statics - tam 210 & tam 211...b 300 mm, and m l 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g)...

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Lecture 41 April 30, 2018 Statics - TAM 211

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Page 1: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

Lecture 41

April 30, 2018

Statics - TAM 211

Page 2: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

Announcements

Check ALL of your grades on Compass2g. Report issues

Exam grades are now posted

There will be Discussion Sections this week

Upcoming deadlines:

Tuesday (5/1)

PL HW 15

Wednesday (5/2)

Written Assignment 6

Quiz 7

CBTF (Thurs-Tues: 5/3-8)

50 minutes

Fluid Pressure -Virtual Workhttp://knowledge.wharton.upenn.edu

Page 3: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

Chapter 11: Virtual Work

Page 4: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

Goals and Objectives

Introduce the principle of virtual work

Show how it applies to determining the equilibrium

configuration of a series of pin-connected members

Page 5: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

Recap: Principle of Virtual WorkThe principle of virtual work states that if a body is in equilibrium, then the

algebraic sum of the virtual work done by all the forces and couple

moments acting on the body is zero for any virtual displacement of the

body. Thus, 𝛿𝑈 = 0

𝛿𝑈 = Σ 𝑭 ∙ 𝛿𝒓 + Σ 𝑴 ∙ 𝛿𝜽 = 0

For 2D:

𝛿𝑈 = Σ 𝑭 ∙ 𝛿𝐫 + Σ 𝑀 𝛿𝜃 = 0

Page 6: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

Procedure for Analysis

1. Draw FBD of the entire system and provide coordinate system

2. Sketch the “deflected position” of the system

3. Define position coordinates measured from a fixed point and select

the parallel line of action component and remove forces that do no

work

4. Differentiate position coordinates to obtain virtual displacement

5. Write the virtual work equation and express the virtual work of each

force/ couple moment

6. Factor out the comment virtual displacement term and solve

Page 7: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

Determine the tension in the cable AC.

The lamp weighs 10 lb.

Page 8: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

Determine the tension in the cable AC.

The lamp weighs 10 lb.

Page 9: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

The scissors jack supports a load P. Determine axial force in the screw necessary

for equilibrium when the jack is in the position shown. Each of the four links has a

length l and is pin-connected at its center. Points B and D can move horizontally.

Page 10: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth
Page 11: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

The scissors jack supports a load P. Determine axial force in the screw necessary

for equilibrium when the jack is in the position shown. Each of the four links has a

length l and is pin-connected at its center. Points B and D can move horizontally.

Page 12: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

Determine the mass of A and B required to hold the 400 g

desk lamp in balance for any angles θ and ϕ. Neglect the

weight of the mechanism and the size of the lamp.

Assume that pins are frictionless.

Page 13: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

Solution

For equilibrium, we require

0L L A A B Bm g y m g y m g y W

Use the principle of virtual work. This problem has two degrees of freedom: and .

Let and be the virtual displacements from the equilibrium positions. We assume that

all the pins are frictionless.

Here,

( )sin sin ( )cos cos

sin cos

sin cos

L L

A A

B B

y a b b y a b b

y a y a

y a y a

Page 14: Statics - TAM 210 & TAM 211...b 300 mm, and m L 400 g, 300 (400 g) 1.6 kg , 75 75 300 (400 g) 2.0 kg. 75 A B m m Title Statics - TAM 210 & TAM 211 Author Hsiao-Wecksler, Elizabeth

Substitution gives

( )cos cos

cos cos 0

L B

L A

m g a b m g a

m g b m g a

W

or

( ) cos

cos 0 .

L B

L A

m g a b m g a

m g b m g a

W

If this relation is to hold for arbitrary and , for the general case where

cos 0 and cos 0 , we require

and .A L B L

b a bm m m m

a a

For 75 mma , 300 mmb , and 400 gLm ,

300(400 g) 1.6 kg ,

75

75 300(400 g) 2.0 kg .

75

A

B

m

m