Ch.07 Internal Forces

2/24/2013

1

07. Internal Forces

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

Engineering Mechanics – Statics 7.01 Internal Forces

Chapter Objectives

• To show how to use the method of sections to determine the

internal loadings in a member

• To generalize this procedure by formulating equations that can

be plotted so that they describe the internal shear and moment

throughout a member

• To analyze the forces and study the geometry of cables

supporting a load



§1. Internal Forces Developed in Structural Members

- The design of any structural member requires to know both the

external loads acting on the member and the internal forces

acting within the member in order to be sure the material can

resist these loading

- The concrete supporting a bridge has fractured: What might

have caused it to do this?

• Is it because of the internal forces?

• If so, what are they and how can we

design these structures to make them

safer?




- If a coplanar force system acts on a member, then in general a

resultant internal normal force 𝑁 (acting perpendicular to the

section), shear force 𝑉 (acting along the surface), and bending

moment 𝑀 will act at any cross section along the member




- Steps for determining internal loadings

Internal loadings can be determined by using the method of section

The following example explains the steps that we should follow

to determine the internal forces acting on the cross section at

point 𝐵




• Step1: Determine the support reactions

• Step 2: Cut the beam at 𝐵 and draw a free-body diagram of

one of the halves of the beam (Method of sections)

• Step 3: Apply the equations of equilibrium to the free-body diagram

of each segment and solve for the unknown internal loads

∑𝐹𝑥 = 0, ∑𝐹𝑦 = 0, ∑𝑀𝐵 = 0



Steps

2/24/2013

2


- Sign convention

• a positive normal force creates tension

• a positive shear force will cause the beam segment on which

it acts to rotate clockwise

• a positive bending moment will tend to bend the segment on

which it acts in a concave upward manner




- Example 7.1 Determine the normal force, shear force, and

bending moment acting just to the left,

point 𝐵, and just to the right, point 𝐶, of the

6𝑘𝑁 force on the beam

Solution

Support reactions

+↺ ∑𝑀𝐷 = 0: 9 + 6 × 6 − 𝐴𝑦 × 9 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐴𝑦 − 6 + 𝐷𝑦 = 0

⟹ 𝐴𝑦 = 5𝑘𝑁

𝐷𝑦 = 1𝑘𝑁




Equations of equilibrium

Section 𝐴𝐵

+→ ∑𝐹𝑥 = 0: 𝑁𝐵 = 0

+ ↑ ∑𝐹𝑦 = 0: 5 − 𝑉𝐵 = 0

+↺ ∑𝑀𝐵 = 0: −5 × 3 + 𝑀𝐵 = 0

⟹ 𝑁𝐵 = 0, 𝑉𝐵 = 5𝑘𝑁 , 𝑀𝐵 = 15𝑘𝑁𝑚

Section 𝐴𝐶

+→ ∑𝐹𝑥 = 0: 𝑁𝐶 = 0

+ ↑ ∑𝐹𝑦 = 0: 5 − 6 − 𝑉𝐵 = 0

+↺ ∑𝑀𝐵 = 0: −5 × 3 + 𝑀𝐶 = 0

⟹ 𝑁𝐶 = 0, 𝑉𝐶 = −1𝑘𝑁 , 𝑀𝐶 = 15𝑘𝑁𝑚

The negative sign indicates that 𝑉𝐶 acts in the

opposite sense to that shown on the free-body diagram HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien




bending moment at 𝐶 of the beam

Solution

Free-body diagrams

It is not necessary to find the support

reactions at 𝐴 since segment 𝐵𝐶 of

the beam can be used to determine

the internal loadings at 𝐶

The intensity of the triangular

distributed load at 𝐶 is determined

using similar triangles from the

geometry

𝑤𝐶 = 12001.5

3= 600𝑁/𝑚





+→ ∑𝐹𝑥 = 0: 𝑁𝐶 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝑉𝐶 −1

2600 × 1.5 = 0

+↺ ∑𝑀𝐶 = 0: −𝑀𝐶

−1

2600 × 1.5 × 0.5 = 0

⟹ 𝑁𝐶 = 0, 𝑉𝐶 = 450𝑁 , 𝑀𝐶 = −225𝑁𝑚





bending moment acting at point 𝐵 of the two-member frame

Solution

Support reactions

+→ ∑𝐹𝑥 = 0: −𝐴𝑥 +4

5𝐹𝐷𝐶 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐴𝑦 − 100 +3

5𝐹𝐷𝐶 = 0

+↺ ∑𝑀𝐴 = 0: −100 × 1 +3

5𝐹𝐷𝐶 × 2 = 0

⟹ 𝐹𝐷𝐶 = 83.33𝑁

𝐴𝑥 = 66.67𝑁

𝐴𝑦 = 50𝑁



2/24/2013

3


Free-body diagrams


Applying the equations of equilibrium

to segment 𝐴𝐵

+→ ∑𝐹𝑥 = 0: 𝑁𝐵 − 66.67 = 0

+ ↑ ∑𝐹𝑦 = 0: 50 − 50 − 𝑉𝐵 = 0

+↺ ∑𝑀𝐵 = 0: −50×1+50×0.5+𝑀𝐵 =0

⟹ 𝑁𝐵 = 66.67𝑁, 𝑉𝐵 = 0, 𝑀𝐵 = 25𝑁𝑚

We can obtain these same results using segment 𝐵𝐶





bending moment acting at point 𝐸 of the frame loaded

Solution

Support reactions

for the pin

+ ↑ ∑𝐹𝑦 = 0: 𝑅𝑠𝑖𝑛450 − 600 = 0 ⟹ 𝑅 = 848.5𝑁

Free-body diagram


+→ ∑𝐹𝑥 = 0: 848.5𝑐𝑜𝑠450 − 𝑉𝐸 = 0

+ ↑ ∑𝐹𝑦 = 0: −848.5𝑠𝑖𝑛450 + 𝑁𝐸 = 0

+↺ ∑𝑀𝐵 = 0: 848.5𝑐𝑜𝑠450 × 0.5 − 𝑀𝐸 = 0



𝑉𝐸 = 600𝑁 ⟹ 𝑁𝐸 = 600𝑁

𝑀𝐸 = 300𝑁𝑚


- Example 7.5 The uniform sign has a mass of

650𝑘𝑔 and is supported on the fixed column.

Design codes indicate that the expected

maximum uniform wind loading that will occur

in the area where it is located is 900𝑃𝑎 .

Determine the internal loadings at 𝐴

Solution

The idealized model for the sign

Free-body diagram

the sign has a weight

𝑊 = 650 × 9.81 = 6.376𝑁

the wind creates a resultant force

𝐹𝑤 = 900 × 6 × 2.5 = 13.5𝑘𝑁





∑𝐹 = 0: 𝐹 𝐴 − 13.5𝑖 − 6.376𝑘 = 0

⟹ 𝐹 𝐴 = 13.5𝑖 + 6.376𝑘

∑𝑀𝐴 = 0: 𝑀𝐴 + 𝑟 × (𝐹 𝑤 + 𝑊) = 0

⟹ 𝑀𝐴 +𝑖 𝑗 𝑘0 3 5.25

−13.5 0 −6.376

= 0

⟹ 𝑀𝐴 = 19.1𝑖 + 70.9𝑗 − 40.5𝑘



Fundamental Problems

- F7.1 Determine the normal force, shear force, and moment

at point 𝐶





at point 𝐶



2/24/2013

4



at point 𝐶





at point 𝐶





at point 𝐶





at point 𝐶. Assume 𝐴 is pinned and 𝐵 is a roller



§2. Shear and Moment Equations and Diagrams

- Beams are designed to support loads perpendicular to their

axes

- The design of a beam requires a detailed

knowledge of the variation of the internal

shear force 𝑉 and bending moment 𝑀 acting

at each point along the axis of the beam

- To construct the shear and moment

diagrams, it is necessary to section the member at an arbitrary

point, located at distance 𝑥 from the left end

- 𝑉 and 𝑀 are functions of the position 𝑥 along the beam’s axis

can be obtained using the method of sections




- The graphical variations of 𝑉 and 𝑀 as functions of 𝑥 are

termed the shear diagram and bending

moment diagram respectively

shear diagram 𝑉(𝑥)

bending moment diagram 𝑀(𝑥)



