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Statics Course Code: CIVL211 Dr. Aeid A. Abdulrazeg

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1 Statics Course Code: CIVL211 Dr. Aeid A. Abdulrazeg
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

2 The force tends to cause the pipe to turn about the z axis.
MOMENT OF A FORCE The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis. Consider the force Fx, which acts perpendicular to the handle of the wrench and is located distance dy from point O The force tends to cause the pipe to turn about the z axis. The larger the force or the distance, the greater the turning effect. Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

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8 In the 2-D case, the magnitude of the moment is Mo = F d
MOMENT IN 2-D In the 2-D case, the magnitude of the moment is Mo = F d As shown, d is the perpendicular distance from point O to the line of action of the force. In 2-D, the direction of MO is either clockwise or counter-clockwise depending on the tendency for rotation. Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

9 The sense of M depends on the direction in which F tends to rotate the body. The right-hand rule is used to identify this sense. We represent the moment of F about O-O as a vector pointing in the direction of the thumb, with the fingers curled in the direction of the rotational tendency.

10 For example, MO = F d and the direction is counter-clockwise.
MOMENT IN 2-D F a b d O For example, MO = F d and the direction is counter-clockwise. Often it is easier to determine MO by using the components of F as shown. a b O F F x F y Using this approach, MO = (FY a) – (FX b). Note the different signs on the terms! The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive. We can determine the direction of rotation by imagining the body pinned at O and deciding which way the body would rotate because of the force. Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

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13 EXAMPLE 2 Given: A 400 N force is applied to the frame and  = 20°. Find: The moment of the force at A. Plan: 1) Resolve the force along x and y axes. 2) Determine MA using scalar analysis.

14 EXAMPLE 1 (continued) Solution +  Fy = cos 20° N +  Fx = sin 20° N + MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m = N·m

15 2. If M = r  F, then what will be the value of M • r ? A) 0 B) 1
CONCEPT QUIZ 1. If a force of magnitude F can be applied in four different 2-D configurations (P,Q,R, & S), select the cases resulting in the maximum and minimum torque values on the nut. (Max, Min). A) (Q, P) B) (R, S) C) (P, R) D) (Q, S) Answers: 1. D 2. A 2. If M = r  F, then what will be the value of M • r ? A) 0 B) 1 C) r 2 F D) None of the above. Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

16 Given: A 40 N force is applied to the wrench.
GROUP PROBLEM SOLVING Given: A 40 N force is applied to the wrench. Find: The moment of the force at O. Plan: 1) Resolve the force along x and y axes. 2) Determine MO using scalar analysis. Solution: +  Fy = - 40 cos 20° N +  Fx = - 40 sin 20° N + MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm = N·mm = N·m Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

17 ATTENTION QUIZ 10 N 5 N 3 m P m 1. Using the CCW direction as positive, the net moment of the two forces about point P is A) 10 N ·m B) 20 N ·m C) N ·m D) 40 N ·m E) N ·m Answers: 1. B 2. A Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

18 MOMENT OF A COUPLE Today’s Objectives: Students will be able to a) define a couple, and, b) determine the moment of a couple. In-Class activities: Check homework, if any Reading quiz Applications Moment of a Couple Concept quiz Group problem solving Attention quiz

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22 READING QUIZ 1. In statics, a couple is defined as __________ separated by a perpendicular distance. A) two forces in the same direction. B) two forces of equal magnitude. C) two forces of equal magnitude acting in the same direction. D) two forces of equal magnitude acting in opposite directions. 2. The moment of a couple is called a _________ vector. A) free B) spin C) romantic D) sliding

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24 Given: Two couples act on the beam and d equals 8 ft.
EXAMPLE Given: Two couples act on the beam and d equals 8 ft. Find: The resultant couple Plan: 1) Resolve the forces in x and y directions so they can be treated as couples. 2) Determine the net moment due to the two couples. Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

25 Solution: The x and y components of the top 60 lb force are: (4/5)(60 lb) = 48 lb vertically up (3/5)(60 lb) = 36 lb to the left Similarly for the top 40 lb force: (40 lb) (sin 30°) up (40 lb) (cos 30°) to the left The net moment equals to + M = -(48 lb)(4 ft) + (40 lb)(cos 30º)(8ft) = = ft·lb

26 GROUP PROBLEM SOLVING - SCALAR
Given: Two couples act on the beam. The resultant couple is zero. Find: The magnitudes of the forces P and F and the distance d. PLAN: 1) Use definition of a couple to find P and F. 2) Resolve the 300 N force in x and y directions. 3) Determine the net moment. 4) Equate the net moment to zero to find d.

27 Solution: From the definition of a couple P = 500 N and F = 300 N. Resolve the 300 N force into vertical and horizontal components. The vertical component is (300 cos 30º) N and the horizontal component is (300 sin 30º) N. It was given that the net moment equals zero. So + M = - (500)(2) + (300 cos 30º)(d) + (300 sin 30º)(0.2) = 0 Now solve this equation for d. d = (1000 – 60 sin 30º) / (300 cos 30º) = 3.96 m

28 EXAMPLE The rigid structural member is subjected to a couple consisting of the two 100-N forces. Replace this couple by an equivalent couple consisting of the two forces (P) and (-P), each of which has a magnitude of 400 N. Determine the proper angle Ɵ.

29 The forces P and -P produce a counterclockwise couple
The original couple is counterclockwise when the plane of the forces is viewed from above, and its magnitude is M = 100(0.1) =10 N. m The forces P and -P produce a counterclockwise couple M = 400(0.040) Cos Ɵ 10 = (400)(0.040) Cos Ɵ

30 . Its moment and force equals _____ N·m. &_____ N.
ATTENTION QUIZ 1. A couple is applied to the beam as shown. Its moment equals _____ N·m. A) 50 B) 60 C) 80 D) 100 50 N 1m 2m 5 3 4 2. Replace the horizontal 80-lb force acting on the lever by an equivalent system consisting of a force at O and a couple. . Its moment and force equals _____ N·m. &_____ N. A) 624, 50 B) 600, 80 C) 624, 80 D) 100, 20

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