Ch.06 Structural Analysis

2/13/2013

1

06. Structural Analysis

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Engineering Mechanics – Statics 6.01 Structural Analysis

Chapter Objectives

• To show how to determine the forces in the members of a truss

using the method of joints and the method of sections

• To analyze the forces acting on the members of frames and

machines composed of pin-connected members



§1. Simple Trusses

- Truss: a structure composed of slender members joined

together at their end points

- The members are usually wooden struts or metal bars

- Planar trusses lie in a single plane and are often used to

support roofs and bridges

- The roof load is transmitted to the truss at the joints by means

of a series of purlins (xà gồ)



§1. Simple Trusses

- The analysis of the forces developed in the truss members will

be two-dimensional

- In the case of a bridge, the load on the deck is first transmitted

to stringers, then to floor beams, and finally to the joints of the

two supporting side trusses

- The bridge truss loading is also coplanar



§1. Simple Trusses

- Force analysis of a truss: determine the forces in each truss

member when the truss is subjected to a given loading

- Assumption for design

• All loadings are applied at the joints

• The weight of the truss members is neglected

• The members are joined together by smooth pins



§1. Simple Trusses

• The joint connections are usually formed by bolting or

welding the ends of the members to a common plate, called

a gusset plate, or by simply passing a large bolt or pin

through each of the members

• The force acting at each end of the member is directed along

the axis of the member

+ If this tends to elongate the member, it is a tensile force

+ If it tends to shorten the member, it is a compressive force



http://www.google.com.vn/url?sa=i&source=images&cd=&cad=rja&docid=Cao6h4QuRvvw-M&tbnid=6xjOrglOuMqVMM:&ved=0CAgQjRwwADjCAQ&url=http://blog.wolfram.com/2007/08/03/the-space-of-all-possible-bridge-shapes/&ei=0jwYUY2ED8itiAetx4Aw&psig=AFQjCNEp4a9ccFufJcuc0Hz-GeUyE2q0Tw&ust=1360629330292810

2/13/2013

2

§1. Simple Trusses

- Simple trusses

• A simple truss is a planar truss which begins with a triangular

element and can be expanded by adding two members and

a joint

• The number of members, 𝑀, and the number of joints, 𝐽, are

related by the equation 𝑀 = 2𝐽– 3



§2. The Method of Joints

- If the entire truss is in equilibrium, then each of its joints is also

in equilibrium

- The method of joints

• Draw a free-body diagram of the whole truss and determine

the support reactions (using scalar equations of equilibrium)

• Draw the free-body diagram diagram of a joint with one or

two unknowns

• Apply the scalar equations of equilibrium

∑𝐹𝑥 = 0

∑𝐹𝑦 = 0

to determine the unknown(s)

• Repeat steps 2 and 3 at each joint in succession until all

forces are determined




- For example, consider the pin at joint 𝐵 of the truss

• Three forces act on the pin: the 500𝑁 force and the forces

exerted by members 𝐵𝐴 and 𝐵𝐶

• 𝐹 𝐵𝐴 is pulling on the pin, which means that member 𝐵𝐴 is in

tension

• 𝐹 𝐵𝐶 is pushing on the pin, and consequently member 𝐵𝐶 is in

compression




- When using the method of joints, always start at a joint having

at least one known force and at most two unknown forces and

can be solved to determine the two unknowns

- The correct sense of an unknown member force can be

determined using one of two possible methods

- The correct sense of direction of an unknown member force

can be determined by inspection

- Assume the unknown member forces acting on the joint’s free-

body diagram to be in tension




- Example 6.1 Determine the force in each member of the

truss and indicate whether the members are in tension or

compression

Solution

Joint 𝐵

+→ ∑𝐹𝑥 = 0: 500 − 𝐹𝐵𝐶𝑠𝑖𝑛450 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐹𝐵𝐶𝑐𝑜𝑠450 − 𝐹𝐵𝐴 = 0

⟹ 𝐹𝐵𝐶 = 707.1𝑁, 𝐹𝐵𝐴 = 500𝑁

Joint 𝐶

+→ ∑𝐹𝑥 = 0: −𝐹𝐶𝐴 + 707.1𝑐𝑜𝑠450 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐶𝑦 − 707.1𝑠𝑖𝑛450 = 0