2/24/2013

5


- Example 7.6 Draw the shear and moment diagrams for the

shaft. The support at 𝐴 is a thrust bearing

and the support at 𝐶 is a journal bearing

Solution

Support reactions

Shear and moment function

0 ≤ 𝑥 < 2𝑚

+ ↑ ∑𝐹𝑦 = 0: 2.5 − V = 0 ⟹ 𝑉 = 2.5

+↺ ∑𝑀 = 0: −2.5𝑥 + 𝑀 = 0 ⟹ 𝑀 = 2.5𝑥





0 ≤ 𝑥 < 2𝑚

+ ↑ ∑𝐹𝑦 = 0: 2.5 − 𝑉 = 0 ⟹ 𝑉 = 2.5

+↺ ∑𝑀 = 0: −2.5𝑥 + 𝑀 = 0 ⟹ 𝑀 = 2.5𝑥

2𝑚 < 𝑥 ≤ 4𝑚

+ ↑ ∑𝐹𝑦 = 0: 2.5−5−𝑉 = 0 ⟹ 𝑉 = −2.5

+↺ ∑𝑀 = 0: 𝑀 + 5 𝑥 − 2 − 2.5 = 0

⟹ 𝑀 = 10 − 2.5𝑥

Shear and moment diagrams




- Example 7.7 Draw the shear and moment

diagrams for the beam

Solution

Support reactions


+ ↑ ∑𝐹𝑦 = 0: 9 −1

3𝑥2 − 𝑉 = 0

+↺ ∑𝑀 = 0: 𝑀 +1

3𝑥2 𝑥

3− 9𝑥 = 0

⟹ 𝑉 = 9 −1

3𝑥2

𝑀 = −1

9𝑥3 + 9𝑥

Shear and moment diagram




- F7.7 Determine the shear and moment as a function of 𝑥,

and then draw shear and moment diagrams













2/24/2013

6








where 0 ≤ 𝑥 < 3𝑚 and 3𝑚 < 𝑥 ≤ 6𝑚 then draw shear and

moment diagrams





where 0 ≤ 𝑥 < 3𝑚 and 3𝑚 < 𝑥 ≤ 6𝑚 then draw shear and

moment diagrams



§3. Relations between Distributed Load, Shear, and Moment

- Distributed Load

𝑤 = 𝑤(𝑥): distributed load (+ ↑)

𝐹 1, 𝐹 2: concentrated forces (+ ↑)

𝑀1, 𝑀2: concentrated moments (+↺)

Consider a free-body diagram for a small segment of the beam

having a length ∆𝑥 is chosen at a point 𝑥 along the beam

which is not subjected to a concentrated force or couple

moment

The distributed loading has been replaced by

a resultant force ∆𝐹 = 𝑤(𝑥)∆𝑥 that acts at a

fractional distance 𝑘(∆𝑥) from the right end,

where 0 < 𝑘 < 1 (if 𝑤 𝑥 is uniform, 𝑘 = 1/2)




