Ch.07 Internal Forces
description
Transcript of 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.02 Internal Forces
§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?
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.03 Internal Forces
§1. Internal Forces Developed in Structural Members
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.04 Internal Forces
§1. Internal Forces Developed in Structural Members
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.05 Internal Forces
§1. Internal Forces Developed in Structural Members
• 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.06 Internal Forces
Steps
2/24/2013
2
§1. Internal Forces Developed in Structural Members
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.07 Internal Forces
§1. Internal Forces Developed in Structural Members
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.08 Internal Forces
§1. Internal Forces Developed in Structural Members
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
Engineering Mechanics – Statics 7.09 Internal Forces
§1. Internal Forces Developed in Structural Members
- Example 7.2 Determine the normal force, shear force, and
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.10 Internal Forces
§1. Internal Forces Developed in Structural Members
Equations of equilibrium
+→ ∑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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.11 Internal Forces
§1. Internal Forces Developed in Structural Members
- Example 7.3 Determine the normal force, shear force, and
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.12 Internal Forces
2/24/2013
3
§1. Internal Forces Developed in Structural Members
Free-body diagrams
Equations of equilibrium
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.13 Internal Forces
§1. Internal Forces Developed in Structural Members
- Example 7.4 Determine the normal force, shear force, and
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
Equations of equilibrium
+→ ∑𝐹𝑥 = 0: 848.5𝑐𝑜𝑠450 − 𝑉𝐸 = 0
+ ↑ ∑𝐹𝑦 = 0: −848.5𝑠𝑖𝑛450 + 𝑁𝐸 = 0
+↺ ∑𝑀𝐵 = 0: 848.5𝑐𝑜𝑠450 × 0.5 − 𝑀𝐸 = 0
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.14 Internal Forces
𝑉𝐸 = 600𝑁 ⟹ 𝑁𝐸 = 600𝑁
𝑀𝐸 = 300𝑁𝑚
§1. Internal Forces Developed in Structural Members
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.15 Internal Forces
§1. Internal Forces Developed in Structural Members
Equations of equilibrium
∑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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.16 Internal Forces
Fundamental Problems
- F7.1 Determine the normal force, shear force, and moment
at point 𝐶
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.17 Internal Forces
Fundamental Problems
- F7.2 Determine the normal force, shear force, and moment
at point 𝐶
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.18 Internal Forces
2/24/2013
4
Fundamental Problems
- F7.3 Determine the normal force, shear force, and moment
at point 𝐶
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.19 Internal Forces
Fundamental Problems
- F7.4 Determine the normal force, shear force, and moment
at point 𝐶
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.20 Internal Forces
Fundamental Problems
- F7.5 Determine the normal force, shear force, and moment
at point 𝐶
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.21 Internal Forces
Fundamental Problems
- F7.6 Determine the normal force, shear force, and moment
at point 𝐶. Assume 𝐴 is pinned and 𝐵 is a roller
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.22 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.23 Internal Forces
§2. Shear and Moment Equations and Diagrams
- 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)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.24 Internal Forces
2/24/2013
5
§2. Shear and Moment Equations and Diagrams
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.25 Internal Forces
§2. Shear and Moment Equations and Diagrams
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
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.26 Internal Forces
§2. Shear and Moment Equations and Diagrams
- Example 7.7 Draw the shear and moment
diagrams for the beam
Solution
Support reactions
Shear and moment function
+ ↑ ∑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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.27 Internal Forces
Fundamental Problems
- F7.7 Determine the shear and moment as a function of 𝑥,
and then draw shear and moment diagrams
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.28 Internal Forces
Fundamental Problems
- F7.8 Determine the shear and moment as a function of 𝑥,
and then draw shear and moment diagrams
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.29 Internal Forces
Fundamental Problems
- F7.9 Determine the shear and moment as a function of 𝑥,
and then draw shear and moment diagrams
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.30 Internal Forces
2/24/2013
6
Fundamental Problems
