PRINCIPLES OF STTICS Statics is the study of how forces act and react on rigid bodies which are at rest or not in motion. This study is the basis for the engineering principles, which guide the design of all structures, since before we can begin to design any structure we must first know the forces applied to it. Newton s third law of motion states that: For every action there is always an equal and opposite reaction. The study of statics has many applications in daily life. nyone who has used a ladder should have a deep appreciation for the laws that govern the stability of the ladder. Hopefully the reactions generated by the ground surface and the wall surface are sufficient to maintain equilibrium. If these forces cannot be developed, an accident will likely occur. Page 1-1
EQUILIRIUM OF STRUCTURES structure is considered to be in equilibrium if it remains at rest when subject to a system of forces and moments. If a structure is in equilibrium, then all its members and parts are also in equilibrium. For a structure to be in equilibrium, all the forces and moments (including support reactions) acting on it must balance each other. For a plane structure subject to forces in its own plane, the conditions for equilibrium can be expressed by the following equations of equilibrium: The third equation above states that the sum of moments of all forces about any point in the plane of the structure is zero. F = 0, x F x 0, Fy = 0, M z = F1 + (-F2) = 0 F1 = F2 F y = 0, (-F3) + (F4) = 0 F3 = F4 = 0 Forces and their resultants Forces are vectors, and because of this they have both magnitude and direction. This is extremely important to realize because an analysis of a structure cannot be completed by simply knowing the magnitude of forces without the direction of application. Page 1-2
Techniques of finding resultants In solving statics problems, often it is an advantageous to replace all the forces, which act on a body with a single force. This single force, called a resultant, must produce the same movement and effects that all the original forces would produce on the body. Vector addition using trigonometric functions Resolution into rectangular components Vector addition using trigonometric functions The resultant of two vectors can be found by placing the two vectors (which act at the same point) in a tip-to-tail fashion and completing the triangle with a vector. The resultant vector (c) is the sum of the original two vectors (a and b) and therefore replaces the original two vectors. y using Cosine Law, c 2 = a 2 +b 2 2*a*b*cos C The resultant of three or more vectors can be found by placing the vectors (which act at the same point) in a tip-to-tail fashion and completing the polygon. Page 1-3
The polygon method requires finding an intermediate resultant, r. This intermediate resultant is to be the sum of the two preceding vectors and will subsequently be added to the next vector to find another intermediate resultant. This process will repeat itself until the last vector has to be added to the last intermediate resultant. Finding the resultant of parallel force systems The resultant of a parallel force system must have the same net effect on the body as the parallel system, which it is replacing. That means the translational characteristics, as well as the rotational characteristics, must be equal. Page 1-4
Example The following body is acted upon by a group of parallel forces. Find the resultant of the resultant. Solution Considering the translational effects, the magnitude of the resultant can be found by simply summing the forces involved. R = 5 + 8 3 N = 10 N This 10 N force has to act at a particular point that gives the resultant the same rotational characteristics as the real system. If point O is chosen as the reference point, the moments from the real system would be as follow: Take moment about point O, R*x = 5*3 + 8*7 3*11 x = 3.8 m Page 1-5