- Relation between the distributed load and shear

Apply the force equation of equilibrium to the

segment

+ ↑ ∑𝐹𝑦 = 0: 𝑉 + 𝑤 𝑥 ∆𝑥 − 𝑉 + ∆𝑉 = 0

⟹ 𝑤 𝑥 =∆𝑉

∆𝑥= lim

∆𝑥→0

∆𝑉

∆𝑥=

𝑑𝑉

𝑑𝑥

: slope of shear diagram = distributed load intensity

: change in shear = area under loading curve



𝑑𝑉

𝑑𝑥= 𝑤 𝑥

∆𝑉 = 𝑤 𝑥 𝑑𝑥


- Relation between shear and moment

Apply the moment equation of equilibrium

about point 𝑂 on the free-body diagram

+↺ ∑𝑀𝑂 = 0: 𝑀 + ∆𝑀 − 𝑤 𝑥 ∆𝑥 𝑘∆𝑥

−𝑉∆𝑥 − 𝑀 = 0

⟹ 𝑉 =∆𝑀

∆𝑥− 𝑘𝑤 𝑥 ∆𝑥 = lim

∆𝑥→0

∆𝑀

∆𝑥=

𝑑𝑀

𝑑𝑥

: slope of moment diagram = shear

: change in moment = area under shear diagram



𝑑𝑀

𝑑𝑥= 𝑉

∆𝑀 = 𝑉𝑑𝑥

2/24/2013

7


- Force

Consider a free-body diagram of a small

segment of the beam, taken from under one

of the forces

Here force equilibrium requires

+ ↑ ∑𝐹𝑦 = 0: 𝑉 + 𝐹 − 𝑉 + ∆𝑉 = 0

⟹ ∆𝑉 = 𝐹

The shear diagram will “jump” upward/downward when 𝐹 acts

upward/downward on the beam




- Couple moment

Consider a segment of the beam that is

located at the couple moment, the free-body

diagram

Letting moment equilibrium requires

+↺ ∑𝑀𝑂 = 0: − 𝑀 +𝑉∆𝑥 −𝑀𝑂 +𝑀 +∆𝑀 = 0

⟹ ∆𝑀 = 𝑀𝑂 (∆𝑥 → 0)

