reading Quiz
• Applications
• Equivalent Force
• Concept Quiz
• Group Problem Solving
• Attention Quiz

Today’s Objectives:

Students will be able to determine an equivalent force for a distributed load.

APPLICATIONS

(continued)

The uniform wind pressure is acting on a triangular sign (shown in light brown).

To be able to design the joint between the sign and the sign post, we need to determine a single equivalent resultant force and its location.

DISTRIBUTED LOADING

In such cases, w is a function of x and has units of force per length.

In many situations, a surface area of a body is subjected to a distributed load. Such forces are caused by winds, fluids, or the weight of items on the body’s surface. We will analyze the most common case of a distributed pressure loading. This is a uniform load along one axis of a flat rectangular body.

LOCATION OF THE RESULTANT FORCE

The total moment about point O is given as

• M (^) RO = ∫L x dF = ∫L x w(x) dx

Assuming that FR acts at 𝑥𝑥 ̅, it will produce the moment about point O as

• M (^) RO = (𝑥𝑥) (FR ) = 𝑥𝑥 ∫L w(x) dx

The force dF will produce a moment of (x)(dF) about point O.

LOCATION OF THE RESULTANT FORCE (continued)

Comparing the last two equations, we get

You will learn more detail later, but FR acts through a point “C,” which is called the geometric center or centroid of the area under the loading curve w(x).

EXAMPLES

The rectangular load: FR = 400 × 10 = 4,000 lb and x= 5 ft. The triangular loading: FR = (0.5) (600) (6) = 1,800 N and = 6 – (1/3) 6 = 4 m. Please note that the centroid of a right triangle is at a distance one third the width of the triangle as measured from its base.

x

Now let’s complete the calculations to find the concentrated loads (which is a common name for the resultant of the distributed load).

READING QUIZ

2. The line of action of the distributed load’s equivalent force passes through the ______ of the distributed load. A) Centroid B) Mid-point C) Left edge D) Right edge
3. The resultant force (FR ) due to a distributed load is equivalent to the _____ under the distributed loading curve, w = w(x). A) Centroid B) Arc length

C) Area D) Volume

x

w

FR

y Distributed load curve

ATTENTION QUIZ

1. FR = ____________

A) 12 N B) 100 N

C) 600 N D) 1200 N

2. x = __________. A) 3 m B) 4 m C) 6 m D) 8 m

100 N/m^ FR

12 m x

GROUP PROBLEM SOLVING

Given : The loading on the beam as shown. Find : The equivalent force and its location from point A. Plan:

For the left triangular loading of height 800 lb/ft and width 12 ft, F (^) R1 = (0.5)(800) x 12 = 4800 lb

𝑥𝑥 � 1 =(2/3)(12) = 8ft from A

For the rectangular loading of height 500 lb/ft and width 9ft,

F (^) R3 = (500)(9) = 4500 lb

and its line of action is as 𝑥𝑥� 3 =(1/2)(9) + 12 = 16.5ft from A

For the top right loading of height 300 lb/ft and width 9 ft,

F (^) R2 = (0.5)(300)(9) = 1350 lb

and its line of action is at 𝑥𝑥� 2 =(1/3)(9) + 12 = 15ft from A

GROUP PROBLEM SOLVING (continued)

GROUP PROBLEM SOLVING (continued)

4800 lb (^) 4500 lb

1350 lb

8 ft

16.5 ft

15 ft

For the combined loading of the three forces, add them.

F (^) R = 4800 + 4500 +1350 = 10650 lb

• M (^) RA = 8(4800) + 15(1350) + 16.5(4500) = 132900 lb-ft

Now, (F (^) R 𝑥𝑥�) has to equal MRA = 132.9 kip-ft

So solve for 𝑥𝑥� to find the equivalent forces locations.

Hence, 𝑥𝑥� = (132.9 kip-ft) / (10.65 kip) = 12.5 ft from A