- F7.10 Determine the shear and moment as a function of 𝑥,
and then draw shear and moment diagrams
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.31 Internal Forces
Fundamental Problems
- F7.11 Determine the shear and moment as a function of 𝑥,
where 0 ≤ 𝑥 < 3𝑚 and 3𝑚 < 𝑥 ≤ 6𝑚 then draw shear and
moment diagrams
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.32 Internal Forces
Fundamental Problems
- F7.12 Determine the shear and moment as a function of 𝑥,
where 0 ≤ 𝑥 < 3𝑚 and 3𝑚 < 𝑥 ≤ 6𝑚 then draw shear and
moment diagrams
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.33 Internal Forces
§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)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.34 Internal Forces
§3. Relations between Distributed Load, Shear, and Moment
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.35 Internal Forces
𝑑𝑉
𝑑𝑥= 𝑤 𝑥
∆𝑉 = 𝑤 𝑥 𝑑𝑥
§3. Relations between Distributed Load, Shear, and Moment
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.36 Internal Forces
𝑑𝑀
𝑑𝑥= 𝑉
∆𝑀 = 𝑉𝑑𝑥
2/24/2013
7
§3. Relations between Distributed Load, Shear, and Moment
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.37 Internal Forces
§3. Relations between Distributed Load, Shear, and Moment
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.38 Internal Forces
§3. Relations between Distributed Load, Shear, and Moment
- Example 7.8 Draw the shear and moment diagrams for the
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.39 Internal Forces
Solution
§3. Relations between Distributed Load, Shear, and Moment
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.40 Internal Forces
§3. Relations between Distributed Load, Shear, and Moment
- Example 7.9 Draw the shear and moment diagrams for the
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.41 Internal Forces
§3. Relations between Distributed Load, Shear, and Moment
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.42 Internal Forces
2/24/2013
8
§3. Relations between Distributed Load, Shear, and Moment
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.43 Internal Forces
§3. Relations between Distributed Load, Shear, and Moment
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.44 Internal Forces
Fundamental Problems
- F7.13 Draw the shear and moment diagrams for the beam
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.45 Internal Forces
Fundamental Problems
- F7.14 Draw the shear and moment diagrams for the beam
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.46 Internal Forces
Fundamental Problems
- F7.15 Draw the shear and moment diagrams for the beam
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.47 Internal Forces
Fundamental Problems
- F7.16 Draw the shear and moment diagrams for the beam
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.48 Internal Forces
2/24/2013
9
Fundamental Problems
- F7.17 Draw the shear and moment diagrams for the beam
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.49 Internal Forces
Fundamental Problems
- F7.18 Draw the shear and moment diagrams for the beam
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.50 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.51 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.52 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.53 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.54 Internal Forces
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.55 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.56 Internal Forces
§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
Engineering Mechanics – Statics 7.57 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.58 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.59 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.60 Internal Forces
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.61 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.62 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.63 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.64 Internal Forces
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.65 Internal Forces
𝑦 =𝐹𝐻
𝑤0𝑐𝑜𝑠ℎ
𝑤0
𝐹𝐻𝑥 − 1
§4. Cables
At 𝑥 = 𝐿/2
𝑦 𝑥=𝐿/2
= ℎ =𝐹𝐻
𝑤0𝑐𝑜𝑠ℎ
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.66 Internal Forces
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.67 Internal Forces
Problems
- Prob.7.90 If each cable segment can support a maximum
tension of 75𝑁 , determine the largest load 𝑃 that can be
applied
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.68 Internal Forces
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.69 Internal Forces
Problems
- Prob.7.92 The cable segments support the loading shown.
Determine the magnitude of the horizontal force 𝑃 so that
𝑥𝐵 = 1.8𝑚
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.70 Internal Forces
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.71 Internal Forces
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.72 Internal Forces
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 7.73 Internal Forces