⟹ 𝐹𝐶𝐴 = 500𝑁, 𝐶𝑦 = 500𝑁




Joint 𝐴

+→ ∑𝐹𝑥 = 0: 500 − 𝐴𝑥 = 0

+ ↑ ∑𝐹𝑦 = 0: 500 − 𝐴𝑦 = 0

⟹ 𝐴𝑥 = 500𝑁, 𝐴𝑦 = 500𝑁

Note: The free-body diagram of each joint

shows the effects of all the connected

members and external forces applied to

the joint



2/13/2013

3


- Example 6.2 Determine the force in each member of the truss

and indicate if the members are in tension or compression

Solution

Joint 𝐶

+→ ∑𝐹𝑥 = 0: 𝐹𝐵𝐶𝑠𝑖𝑛450 − 400 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐹𝐶𝐷 − 𝐹𝐵𝐶𝑐𝑜𝑠450 = 0

⟹ 𝐹𝐵𝐶 = 565.69𝑁, 𝐹𝐶𝐷 = 400𝑁

Joint 𝐷

+↘ ∑𝐹𝑥′ = 0: 𝐹𝐵𝐷 − 𝐹𝐴𝐷𝑐𝑜𝑠150

−400𝑐𝑜𝑠300 = 0

+↗ ∑𝐹𝑦′ = 0: −𝐹𝐴𝐷𝑠𝑖𝑛150 −400𝑠𝑖𝑛300 = 0

⟹ 𝐹𝐴𝐷 = −772.74𝑁, 𝐹𝐵𝐷 = 1092.82𝑁




𝐹𝐵𝐶 = 565.69𝑁

𝐹𝐶𝐷 = 400𝑁

𝐹𝐴𝐷 = −772.74𝑁

𝐹𝐵𝐷 = 1092.82𝑁

Joint 𝐴

+→ ∑𝐹𝑥 = 0: 772.74𝑐𝑜𝑠450 − 𝐹𝐴𝐵 = 0

+ ↑ ∑𝐹𝑦 = 0: −772.74𝑠𝑖𝑛450 + 𝐴𝑦 = 0

⟹ 𝐹𝐴𝐵 = 546𝑁, 𝐴𝑦 = 546𝑁




- Example 6.3 Determine the force in

each member of the truss and indicate

whether the members are in tension or

compression

Solution

Free-body diagram

Support reactions

+→ ∑𝐹𝑥 = 0: −𝐶𝑥 + 600 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐴𝑦 − 400 − 𝐶𝑦 = 0

+↺ ∑𝑀𝐶 = 0: −𝐴𝑦 × 6 + 400 × 3

+600 × 4 = 0

⟹ 𝐶𝑥 = 600𝑁, 𝐶𝑦 = 200𝑁, 𝐴𝑦 = 600𝑁




Joint 𝐴

+→ ∑𝐹𝑥 = 0: 600 −4

5𝐹𝐴𝐵 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐹𝐴𝐷 −3

5750 = 0

⟹ 𝐹𝐴𝐵 = 750𝑁 (C), 𝐹𝐴𝐷 = 450𝑁 (T)

Joint 𝐷

+→ ∑𝐹𝑥 = 0: −450 +3

5𝐹𝐷𝐵 + 600 = 0

+ ↑ ∑𝐹𝑦 = 0: −𝐹𝐷𝐶 −4

5(−250) = 0

⟹ 𝐹𝐷𝐵 = 250𝑁 (C), 𝐹𝐷𝐶 = 200𝑁 (C)

Joint 𝐶

+→ ∑𝐹𝑥 = 0: 𝐹𝐶𝐵 − 600 = 0

⟹ 𝐹𝐶𝐵 = 600𝑁 (C)






§3. Zero-Force Members

- Zero-force member: member support no loading

- If they are not needed why do we see them in structures?

