理论力学教案8
![理论力学教案8](https://img.360docs.net/img2e/01dygmb33bkk1ifqo6kv-e1.webp)
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
4.8 Equilibrium Analysis of Composite Bodies
The three steps in the equilibrium analysis of a composite body and its various parts are:
1. Draw the appropriate free-body diagrams.
2. Write the equilibrium equations.
3. Solve the equilibrium equations for the unknowns.
The primary difference between one-body and composite-body problems is that the latter often require that you analyze more than one free-body diagram.
Importance:
1.Begin by drawing the FBD of the entire body and, if possible, calculate the external reactions.
2. Then, and only then, should you consider the analysis of one or more parts of the composite body.
Sample problem 4.13
The structure is loaded by the 240-lb·in. counterclockwise couple applied to member AB. Neglecting the weights of the members, determine all forces acting on member BCD.
Solution: step 1: analysis
· The FBD of the entire frame ?
· The unknowns: A x , A y , T C and N D . (four)
· The independent equilibrium equations. (three)
Statically indeterminate ?!
· The FBD of member BCD ?
6 = 6 !!! So the problem is statically determinate!!
step 2: mathematical details
From the FBD of entire structure,
ΣM
A = 0T C cos30º×8– 12N D + 240 = 0 N D = 0.5774T C + 20
From the FBD of member BCD,
ΣM
B = 0T
C cos30º×4+ T C sin30º×3– 8N
D = 0 N D = 0.6205T C
Solving the above two equations yields,
T C = 464 lb, N D = 288 lb
Also from the FBD of member BCD,
ΣF x = 0N D–T C cos30º + B x = 0 B x = 114 lb
ΣF y = 0B y–T C sin30º = 0 B y = 232 lb
4.9 Special cases: two-force and three-force bodies
a. Two-force bodies (二力体)
Two-force principle: If a body is held in equilibrium by two forces, the forces must be equal in magnitude and oppositely directed along the same line of action.
ΣF= 0 ΣM O = 0
An example for the application of the
Two-force principle:
Consider the frame as shown in the following
(neglect the weights of the members):
· The FBD of the whole structure: ·T he FBD of each component:
· The number of unknowns: A x , A y , C x , C y , D x , and D y . (six)
· The number of independent equilibrium equations. (six)
Statically determinate !
· Consider that the member AC · And the FBD of the whole structure
is a two-force body. So its FBD can also be reduced as:
can be reduced as:
· The number of unknowns: P AC, D x , and D y . (three)
· The number of independent equilibrium equations. (three)
Simplification !
b. Three-force bodies (三力体)
Three-force principle: Three nonparallel,
coplanar forces that hold a body in equilibrium
must be concurrent.
ΣM O = 0