The moment diagram will “jump” upward/downward if 𝑀𝑂 is

clockwise/counterclockwise





cantilever beam

Support reactions

Shear diagram

Segment 𝐴𝐶

no distributed load ⟹ constant shear

at 𝐴, 𝑉 0 = −2𝑘𝑁

at 𝐶, 𝑉 2 = −2𝑘𝑁

Segment 𝐶𝐵

constant distributed load ⟹ linear shear

at 𝐶, 𝑉 2 = −2𝑘𝑁

at 𝐵, 𝑉 4 = 𝑉 2 + ∆𝑉

= −2+ −1.5 ×2 = −5𝑘𝑁



Solution


Moment diagram

Segment 𝐴𝐶

constant shear ⟹ linear moment

at 𝐴, 𝑀 0 = 0

at 𝐶, 𝑀 2 = 𝑀 0 + ∆𝑀

= 0 + −2 × 2

= −4𝑘𝑁𝑚

Segment 𝐶𝐵

linear shear ⟹ parabolic moment

at 𝐶, 𝑀 2 = −4𝑘𝑁𝑚

at 𝐶, 𝑀 4 = 𝑀 2 + ∆𝑀

= −4+[ −2 + −5 ]×2/2

= −11𝑘𝑁𝑚





overhang beam

Solution

Support reactions

Shear diagram

Segment 𝐴𝐵

no distributed load ⟹ constant shear

at 𝐴, 𝑉 0 = −2𝑘𝑁

at 𝐵, 𝑉 4 = −2𝑘𝑁

Segment 𝐵𝐶

constant distributed load ⟹ linear shear

at 𝐵, 𝑉 4 = −2𝑘𝑁 + 10𝑘𝑁 = 8𝑘𝑁

at 𝐶, 𝑉 6 = 0




Moment diagram

Segment 𝐴𝐵

constant shear ⟹ linear moment

at 𝐴, 𝑀 0 = 0

at 𝐵, 𝑀 4 = 𝑀 0 + ∆𝑀

= 0 + −2 × 4

= −8𝑘𝑁𝑚

Segment 𝐵𝐶

linear shear ⟹ parabolic moment

at 𝐵, 𝑀 4 = −8𝑘𝑁𝑚

at 𝐶, 𝑀 6 = 0



2/24/2013

8


- Example 7.10 The shaft is supported by

a thrust bearing at 𝐴 and a journal

bearing at 𝐵 . Draw the shear and

moment diagrams

Solution

Support reactions

Shear diagram

linear load ⟹ parabolic shear diagram

at 𝐴, 𝑉 0 = 120𝑁

at 𝐵, 𝑉 3.6 = −240𝑁

at 𝐶, 𝑉 𝑥 = 0 = 120−250𝑥2/9 ⟹ 𝑥 = 2.08




- Moment diagram

parabolic shear ⟹ cubic moment diagram

at 𝐴, 𝑀 0 = 0𝑁𝑚

at 𝐵, 𝑀 0 = 0𝑁𝑚

at 𝐶, 𝑉 2.08 = 0 ⟹ 𝑀(2.08) = 𝑀𝑚𝑎𝑥

+↺ ∑𝑀 = 0: −120𝑥 +250

27𝑥3 + 𝑀 = 0

⟹ 𝑀𝑚𝑎𝑥 = 120×2.08−250

272.083 = 258.86𝑁𝑚




- F7.13 Draw the shear and moment diagrams for the beam















2/24/2013

9









§4. Cables

- Flexible cables and chains are often used in structures for

support or to transmit loads from one member to another

- Assumptions are considered in the force analysis of these systems

• The weight of the cable is negligible

• The cable is perfectly flexible and inextensible



§4. Cables

- Cable subjected to concentrated loads

• When a cable of negligible weight

supports several concentrated

loads, the cable takes the form of

several straight line segments,

each of which is subjected to a

constant tensile

• The equilibrium analysis is performed by writing down a

sufficient number of equilibrium equations and equations

describing the geometry of the cable to solve for all the

unknowns leading to a description of the tension in each

segment of the cable



§4. Cables

- Example: Consider the cable shown in the figure, where ℎ, 𝐿1,

𝐿2, 𝐿3, 𝑃1 and 𝑃2 are known

The problem here is to determine the nine unknowns

• The three tensions in each of the three segments

• The four components of reaction at 𝐴 and 𝐵

• The two sags 𝑦𝐶 and 𝑦𝐷 at point 𝐶 and 𝐷



§4. Cables

- Example 7.11 Determine the tension in

each segment of the cable

Solution

There are ten unknowns

• four external reactions 𝐴𝑥, 𝐴𝑦, 𝐸𝑥, 𝐸𝑦

• four cable tensions 𝑇𝐴𝐵, 𝑇𝐵𝐶, 𝑇𝐶𝐷, 𝑇𝐷𝐸

• two sags 𝑦𝐵, 𝑦𝐷

Consider the free-body diagram for the

entire cable

+→ ∑𝐹𝑥 = 0: −𝐴𝑥 + 𝐸𝑥 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐴𝑦 − 4 − 15 − 3 + 𝐸𝑦 = 0

+↺ ∑𝑀𝐸 = 0: −𝐴𝑦 × 18 − 4 × 15

−15 × 10 − 3 × 2 = 0



2/24/2013

10

§4. Cables

Since the sag 𝑦𝐶 = 12𝑚 is known, we

will now consider the leftmost section,

which cuts cable 𝐵𝐶

+→ ∑𝐹𝑥 = 0: 𝑇𝐵𝐶𝑐𝑜𝑠𝜃𝐵𝐶 − 𝐴𝑥 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐴𝑦 − 4 − 𝑇𝐵𝐶𝑠𝑖𝑛𝜃𝐵𝐶 = 0

+↺ ∑𝑀𝐶 = 0: 𝐴𝑥 ×12−𝐴𝑦 ×8+4×5 = 0

Point 𝐴

+→ ∑𝐹𝑥 = 0: −𝐴𝑥 + 𝑇𝐴𝐵𝑐𝑜𝑠𝜃𝐴𝐵 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐴𝑦 − 𝑇𝐴𝐵𝑠𝑖𝑛𝜃𝐴𝐵 = 0

Point 𝐶

+→ ∑𝐹𝑥 = 0: 𝑇𝐶𝐵𝑐𝑜𝑠𝜃𝐶𝐵 + 𝑇𝐶𝐷𝑐𝑜𝑠𝜃𝐶𝐷 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝑇𝐶𝐵𝑠𝑖𝑛𝜃𝐶𝐵 + 𝑇𝐶𝐷𝑠𝑖𝑛𝜃𝐶𝐷

−15 = 0



§4. Cables

Point 𝐸

+→ ∑𝐹𝑥 = 0: 𝐸𝑥 + 𝑇𝐸𝐷𝑐𝑜𝑠𝜃𝐸𝐷 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐸𝑦 − 𝑇𝐸𝐷𝑠𝑖𝑛𝜃𝐸𝐷 = 0