Used to increase the stability of the truss during construction

and to provide added support for various different loading

conditions

- Can generally be found by inspection of each of the joints



2/13/2013

4


- Example

Consider the truss

• At Joint 𝐴, it is seen that 𝐴𝐵 and 𝐴𝐹

are zero-force members

+→ ∑𝐹𝑥 = 0: 𝐹𝐴𝐵 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐹𝐴𝐹 = 0

• At Joint 𝐷, it is seen that 𝐷𝐶 and 𝐷𝐸

are zero-force members

+→ ∑𝐹𝑥 = 0: 𝐹𝐷𝐸 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐹𝐷𝐶𝑠𝑖𝑛𝜃 = 0

⟹ If only two members form a truss joint and no external load

or support reaction is applied to the joint, the two members

must be zero-force members




- Example: Now consider the truss

• At Joint 𝐷, it is seen that 𝐷𝐴 is zero-

force members

+→ ∑𝐹𝑥 = 0: 𝐹𝐷𝐴 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐹𝐷𝐶 = 𝐹𝐷𝐸

• At Joint 𝐶, it is seen that 𝐶𝐴 is zero-

force members

+↙ ∑𝐹𝑥 = 0: 𝐹𝐶𝐴𝑠𝑖𝑛𝜃 = 0

+↘ ∑𝐹𝑦 = 0: 𝐹𝐶𝐵 = 𝐹𝐶𝐷

⟹ If three members form a truss joint for

which two of the members are collinear,

the third member is a zero-force

member provided no external force or

support reaction is applied to the joint




- Example 6.4 Using the method of joints, determine all the

zero-force members of the Fink roof truss. Assume all joints

are pin connected

Solution The joint geometries that have three

members for which two are collinear: 𝐺

Joint 𝐺

+ ↑ ∑𝐹𝑦 = 0: 𝐹𝐺𝐶 = 0

𝐺𝐶 is a zero-force member ⟹

the 5𝑘𝑁 load at 𝐶 must be

supported by 𝐶𝐵, 𝐶𝐻, 𝐶𝐹, 𝐶𝐷

Joint 𝐷

+↙ ∑𝐹𝑥 = 0: 𝐹𝐷𝐹 = 0

Joint 𝐹

+ ↑ ∑𝐹𝑦 = 0: 𝐹𝐹𝐶𝑐𝑜𝑠𝜃 = 0



Fundamental Problems

- F6.1 Determine the force in each member of the truss. State

if the members are in tension or compression









- F6.3 Determine the force in members 𝐴𝐸 and 𝐷𝐶. State if the

members are in tension or compression



2/13/2013

5


- F6.4 Determine the greatest load 𝑃 that can be applied to the

truss so that none of the members are subjected to a force

exceeding either 2𝑘𝑁 in tension or 1.5𝑘𝑁 in compression




- F6.5 Identify the zero-force members in the truss










§4. The Method of Sections

- When we need to find the force in only a few members of a

truss we can analyze the truss using the method of sections

- This method is based on the principle that if the truss is in

equilibrium then any segment of the truss is also in equilibrium

- If the forces within the members are to be determined, then an

imaginary section can be used to cut each member into two

parts




- The method of sections can also be used to cut or section the

members of an entire truss

- For example, consider the truss

- The free-body diagrams of the two segments are then

If the forces in members 𝐵𝐶 ,

𝐺𝐶, 𝐺𝐹 are to be determined,

then section 𝑎 − 𝑎 would be

appropriate




- Members 𝐵𝐶 and 𝐺𝐶 are assumed to be in tension since they

are subjected to a pull, whereas 𝐺𝐹 in compression since it is

subjected to a push

- The three unknown member forces 𝐹 𝐵𝐶, 𝐹 𝐺𝐶, and 𝐹 𝐺𝐹 can be

obtained by applying the three equilibrium equations of the

segments

- The three support reactions 𝐷𝑥, 𝐷𝑦 and 𝐸𝑥 can be obtained by

considering a free-body diagram of the entire truss

2/13/2013

6




- Method of sections

• Where to make the cut?

+ where you need to determine forces, and

+ where the total number of unknowns does not exceed three

• Decide which side of the cut truss will be easier to work with

(minimize the number unknowns)

• Determine any necessary support reactions by drawing the

free-body diagram of the entire truss

• Draw the free-body diagram of the selected part of the cut

truss. We need to indicate the unknown forces at the cut

members

• Apply the equations of equilibrium to the selected cut section

to solve for the unknown member forces




- Example 6.5 Determine the force in members 𝐺𝐸, 𝐺𝐶, and

𝐵𝐶 of the truss. Indicate whether the members are in tension

or compression

Solution

Support Reactions

+→ ∑𝐹𝑥 = 0: 400 − 𝐴𝑥 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐴𝑦 − 1200 + 𝐷𝑦 = 0

