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Unformatted text preview: FORCE SYSTEM RESULTANTS RESULTANTS
Reduction of a Simply Distributed Load Reduction LEARNING OBJECTIVES LEARNING
Be able to find an equivalent force for a Be simply distributed load simply PREREQUISITE KNOWLEDGE PREREQUISITE Units of measurement Units Integration of functions over an area LINEAR LOAD DISTRIBUTION LINEAR Swimming Pool
Water Z
Floor The unit weight of water (γ) is 62.4 pcf , in SI units it is 1 gm/cm3. The pressure (p) in the water is the same in all direction and is equal to the unit weight (γ) multiplied by the depth (Z); p = (γZ) p = (γZ)
The total force due to the water pressure is equal to the integral of the pressure {p = (γZ)} over the area of the swimming pool wall. LINEAR LOAD DISTRIBUTION LINEAR
Water Z
Floor For a swimming pool depth of 6 feet and width of 50 feet, the total force can be calculated as follows:
F = ∫ pdA =
A 6 p = (γZ) z =0 y =0 ∫ 50 12 ∫ γZdydz = 2 γyz F = 0.5(62.4)(50)(6) 2 = 56160 lb = 28.08 tons NONLINEAR LOAD DISTRIBUTION NONLINEAR
In general, a distributed load can be replaced by a resultant force as follows: The given distributed load Step 1 Reduce the ydimension w = w(x) = (A)[p(x)] NONLINEAR LOAD DISTRIBUTION NONLINEAR
A distributed load can be replaced by a resultant force as follows Step 2  Determine the resultant force and its location FR = ∫ w( x ) dx
L Step 3  Replace the distributed load with the resultant force FR INTEGRATION INTEGRATION
1. The magnitude of the resultant force FR = ∫ w ( x ) dx
L
2. The resultant moment at point O 2. M R O = ∑M O
L 3. The location of the resultant 3. force measured from point O force x ⋅ FR = ∫ x ⋅ w ( x ) dx x =L ∫ x ⋅w ( x ) dx ∫w ( x ) dx
L x=L ∫ x ⋅ w ( x ) dx
FR EXAMPLE 1 EXAMPLE
x FR Replace the distributed load with a single resulting load and identify its location SOLUTION 1
F R x Step 1: Identify the loading equation w( x ) = 0.5 ⋅ x 3 N/m
Step 2: Compute the magnitude of the resultant force 0.5 4 FR = ∫ w( x ) dx = ∫ 0.5 ⋅ x dx = ⋅x L 4 0
3 0 10 10 0.5 = ⋅ (10000 − 0 ) = 1250 N 4 SOLUTION 1  continued continued
x F R Step 3: Compute the moment of the distributed load
M = ∫ x ⋅ w( x ) dx = ∫ x ⋅ 0.5 ⋅ x 3 dx =
L 0 10 ( ) 0.5 5 ⋅x 5 0 10 0.5 ⋅ (100000 − 0 ) = 10000 N  m 5 Step 4: Compute the location of the resultant = M 10000 x= = =8m FR 1250 EXAMPLE 2 EXAMPLE Replace the distributed load by an equivalent resultant Replace force and specify its location from point C force SOLUTION 2 SOLUTION
Y w ( x ) = 800 lb/ft; for 0 < x ≤ 15 ft
X Function of distributed load Function 800 ( x − 15) lb/ft; for 15 < x ≤ 30 ft w ( x) = 800 − 15 FR = ∫ w ( x ) dx
L The magnitude of The the resultant force the 800 ( x  15) dx = ∫ 800 dx + ∫ 800 15 0 15 800 x = [800 x ] + 800 x × + 800 x 15 2 15 = [12000  0] + [ 24000  18000] = 18000 lb