Equilibrium equations: 12

Variables: 12

𝐴𝑥, 𝐴𝑦, 𝐸𝑥, 𝐸𝑦, 𝑇𝐴𝐵, 𝑇𝐵𝐶, 𝑇𝐶𝐷, 𝑇𝐷𝐸

𝜃𝐴𝐵, 𝜃𝐵𝐶, 𝜃𝐴𝐵, 𝜃𝐶𝐷

𝑦𝐵 = 3𝑡𝑎𝑛𝜃𝐴𝐵, 𝑦𝐷 = 2𝑡𝑎𝑛𝜃𝐸𝐷

Result 𝐴𝑥 = 6.33𝑘𝑁, 𝐴𝑦 = 12𝑘𝑁

𝐸𝑥 = 6.33𝑘𝑁, 𝐸𝑦 = 10𝑘𝑁

𝑇𝐴𝐵 = 13.6𝑘𝑁, 𝑇𝐵𝐶 = 10.2𝑘𝑁

𝑇𝐶𝐷 = 9.44𝑘𝑁, 𝑇𝐷𝐸 = 11.8𝑘𝑁

𝑦𝐵 = 5.69𝑚, 𝑦𝐷 = 3.16𝑚



§4. Cables

- Cable Subjected to a Distributed Load

Consider the small segment of the cable having a length ∆𝑠

+→ ∑𝐹𝑥 = 0: −𝑇𝑐𝑜𝑠𝜃 + 𝑇 + ∆𝑇 𝑐𝑜𝑠 (𝜃 + ∆𝜃) = 0

+ ↑ ∑𝐹𝑦 = 0: −𝑇𝑠𝑖𝑛𝜃 − 𝑤(𝑥) × ∆𝑥 + 𝑇 + ∆𝑇 𝑠𝑖𝑛 (𝜃 + ∆𝜃) = 0

+↺ ∑𝑀𝑂 = 0: 𝑤 𝑥 × ∆𝑥 × 𝑘 ∆𝑥 − 𝑇𝑐𝑜𝑠𝜃∆𝑦 + 𝑇𝑠𝑖𝑛𝜃∆𝑥 = 0 HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien


§4. Cables

+→ ∑𝐹𝑥 = 0: −𝑇𝑐𝑜𝑠𝜃 + 𝑇 + ∆𝑇 𝑐𝑜𝑠 (𝜃 + ∆𝜃) = 0

+ ↑ ∑𝐹𝑦 = 0: −𝑇𝑠𝑖𝑛𝜃 − 𝑤(𝑥) × ∆𝑥 + 𝑇 + ∆𝑇 𝑠𝑖𝑛 (𝜃 + ∆𝜃) = 0

+↺ ∑𝑀𝑂 = 0: 𝑤 𝑥 × ∆𝑥 × 𝑘 ∆𝑥 − 𝑇𝑐𝑜𝑠𝜃∆𝑦 + 𝑇𝑠𝑖𝑛𝜃∆𝑥 = 0

Dividing the above equations by ∆𝑥 and taking the limit as

∆𝑥 → 0 and therefore ∆𝑦 → 0, ∆𝜃 → 0, ∆𝑇 → 0

𝑑(𝑇𝑐𝑜𝑠𝜃)

𝑑𝑥= 0 ⟹ 𝑇𝑐𝑜𝑠𝜃 = 𝑐𝑜𝑛𝑠𝑡 = 𝐹𝐻

𝑑(𝑇𝑠𝑖𝑛𝜃)

𝑑𝑥− 𝑤 𝑥 = 0 ⟹ 𝑇𝑠𝑖𝑛𝜃 = 𝑤 𝑥 𝑑𝑥

𝑑𝑦

𝑑𝑥= 𝑡𝑎𝑛𝜃 ⟹ 𝑡𝑎𝑛𝜃 =

𝑑𝑦

𝑑𝑥=

1

𝐹𝐻 𝑤 𝑥 𝑑𝑥

⟹ 𝑦 =1

𝐹𝐻 𝑤 𝑥 𝑑𝑥 𝑑𝑥



§4. Cables

- Example 7.12 The cable

of a suspension

bridge supports half

of the uniform road

surface between the

two towers at 𝐴 and 𝐵.