Equilibrium of a structure loaded building shown in the following figure is in equilibrium if it does not move as a rigid body. Rigid body movement can be either a translation (movement in a straight line) or a rotation or a combination of both. In this case we must have some physical means to tie up the building on to the ground in the first place, and through these ties the load can be transmitted from the structural to the ground. These ties are called restraints, or constraints or supports. F x 0, Fy = 0, M z = = 0 Tie up a body or a structure body in a plane has three degrees of freedom, which means that the body is free to move about in x-direction, in y-direction as well as rotation. If we want to maintain this body in equilibrium and prevent it from moving away under the loads, we must tie it up with some ties to the foundation. Page 1-6
It is seen that we have used two ties to tie up the body of. This body now cannot have translated displacements in both x and y directions. ut it can still rotate about. In order to prevent it from rotation too, we used another tie to tie up the body. In conclusion, at least three ties are necessary in order to tie up a body onto a foundation so that it cannot disappear or rotate under the action of external loads, and it can be maintained in the conditions of equilibrium. Page 1-7
Note that the above arrangements of the three ties or restraints cannot maintain the body, or the structure, in equilibrium. Page 1-8
TYPES OF STRUCUTRES Page 1-9
SUPPORTS Pin or Hinge Support pin or hinge support is represented by the symbol H or H V V Prevented: llowed: Horizontal translation and vertical translation Rotation Roller Support roller support is represented by the symbol or V V Prevented: llowed: Vertical translation Horizontal translation and Rotation Page 1-10
Fixed Support fixed support is represented by the symbol M H or H V V M Prevented: llowed: Horizontal translation, Vertical translation and Rotation None Page 1-11
Example 1 eam C has a pinned support at and a roller support at C. It carries two concentrated loads of 20 kn each and a uniformly distributed load of 4 kn/m over the right hand half as shown. etermine the reactions. 20 kn 20 kn 4 kn/m C 3m 1.5m 1.5m 3m Solution: 20 kn 20 kn 4 kn/m C H V V C 3m 1.5m 1.5m 3m X = 0, H = 0 Take moment about, 20*3 + 20*9 + 4*(4.5)*(4.5 + 4.5/2) V c *6 = 0 V c = 60.25 kn Y = 0, 20 + 20 + 4*4.5 = V + V C V = -2.25 kn, (-ve sign indicates V acts in opposite direction) V = 2.25 kn ( ) Page 1-12
Example 2 Find the support reactions for the simple beam shown. 5 40 kn 50 kn 3 4 C 5m 2.5m 2.5m Solution: 5 4 40 kn 40 kn 3 50 kn C 30 kn H V V 5m 2.5m 2.5m Resolve the 50 kn inclined external load into horizontal and vertical components as shown. X = 0, H = 30 kn Take moment about, 40*5 + 40*7.5 V *10 = 0 V = 50 kn Y = 0, 40 + 40 = V + V V = 30 kn Page 1-13
Example 3 etermine the truss reaction forces. 30 kn 30 kn 20 kn 5m 5m 5m 5m 5m Solution: 30 kn 30 kn 20 kn 5m H V V 5m 5m 5m 5m X = 0, H = 20 kn Take moment about, 30*10 + 30*15 20*2.5 V *20 = 0 V = 35 kn Y = 0, 30 + 30 = V + V V = 25 kn Page 1-14
Example 4 etermine the support reactions for the frame shown. 20 kn 25 kn 5 3 4 4m 4m 10 kn C E 12 m Page 1-15
Solution: 20 kn 20 kn 5 3 25 kn 4 4m 4m 10 kn H V V E C E 15 kn 12 m Resolve the 25 kn inclined external load into horizontal and vertical components as shown. X = 0, H + 10 = 15 kn H = 5 kn Take moment about, 10*4 + 20*12 15*8 V E *12 = 0 V E = 13.3 kn Y = 0, 20 + 20 = V + V E V = 26.7 kn Page 1-16
Example 5 Find the reactions for the cantilever beam shown. 6 kn 4 knm 4 kn/m 1.5m 1.5m 1.5m 1.5m Solution: M H V 6 kn 4 kn/m 4 knm 1.5m 1.5m 1.5m 1.5m X = 0, H = 0 kn Y = 0, 6 + 4*1.5/2 = V V = 9 kn Take moment about, 6*1.5 + 4*(1.5/2)*(6-1.5*1/3) 4 M = 0 M = 21.5 knm Page 1-17
Example 6 etermine the support reactions for the frame shown. 3 kn/m 8 kn/m C 3 kn/m 12m 8m Page 1-18
Solution: 3 kn/m 5 kn/m C 3 kn/m 3 kn/m 12m H M V 8m X = 0, H = 3*12 kn = 36 kn Y = 0, 3*8 + 5*8/2 = V V = 44 kn Take moment about, 3*12*6 + 3*8*4 + 5*(8/2)*(8*2/3) M = 0 M = 418.7 knm Page 1-19
Problems etermine the support reactions for the following structures. Q1. 8 KN/m 3 m 2 m C Q2. 40 KN 12 KN/m C 2 m 2 m 5 m Q3. 4 MN 2 MN/m C 2 m 2 m 5 m Page 1-20
Q4. 40 KN/m 500 KNm C 6 m 2 m 2 m Q5. 500 Kg 800 kg/m C 500 mm 200 mm 100mm Q6. 500 N 800 N 200 Nm C 500 mm 300 mm 200mm Page 1-21
Q7. 250 N/m 700 mm 500 mm C Q8. 1200 N 1500N/m C 200mm 600 mm 500 mm Q9. 800 N 600 N/m C 700 mm 200 mm 300 mm Page 1-22
Q10. 150 KNm 60 KN C 4 m 3 m 2 m Q11. 800 kg 300 kg/m 2 m 3 m C Q12. 500 Nm 1500 N/m 1200m 700 mm C Page 1-23