+↺ ∑𝑀𝐴 = 0: −1200 × 8 − 400 × 3 +𝐷𝑦 × 12 = 0

⟹ 𝐴𝑥 = 400𝑁, 𝐴𝑦 = 300𝑁, 𝐷𝑦 = 900𝑁

To determine the force in members 𝐺𝐸,

𝐺𝐶, and 𝐵𝐶 of the truss, section 𝑎 − 𝑎 has been chosen since

it cuts through the three members whose forces are to be

determined




Free-Body Diagram: for the analysis the free-body diagram of

the left portion of the sectioned truss will be used, since it

involves the least number of forces

Equations of equilibrium

+ ↑ ∑𝐹𝑦 = 0: 300 −3

5𝐹𝐺𝐶 = 0

+↺ ∑𝑀𝐺 = 0: −300 × 4 − 400 × 3

+𝐹𝐵𝐶 × 3 = 0

+↺ ∑𝑀𝐶 = 0: −300 × 8 + 𝐹𝐺𝐸 × 3 = 0

⟹ 𝐹𝐵𝐶 = 800𝑁(T)

𝐹𝐺𝐸 = 800𝑁(C)

𝐹𝐺𝐶 = 500𝑁(T)




- Example 6.6 Determine the force in member 𝐶𝐹 of the truss.

Indicate whether the member is in tension or compression.

Assume each member is pin connected

Solution

Free-body diagram


+↺ ∑𝑀𝑂 = 0: −𝐹𝐶𝐹𝑠𝑖𝑛450 × 12

+3 × 8 − 4.75 × 4 = 0

⟹ 𝐹𝐶𝐹 = 0.589𝑁(C)




- Example 6.7 Determine the force in member 𝐸𝐵 of the roof

truss. Indicate whether the member is in tension or

compression

Solution

Free-body diagram


+→ ∑𝐹𝑥 = 0: 𝐹𝐸𝐹𝑐𝑜𝑠300 −3𝑐𝑜𝑠300 = 0

+ ↑ ∑𝐹𝑦 = 0: 2 3𝑠𝑖𝑛300 −1−𝐹𝐸𝐵 = 0

+↺ ∑𝑀𝐵 = 0: 1 × 4 + 3 × 2 − 4 × 4

+𝐹𝐸𝐷𝑠𝑖𝑛300 × 4 = 0

⟹ 𝐹𝐸𝐹 = 3𝑘𝑁(C)

𝐹𝐸𝐵 = 2𝑘𝑁(T)

𝐹𝐸𝐷 = 3𝑘𝑁(C)




- F6.7 Determine the force in members 𝐵𝐶, 𝐶𝐹, and 𝐹𝐸. State


2/13/2013

7




- F6.8 Determine the force in members 𝐿𝐾, 𝐾𝐶, and 𝐶𝐷 of the

Pratt truss. State if the members are in tension or compression




- F6.9 Determine the force in members 𝐾𝐽, 𝐾𝐷, and 𝐶𝐷 of the

Pratt truss. State if the members are in tension or compression




- F6.10 Determine the force in members 𝐸𝐹, 𝐶𝐹, and 𝐵𝐶 of the

truss. State if the members are in tension or compression




- F6.11 Determine the force in members 𝐺𝐹, 𝐺𝐷, and 𝐶𝐷 of the





- F6.12 Determine the force in members 𝐷𝐶, 𝐻𝐼, and 𝐽𝐼 of the




§5. Space Trusses

- A space truss consists of members joined together at their

ends to form a stable three-dimensional structure

For economic reasons, large

electrical transmission towers

are often constructed using

space trusses

Typical roof-supporting

space truss. Notice the

use of ball-and socket

joints for the connections

The simplest form of a space

truss is a tetrahedron, formed

by connecting six members

together

2/13/2013

8



§5. Space Trusses

- Assumptions for Design

• The members of a space truss may be treated as two-force

members provided the external loading is applied at the

joints and the joints consist of ball-and-socket connections

• These assumptions are justified if the welded or bolted

connections of the joined members intersect at a common

point and the weight of the members can be neglected

• In cases where the weight of a member is to be included in

the analysis, it is generally satisfactory to apply it as a

vertical force, half of its magnitude applied at each end of the

member



§5. Space Trusses

- Example 6.8 Determine the forces acting in the members of

the space truss. Indicate whether the

members are in tension or compression

Solution

Since there are one known force and

three unknown forces acting at joint 𝐴,

the force analysis of the truss will begin

at this joint

𝑃 = −4𝑗

𝐹 𝐴𝐵 = 𝐹𝐴𝐵𝑗

𝐹 𝐴𝐶 = −𝐹𝐴𝐶𝑘

𝐹 𝐴𝐸 = 𝐹𝐴𝐸

𝑟 𝐴𝐸

𝑟𝐴𝐸

= 𝐹𝐴𝐸(0.577𝑖 + 0.577𝑗 − 0.577𝑘)



