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B. C. a. A. D. form of a balanced force system. 2a. B. C. A. D. 2-3 Solution. take the rigid frame as object to study. 习 题 (静力学). According to the graphical condition of equilibrium:. By the geometrical relation:. Therefore:. B. C. a. A. D. 2a. B.
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B C a A D form of a balanced force system. 2a B C A D 2-3 Solution take the rigid frame as object to study 习 题(静力学) According to the graphical condition of equilibrium: By the geometrical relation: Therefore:
B C a A D 2a B C A D y x 2-3 Solution take the rigid frame as object to study 习 题(静力学) Solving them, we obtain:
D C l According to the character of two couples in equilibrium, and make up a new force A B l l l couple. C B D C y A x 2-6 Solution 1.Study the rod CB 习 题(静力学) 2.Take the rod DC to study
B 4m 3m y M q x A 2-12Solution: 1. take the rigid structure as object to study 习 题(静力学) 2. draw force diagram (make analysis of force) 3. set equilibrium equation. MA free from constraints Solving them, we obtain:
y x 2-14(a)Solution: M B 1. take beam AB to study A C 2. draw force diagram 习 题(静力学) a 2a 3. set equilibrium equation. Solving them, we obtain:
q M A B D a a 2a y x 2-14(b) 1. take beam AB to study 习 题(静力学) 2. draw force diagram 3. set equilibrium equation. Solving them, we obtain:
10 m 1.5m x (It should be under the action of which the crane can not turn over 3 m around point B.) 2-16Solution: take crane to study, it is acted upon by a C.P.F.S. 习 题(静力学) 1. Firstly, let us examine the full-loaded position of the crane. free from constraints A B set equilibrium equation
due to 10 m 1.5m x (The crane will turn over around point A unless we can ensure .) 3 m we get 习 题(静力学) ① 2. Secondly, let us examine the empty position of the crane. A B set equilibrium equation we get due to ②
① ② From ① and ②, we find the range of equilibrium 习 题(静力学) Hence
y x 2-18Solution: C 1. take the whole system as object to study D 习 题(静力学) 2. draw force diagram A r B (make analysis of force) 3. set equilibrium equation. noting that: free from constraints Solving them, we obtain: (tensile force)
q M A D B C M q D C y x 2-21Solution: 1. take rod CD to study 习 题(静力学) 2m 2m 2m 2m Solving them, we obtain: 2. take the whole system as object to study Solving them, we get:
C 2m 2m 1.5m A D B 1.5m E y x 2-32Solution: • Take the whole system as • object to study 习 题(静力学) free from constraints
C 2m 2m 1.5m A D B 1.5m E C C E D B E B A D (2)Take the rod AB to study 习 题(静力学) (compressive force) (we also can study the rod CE)
2m 4m C D D C 3m 1.8KN B 1.5KN 3m 1.8KN A y The direction of is opposite to the direction indicated in the diagram x 1.8KN 2-38Solution: 1.Take the whole rigid structure to study 习 题(静力学) free from constraints Solving them, we obtain: 2.Then study the rod CD
C B 1.8KN A 3. Study the rod CA 习 题(静力学)
y x 2-57Solution: 6 10 B A 习 题(静力学) 7 5 4 free from constraints • Take the whole structure as object to study
I I 6 10 B A 习 题(静力学) c 7 5 4 (tensile force) free from constraints (2) we assume a virtual section I-I pass through the truss, and consider the equilibrium of the left part: (tensile force) (compressive force)
6 10 A B 习 题(静力学) D 7 5 4 (3) Take the joint D as object to study (compressive force) (compressive force)
Method of joints Method of cuts 习 题(静力学) A question to think about: Which one is more suitable for programming? And why?
span D B A f C x y Suspended cable l sag 习 题(静力学) (parabola) end condition: when : get the maximal tensile force :
span l D B A f C 习 题(静力学)