3-Statics Sample Problems - Teacher-1 PDF

Preview

Summary

Practical sample problems on statics. Know how to do statics for the exam!!!...

Description

Sample Statics Problems - Teacher Name________________________ Period______ 1. Resolve the forces below C

A 0

Fy B 300lbs  AB sin 30  BC sin 60

0

30

60

 Fx B  AB cos 30  BC cos 60 .87 AB .5 BC  AB .57 BC Fy B 300lbs .5 AB  .87 BC Fy B 300lbs .5(.57 BC )  .87 BC

B

 Fy B  300lbs  .285BC  .87BC  FyB 300 lbs 1.15 BC BC 260.9 261lbs

300 lbs

.87 AC .5( 261)  AC 150lbs

2. Determine the force in each member of the loaded truss: B 300 lbs 10’ A

C 24’ 24/26BC = 300  BC = 325 lbs 10/26BC = AB  10/26(325 lbs) = AB = 125 lbs 24/26BC = AC = 300 lbs

3. Determine the forces at AB and BC

A

B

600 C

250

BV: 250 = BCcos60  250lbs = .5BC  BC = 500 lbs BH: AB = BCsin60  AB = .87BC  AB = .87(500lbs)  AB = 435 lbs

4. Determine the forces on each member of the truss. Notice that lengths are not necessary to solve the problem. 4000 lb B

30o A

B V : 4000lbs  AB sin 30  BC sin 60 BH : AB cos 30 BC cos 60 B B : .87 AB .5 BC AB .57 BC 4000lbs .5AB  .87BC 4000lbs .5(.57BC )  .87BC 4000lbs .285 BC  .87 BC 4000 lbs 1.15 BC BC 3478lbs .87 AB .5(3478lbs )  AB 1998lbs AH : AB cos 30 AC A H : .87 AB  AC  .87 (1998)  AC AC 1730lbs

60o C

5.

Determine the forces on the beam. What force is needed at C to put the beam in equilibrium? (A is 2 m from the fixed wall, B is 8 m from the wall. C is 12 m from the wall and free in the vertical.) A 2m

300 @45 N

B 8m

C 0

20 @75 N

Torque @ A = 2m(45Nsin(30) = 45Nm up Torque @ B = 8m(75Nsin(20) = 205.21Nm down Resultant torque = 205.21 Nm - 45 Nm = -160.21 Nm (neg = down) Torque @ C = -160Nm = C@12m Fmoments = 0 = -160.21Nm/12m + C = 0 C = 13.35 N up...