§5. Space Trusses

Joint 𝐴

For equilibrium ∑𝐹 = 0

⟹ 𝑃 + 𝐹 𝐴𝐵 + 𝐹 𝐴𝐶 + 𝐹 𝐴𝐸 = 0

⟹ 0.577𝑖 + −4 + 𝐹𝐴𝐵 + 0.557𝐹𝐴𝐸 𝑗

+ −𝐹𝐴𝐶 − 0.557𝐹𝐴𝐸 𝑘 = 0

⟹ 𝐹𝐴𝐶 = 0, 𝐹𝐴𝐸 = 0, 𝐹𝐴𝐵 = 4𝑘𝑁(T)

Joint 𝐵


+→ ∑𝐹𝑥 = 0: −𝑅𝐵𝑐𝑜𝑠450 +0.707𝐹𝐵𝐸 = 0

+ ↑ ∑𝐹𝑦 = 0: −4 − 𝑅𝐵𝑠𝑖𝑛450 = 0

+↺ ∑𝑀𝐵 = 0: 2 + 𝐹𝐵𝐷 − 0.707𝐹𝐵𝐸 = 0

⟹ 𝑅𝐵 = 5.66𝑘𝑁(T), 𝐹𝐵𝐸 = 5.66𝑘𝑁(T)

𝐹𝐵𝐷 = 2𝑘𝑁(C)



§6. Frames and Machines

- Frames are generally stationary and support external loads

- Machines contain moving parts and are designed to alter the

effect of forces

- Frames and machines have at least one multi-force member.

(Recall that trusses have two-force members)




- Analysis of Frames and Machines

• Draw the free-body diagram of the frame or machine and its

members

• Develop a strategy to apply the equations of equilibrium to

solve for the unknowns




- Example 6.9 Draw the free-body diagram of (a) each

member, (b) the pin at 𝐵 , and (c) the two

members connected together

Solution (b)

(c) (a)

2/13/2013

9




- Example 6.10 A constant tension in the conveyor belt is

maintained by using the device. Draw

the free-body diagrams of the frame

and the cylinder that the belt

surrounds. The suspended block has a

weight of 𝑊

Solution

or




- Example 6.11 Draw the free-body diagrams of (a) the entire

frame including the pulleys and cords, (b) the frame without

the pulleys and cords, and (c) each of the pulleys

Solution




- Example 6.12 Draw the free-body diagrams of

the bucket and the vertical boom of the

backhoe shown in the photo. The bucket and

its contents have a weight 𝑊 . Neglect the

weight of the members

Solution




- Example 6.13 Draw the free-body diagram of each

part of the smooth piston and link mechanism used to

crush recycled cans

Solution




- Example 6.14 Determine the horizontal and vertical

components of force which the pin at 𝐶

exerts on member 𝐵𝐶 of the frame

Solution

Free-body diagram


+→ ∑𝐹𝑥 = 0: 𝐹𝐴𝐵𝑐𝑜𝑠600 − 𝐶𝑥 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐹𝐴𝐵𝑠𝑖𝑛600 − 2000 + 𝐶𝑦 = 0

+↺ ∑𝑀𝐵 = 0: 2000 × 2 − 𝐹𝐴𝐵𝑠𝑖𝑛600 × 4 = 0

⟹ 𝐹𝐴𝐵 = 1154.7𝑘𝑁, 𝐶𝑥 = 577𝑁, 𝐶𝑦 = 1000𝑁

Note: we can applying

equilibrium equations for

each member




- Example 6.15 The compound beam is pin connected at 𝐵.