15 0 2 30 15 30 SOLUTION 2  continued continued
Y X Location of the resultant Location force relative to point C x= ∫ x ⋅ w( x ) dx
L FR
2 15 800 ( x  15) dx x( 800 )dx + ∫ x ⋅ 800 ∫ 15 15 =0 18000
2 3 2 30 15 30 x x 800 x x 800 2 + 800 2  15 × 3 + 800 2 0 15 = 18000 ( 90000  0) + ( 240000  120000) = 210000 = 11.67 ft = 18000 18000 EXAMPLE 3 EXAMPLE A 5m long concrete wall was formed as shown. Determine 5m the resultant force and the distance h where the bracing strut should be placed to counter the concrete pressure. strut SOLUTION 2 SOLUTION
Y Z Function of distributed load Function concrete wall length = 5 m
1 1 w ( z ) = 4 z 2 5 = 20 z 2 kN/m FR = ∫ w ( z ) dz = ∫ 20 z dz
L 0 3 z2 = 20 = 106.67 kN 3 The magnitude of the resultant The force 2 0 4 4 1 2 SOLUTION 2  continued continued
The location of the resultant force relative to the top of the wall The
Y Z z=L ( z )(5)[4 z 0.5 ] dz ∫ FR
5 z2 4 1 4 20 ⋅ 2 dz 5 ∫ z 20 ⋅ z 2 0 0 = = 106.67 106.67 256 = = 2.4 m 106.67 h = 4 − z = 4 − 2.4 = 1.6 m QUESTION
The resultant force (FR) due to a distributed load is equivalent to the _____ under the distributed load curve, w = w(x). A) centroid B) arc length C) area D) volume
w
Distributed load curve x
FR QUESTION
The line of action of the resultant of a distributed load pass through the ______ of the distributed load. A) centroid B) midpoint C) left edge D) right edge QUESTION
FR A 3m 3m B A d B What is the distance d ? A) 2 m C) 4 m E) 6 m B) 3 m D) 5 m QUESTION
x2 x1 F F x F If F1 = 1 N, x1 = 1 m, F2 = 2 N and x2 = 2 m then what is the distance x ? A) 1 m C) 1.5 m E) 2 m B) 1.33 m D) 1.67 m QUESTION
100 N/m 12 m FR = ___________ A) 12 N C) 600 N B) 100 N D) 1200 N QUESTION
FR
100 N/m x 12 m x = __________ A) 3 m C) 6 m B) 4 m D) 8 m QUESTIONS
FR
100 N/m FR 12 m 1. FR = _________ A) 12 N B) 100 N C) 600 N D) 1200 N Approach I: Geometry x FR = Area Under Distributed Load 1 = ⋅ (12 ) ⋅ (100 ) = 600 N 2 QUESTIONS
FR
100 N/m FR x 12 m B) 4 m C) 6 m D) 8 m 2. x = __________. A) 3 m x = Centroid of the Distributed Load 12 = one third from the base of the triangle = =4m 3 QUESTIONS
FR
100 N/m FR 12 m 1. FR = _________ A) 12 N B) 100 N Approach II: Integration C) 600 N D) 1200 N x 25 12 − x w( x ) = 100 ⋅ = 100 − ⋅ x 3 12 25 25 2 FR = ∫ w( x ) dx = ∫ 100 − ⋅ x dx = 100 ⋅ x − ⋅ x = 1200 − 600 = 600 N 3 6 0
L 0 12 12 QUESTIONS
FR
100 N/m FR 12 m 1. FR = _________ A) 12 N B) 100 N
12 0 x C) 600 N D) 1200 N
12 25 100 2 25 3 M = ∫ x ⋅ w( x ) dx = ∫ x ⋅ 100 − ⋅ x dx = ⋅x − ⋅x 3 2 9 0
L = 7200 − 4800 = 2400 N − m M 2400 x= = =4 m FR 600 ...
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This note was uploaded on 10/31/2010 for the course CEE CE 221 taught by Professor Baladi during the Fall '10 term at Michigan State University.
 Fall '10
 Baladi

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