If this distributed loading is 𝑤0, determine the maximum force

developed in the cable and the cable’s required length. The

span length 𝐿 and sag ℎ are known

Solution

Finding the equation of the shape of the cable

Choose the origin of coordinates at the cable’s center

𝑦 =1

𝐹𝐻 𝑤 𝑥 𝑑𝑥 𝑑𝑥 ⟹ 𝑦 =

1

𝐹𝐻

𝑤0𝑥2

2+ 𝐶1𝑥 + 𝐶2



§4. Cables

Boundary conditions

𝑦 𝑥=0

= 0,𝑦 𝑥=𝐿/2

= ℎ,𝑑𝑦

𝑑𝑥 𝑥=0

= 0 ⟹ 𝐶1 = 0,𝐶2 = 0,𝐹𝐻 =𝑤0𝐿

2

8ℎ

⟹ 𝑦 =1

𝐹𝐻

𝑤0𝑥2

2+ 𝐶1𝑥 + 𝐶2 =

4ℎ

𝐿2𝑥2

Determine the maximum tension in the cable

𝑇𝑐𝑜𝑠𝜃 = 𝑐𝑜𝑛𝑠𝑡 = 𝐹𝐻, 0 ≤ 𝜃 ≤ 𝜋/2

with

𝑑𝑦

𝑑𝑥 𝑥=𝐿/2

= 𝑡𝑎𝑛𝜃𝑚𝑎𝑥 =𝑤0𝑥

𝐹𝐻 𝑥=𝐿/2

⟹ 𝜃𝑚𝑎𝑥 = 𝑡𝑎𝑛−1𝑤0𝑥

𝐹𝐻, 𝑇𝑚𝑎𝑥 =

𝐹𝐻

𝑐𝑜𝑠𝜃𝑚𝑎𝑥



2/24/2013

11

§4. Cables

Using the triangular relationship

𝑇𝑚𝑎𝑥 =4𝐹𝐻

2 + 𝑤02𝐿2

2=

𝑤0𝐿

21 +

𝐿

4ℎ

2

For a differential segment of cable length 𝑑𝑠

𝑑𝑠 = (𝑑𝑥)2+(𝑑𝑦)2= 1 +𝑑𝑦

𝑑𝑥

2

𝑑𝑥

The total length of the cable can be determined by integration

ℒ = 𝑑𝑠 =2 1+8ℎ

𝐿2𝑥

2

𝑑𝑥

𝐿2

0

=𝐿

21+

4ℎ

𝐿

2

+𝐿

4ℎ𝑠𝑖𝑛ℎ−1

4ℎ

𝐿



§4. Cables

- Cable subjected to its own weight

• In some cases the self weight of the cable is relevant to the

analysis (e.g. electrical transmission line)