Determine the components of reaction at its supports. Neglect

its weight and thickness

Solution

Free-body diagram


Segment 𝐵𝐶

+→ ∑𝐹𝑥 = 0: 𝐵𝑥 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐵𝑦 − 8 + 𝐶𝑦 = 0

+↺ ∑𝑀𝐵 = 0: −8 × 1 + 𝐶𝑦 × 2 = 0

2/13/2013

10




Segment 𝐵𝐶

+→ ∑𝐹𝑥 = 0: 𝐵𝑥 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐵𝑦 − 8 + 𝐶𝑦 = 0

+↺ ∑𝑀𝐵 = 0: −8 × 1 + 𝐶𝑦 × 2 = 0

Segment 𝐴𝐵

+→ ∑𝐹𝑥 = 0: 𝐴𝑥 − 103

5+ 𝐵𝑥 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐴𝑦 − 104

5− 𝐵𝑦 = 0

+↺ ∑𝑀𝐴 = 0: 𝑀𝐴 −104

5×2+𝐵𝑦 ×4=0

⟹ 𝐴𝑥 = 6𝑘𝑁, 𝐴𝑦 = 12𝑘𝑁

𝐵𝑥 = 0, 𝐵𝑦 = 4𝑘𝑁

𝐶𝑥 = 4𝑘𝑁

𝑀𝐴 = 32𝑘𝑁𝑚




- Example 6.16 A 500𝑘𝑔 elevator car is being hoisted by motor

𝐴 using the pulley system shown. If the car is

traveling with a constant speed, determine

the force developed in the two cables.

Neglect the mass of the cable and pulleys

Solution

Free-body diagram


for pulley 𝐶

+ ↑ ∑𝐹𝑦 = 0: 𝑇2 − 2𝑇1 = 0

for the elevator car

+ ↑ ∑𝐹𝑦 = 0: 3𝑇1+2𝑇2−500×9.81=0

⟹ 𝑇1 = 701.71𝑁, 𝑇2 = 1401𝑁




- Example 6.17 The smooth disk is pinned at 𝐷 and has a

weight of 20𝑁. Neglecting the weights of the other members,

determine the horizontal and vertical components of reaction

at pins 𝐵 and 𝐷

Solution

Free-body diagram

Support Reactions

+→ ∑𝐹𝑥 = 0: 𝐴𝑥 − 𝐶𝑥 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐴𝑦 − 90 = 0

