Using thethod of joints, me determine the force in each ...

statics Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.

SOLUTION

Free body: Entire truss:

Fy = 0: By = 0 B y = 0

MC = 0: - Bx (3.2 m) - (48 kN)(7.2 m) = 0 Bx = -108 kN Bx = 108 kN

Fx = 0: C -108 kN + 48 kN = 0

C = 60 kN

C = 60 kN

Free body: Joint B:

FAB = FBC = 108 kN

54

3

FAB = 180.0 kN T FBC = 144.0 kN T

Free body: Joint C:

FAC = FBC = 60 kN 13 12 5 FBC = 144 kN (checks)

FAC = 156.0 kN C

statics SOLUTION

Reactions: Joint B: Joint A:

Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.

M A = 0: C = 1260 lb Fx = 0: Ax = 0 Fy = 0: A y = 960 lb

FAB = FBC = 300 lb 12 13 5

FAB = 720 lb T FBC = 780 lb C

Fy = 0:

- 960

lb

-

4 5

FAC

=

0

FAC = 1200 lb

FAC = 1200 lb C

Fx = 0:

720 lb - (1200 lb) 3 = 0 5

(checks)

statics Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.

SOLUTION Reactions:

Joint A: Joint C:

AB = 32 + 1.252 = 3.25 m BC = 32 + 42 = 5 m

M A = 0: (84 kN)(3 m) - C(5.25 m) = 0 C = 48 kN

Fx = 0: Ax - C = 0 Ax = 48 kN

Fy = 0: Ay = 84 kN = 0 A y = 84 kN

Fx = 0:

48

kN

-

12 13

FAB

=

0

FAB = +52 kN

Fy = 0:

84 kN

- 5 (52 kN) - 13

FAC

=

0

FAC = +64.0 kN

FAB = 52.0 kN T FAC = 64.0 kN T

FBC = 48 kN

5

3

FBC = 80.0 kN C

statics Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.

SOLUTION Free body: Truss: From the symmetry of the truss and loading, we find

C = D = 600 lb

Free body: Joint B: Free body: Joint C:

From symmetry:

FAB = FBC = 300 lb

52

1

FAB = 671 lb T

FBC = 600 lb C

Fy = 0:

3 5

FAC

+

600

lb

=

0

FAC = -1000 lb

Fx = 0:

4 (-1000 5

lb)

+

600

lb

+

FCD

=

0

FAC = 1000 lb C FCD = 200 lb T

FAD = FAC = 1000 lb C, FAE = FAB = 671 lb T , FDE = FBC = 600 lb C

statics SOLUTION Reactions:

Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.

Joint A: Joint D:

M D = 0: Fy (24) - (4 + 2.4)(12) - (1)(24) = 0

Fx = 0: Fx = 0

Fy = 4.2 kips

Fy = 0: D - (1 + 4 + 1 + 2.4) + 4.2 = 0

D = 4.2 kips

Fx = 0: FAB = 0

Fy = 0 : -1 - FAD = 0

FAD = -1 kip

Fy = 0:

-1+

4.2

+

8 17

FBD

=

0

FBD = -6.8 kips

Fx = 0:

15 (-6.8) + 17

FDE

=0

FDE = +6 kips

FAB = 0 FAD = 1.000 kip C FBD = 6.80 kips C FDE = 6.00 kips T

statics Joint E:

PROBLEM (Continued)

Fy = 0 : FBE - 2.4 = 0 FBE = +2.4 kips

Truss and loading symmetrical about cL.

FBE = 2.40 kips T

statics SOLUTION

Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.

Reactions: Joint D:

AD = 52 + 122 = 13 ft BCD = 122 + 162 = 20 ft

Fx = 0: Dx = 0 M E = 0: Dy (21 ft) - (693 lb)(5 ft) = 0

Fy = 0: 165 lb - 693 lb + E = 0

Fx = 0:

5 13

FAD

+

4 5

FDC

=

0

Fy = 0:

12 13

FAD

+

3 5

FDC

+ 165

lb

=

0

Solving Eqs. (1) and (2) simultaneously, FAD = -260 lb FDC = +125 lb

Dy = 165 lb E = 528 lb

(1) (2)

FAD = 260 lb C FDC = 125 lb T

statics PROBLEM (Continued)

Joint E:

Fx = 0:

5 13

FBE

+

4 5

FCE

=

0

Fy = 0:

12 13

FBE

+

3 5

FCE

+

528

lb

=

0

Solving Eqs. (3) and (4) simultaneously, Joint C:

FBE = -832 lb FCE = +400 lb

(3) (4)

FBE = 832 lb C FCE = 400 lb T

Force polygon is a parallelogram (see Fig. 6.11, p. 209).

Joint A:

Fx = 0:

5 (260 13

lb)

+

4 5

(400

lb)

+

FAB

=

0

FAB = -420 lb

Fy = 0:

12 (260 lb) - 3 (400 lb) = 0

13

5

0 = 0 (Checks)

FAC = 400 lb T FBC = 125.0 lb T

FAB = 420 lb C

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