• Since the self weight is distributed uniformly along the length

of the cable, the cable will have the shape of a catenary



§4. Cables

- Consider a generalized loading function 𝑤 = 𝑤 𝑠 acting

along the cable

It can be shown that

𝑥 = 𝑑𝑠

1 +1𝐹𝐻

2 𝑤 𝑠 𝑑𝑠2



§4. Cables

- Example 7.13 Determine the deflection curve, the length, and

the maximum tension in the uniform

cable. The cable has a weight per

unit length of 𝑤0 = 5𝑁/𝑚

Solution

𝑥 = 𝑑𝑠

1 +1𝐹𝐻

2 𝑤0𝑑𝑠2

= 𝑑𝑠

1 +1𝐹𝐻

2 (𝑤0𝑠 + 𝐶1)2

Let 𝑢 ≡ (1/𝐹𝐻)(𝑤0𝑠 + 𝐶1), 𝑑𝑢 = 𝑤0/𝐹𝐻 𝑑𝑠

⟹ 𝑥 =𝐹𝐻

𝑤0𝑠𝑖𝑛ℎ−1𝑢 + 𝐶2

=𝐹𝐻

𝑤0𝑠𝑖𝑛ℎ−1

1

𝐹𝐻(𝑤0𝑠 + 𝐶1) + 𝐶2



§4. Cables

𝑑𝑦

𝑑𝑥=

1

𝐹𝐻 𝑤0𝑑𝑠 =

1

𝐹𝐻𝑤0𝑠 + 𝐶1

𝑥 =𝐹𝐻

𝑤0𝑠𝑖𝑛ℎ−1

1

𝐹𝐻(𝑤0𝑠 + 𝐶1) + 𝐶2

Boundary conditions 𝑠 𝑥=0 = 0, 𝑑𝑦

𝑑𝑥 𝑥=0

= 0 ⟹ 𝐶1 = 0, 𝐶2 = 0

⟹ 𝑠 =𝐹𝐻

𝑤0𝑠𝑖𝑛ℎ

𝑤0

𝐹𝐻𝑥 ,

𝑑𝑦

𝑑𝑥=

𝑤0𝑠

𝐹𝐻= 𝑠𝑖𝑛ℎ

𝑤0

𝐹𝐻𝑥

𝑦 =𝐹𝐻

𝑤0𝑐𝑜𝑠ℎ

𝑤0

𝐹𝐻𝑥 + 𝐶3

Apply the boundary condition 𝑦 𝑥=0 = 0 ⟹ 𝐶3 = −𝐹𝐻/𝑤0. The

deflection curve



𝑦 =𝐹𝐻


𝑤0

𝐹𝐻𝑥 − 1

§4. Cables

At 𝑥 = 𝐿/2

𝑦 𝑥=𝐿/2

= ℎ =𝐹𝐻


𝑤0𝐿

2𝐹𝐻− 1

Replace the numerical values 𝑤0 = 5𝑁/𝑚, ℎ = 6𝑚, 𝐿 = 20𝑚

⟹ 𝐹𝐻 = 45.9𝑁 and the deflection curve

𝑦 = 9.19[𝑐𝑜𝑠ℎ 0.109𝑥 − 1]

The length of cable

ℒ = 2𝑠 = 2𝐹𝐻

𝑤0𝑠𝑖𝑛ℎ

𝑤0

𝐹𝐻𝑥 = 2

45.9

5𝑠𝑖𝑛ℎ

5

45.9×10 = 24.2𝑚

The maximum tension in the cable

𝑑𝑦

𝑑𝑥 𝑠=12.1

= 𝑡𝑎𝑛𝜃𝑚𝑎𝑥 =5 × 12.1

45.9= 1.32 ⟹ 𝜃𝑚𝑎𝑥 = 52.80

𝑇𝑚𝑎𝑥 = 𝐹𝐻/𝑐𝑜𝑠𝜃𝑚𝑎𝑥 = 45.9/𝑐𝑜𝑠52.80 = 75.9𝑁



2/24/2013

12

Problems

- Prob.7.89 Determine the tension in each segment of the cable

and the cable’s total length. Set 𝑃 = 80𝑁



Problems

- Prob.7.90 If each cable segment can support a maximum

tension of 75𝑁 , determine the largest load 𝑃 that can be

applied



Problems

- Prob.7.91 The cable supports the loading shown. Determine

the horizontal distance 𝑥𝐵 the force at point 𝐵 acts from 𝐴. Set

𝑃 = 40𝑁



Problems

- Prob.7.92 The cable segments support the loading shown.

Determine the magnitude of the horizontal force 𝑃 so that

𝑥𝐵 = 1.8𝑚



Problems

- Prob.7.93 Determine the force 𝑃 needed to hold the cable in

the position shown, i.e., so segment 𝐵𝐶 remains horizontal.

Also, compute the sag 𝑦𝐵 and the maximum tension in the

cable



Problems

- Prob.7.99 Determine the maximum uniform distributed loading

𝑤0𝑁/𝑚 that the cable can support if it is capable of sustaining

a maximum tension of 60𝑘𝑁



2/24/2013

13

Problems

- Prob.7.104 The bridge deck has a weight per unit length of

80𝑘𝑁/𝑚. It is supported on each side by a cable. Determine

the tension in each cable at the piers 𝐴 and 𝐵



Ch.07 Internal Forces

Documents

Transcript of Ch.07 Internal Forces