+↺ ∑𝑀𝐴 = 0: −90 × 0.9 + 𝐶𝑥 × 1 = 0

⟹ 𝐴𝑥 = 81𝑁, 𝐴𝑦 = 90𝑁, 𝐶𝑥 = 81𝑁




Member 𝐴𝐵 +→ ∑𝐹𝑥 = 0: 81 − 𝐵𝑥 = 0

+ ↑ ∑𝐹𝑦 = 0: 90 − 𝑁𝐷 + 𝐵𝑦 = 0

+↺ ∑𝑀𝐵 = 0: −90 × 1.8 + 𝑁𝐷 × 0.9 = 0

⟹ 𝐵𝑥 = 81𝑁, 𝐵𝑦 = 90𝑁, 𝑁𝐷 = 180𝑁

+→ ∑𝐹𝑥 = 0: 𝐷𝑥 = 0

+ ↑ ∑𝐹𝑦 = 0: 180 − 90 − 𝐷𝑦 = 0

⟹ 𝐷𝑥 = 0, 𝐷𝑦 = 90𝑁

Disk




- Example 6.18 Determine the tension in the cables and also

the force 𝑃 required to support the 600𝑁 force

using the frictionless pulley system

Solution

Free-body diagram


for pulley 𝐴

+ ↑ ∑𝐹𝑦 = 0: 3𝑃 − 600 = 0

for pulley 𝐵

+ ↑ ∑𝐹𝑦 = 0: 𝑇 − 2𝑃 = 0

for pulley 𝐶

+ ↑ ∑𝐹𝑦 = 0: 𝑅 − 2𝑃 − 𝑇 = 0

⟹ 𝑃 = 200𝑁, 𝑇 = 400𝑁, 𝑅 = 800𝑁




- Example 6.19 The two planks are connected together by

cable 𝐵𝐶 and a smooth spacer 𝐷𝐸. Determine the reactions at

the smooth supports 𝐴 and 𝐹 , and also find the force

developed in the cable and spacer

Solution

Free-body diagram


for plank 𝐴𝐷

+ ↑ ∑𝐹𝑦 = 0: 𝑁𝐴 −100−𝐹𝐵𝐶 +𝐹𝐷𝐸 = 0

+↺ ∑𝑀𝐴 = 0: −100 × 0.6 − 𝐹𝐵𝐶 × 1.2

+𝐹𝐷𝐸 × 1.8 = 0

2/13/2013

11




for plank 𝐴𝐷

+ ↑ ∑𝐹𝑦 = 0: 𝑁𝐴 −100−𝐹𝐵𝐶 +𝐹𝐷𝐸 = 0

+↺ ∑𝑀𝐴 = 0: −100 × 0.6 − 𝐹𝐵𝐶 × 1.2

+𝐹𝐷𝐸 × 1.8 = 0

for plank 𝐶𝐹

+ ↑ ∑𝐹𝑦 = 0: 𝐹𝐵𝐶 −𝐹𝐷𝐸 −200+𝑁𝐹 = 0

+↺ ∑𝑀𝐹 = 0: −𝐹𝐵𝐶 × 1.8 + 𝐹𝐷𝐸 × 1.2

+200 × 0.6 = 0

⟹ 𝑁𝐴 = 120𝑁

𝑁𝐹 = 180𝑁

𝐹𝐵𝐶 = 160𝑁

𝐹𝐷𝐸 = 140𝑁




- Example 6.20 The 75𝑘𝑔 man attempts to

lift the 40𝑘𝑔 uniform beam off the roller

support at 𝐵 . Determine the tension

developed in the cable attached to 𝐵 and

the normal reaction of the man on the beam

when this is about to occur

Solution

Free-body diagram


for pulley 𝐸

+ ↑ ∑𝐹𝑦 = 0: 2𝑇1 −𝑇2 = 0

for the man

+ ↑ ∑𝐹𝑦 = 0: 𝑁𝑚 +2𝑇1 −75×9.81 = 0




for pulley 𝐸

+ ↑ ∑𝐹𝑦 = 0: 2𝑇1 −𝑇2 = 0

for the man

+ ↑ ∑𝐹𝑦 = 0: 𝑁𝑚 +2𝑇1 −75×9.81 = 0

for the beam

+↺ ∑𝑀𝐴 = 0: 𝑇1 × 3 − 40 × 9.81 × 1.5

−𝑁𝑚 × 0.8 = 0

⟹ 𝑇1 = 256𝑁

𝑇2 = 512𝑁

𝑁𝑚 = 224𝑁




- Example 6.21 The frame supports the

50𝑘𝑔 cylinder. Determine the horizontal and

vertical components of reaction at 𝐴 and the

force at 𝐶

Solution

Free-body diagram


for pulley

+→ ∑𝐹𝑥 = 0: 𝐷𝑥 −50×9.81 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐷𝑦 −50×9.81 = 0




for pulley

+→ ∑𝐹𝑥 = 0: 𝐷𝑥 −50×9.81 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐷𝑦 −50×9.81 = 0

for member 𝐴𝐵𝐷

+→ ∑𝐹𝑥 = 0: 𝐴𝑥 −𝐹𝐵𝐶 −𝐷𝑥 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐴𝑦 −𝐷𝑦 = 0

+↺ ∑𝑀𝐹 = 0: 𝐹𝐵𝐶 × 0.6 + 𝐷𝑥 × 0.9 +𝐷𝑦 × 1.2 = 0

⟹ 𝐴𝑥 = 736𝑁

𝐴𝑦 = 490.5𝑁

𝐷𝑥 = 490.5𝑁

𝐷𝑦 = 490.5𝑁

𝐹𝐵𝐶 = 245.25𝑁




- F6.13 Determine the force 𝑃 needed to hold the 60𝑁 weight in

equilibrium

2/13/2013

12




- F6.14 Determine the horizontal and vertical components of

reaction at pin 𝐶




- F6.15 If a 100𝑁 force is applied to the handles of the pliers,

determine the clamping force exerted on the smooth pipe 𝐵

and the magnitude of the resultant force at pin 𝐴




- F6.16 Determine the horizontal and vertical components of

reaction at pin 𝐶




- F6.17 Determine the normal force that the 100𝑁 plate 𝐴 exerts

on the 30𝑁 plate 𝐵




- F6.18 Determine the force 𝑃 needed to lift the load. Also,

determine the proper placement 𝑥 of the hook for equilibrium.

Neglect the weight of the beam

Documents

Ch.06 Structural Analysis