PPLIED MECHNICS I 1. Introduction to Mechanics Mechanics is a science that describes and predicts the conditions of rest or motion of bodies under the action of forces. It is divided into three parts 1. Mechanics of rigid bodies 2. Mechanics of Deformable bodies 3. Mechanics of Fluids Mechanics of Rigid bodies is further divided into statics and dnamics. In this part of the course, bodies are assumed to be perfectl rigid, although all bodies deform under the application of load. But these deformations are usuall small and do not adversel affect the conditions of equilibrium or motion of the structure. Mechanics is the foundation of most Engineering Science. The purpose is to eplain and predict phsical phenomena and thus la the foundations for Engineering applications 1.1 Force Force is a push or pull on a bod. It is generall characterized b its point of application, its magnitude and its direction. The direction of a force is defined b the line of action and the sense of the force. 1.1.1 Resultant of Concurrent Forces Consider a bod acted upon b co-planar forces as shown in Fig 1.1(a). P P Q S Q S Fig 1.1 (a) Fig 1.1 (b) Since all the forces pass through point O, the are said to be concurrent forces. The resultant R of the concurrent forces can be obtained b using the polgon rule which is equivalent to the repeated use of the parallelogram law as shown in Fig 1.1(b) 1.1.2 Resolution of a Force s two or more concurrent forces can be replaced b a single resultant force, similarl a single force acting on a bod ma be resolved into two or more forces which togther have the same effect on the bod. These forces are referred to as components of the force. Eample 1.1 Determine the resultant force for the sstem of forces P and Q shown in Fig Q.1
C 2 R 155 25 B B 4N C 6N C 6N B 4N 25 2 Fig. E1.1 Solution The triangle BC is drawn Using the triangle rule, and appling the Cosine law, we have, R 2 B 2 + C 2-2BC cos B R 97.73N ppling Sine law s in s in B C R s in s in 1 5 5 6 9 7. 7 3 1 5. 4 α 2 + 1 5. 4 3 5. 4 1.1.3 Rectangular Components of a Force force can be resolved into two rectangular components b representing the orthogonal forces as F and F in the and aes respectivel. If the unit vectors along and aes are and, then vector F can be written as F F + F If F is the magnitude of force F and the angle between F and ais is denoted b, then F F cos F F sin Eample 1.2 man pulls with a force of 3N on a rope attached to a 6m tall building as shown in Fig 1.2. Compute the horizontal and vertical components of the force eerted on the rope at point.
6m 3N C Fig E1.2(a) 8m Solution From Fig. E1.2(b) B 24N F 3 cos F -3 sin 8 cosθ 1 6 cosθ 1 F 24-18 Fig 1.2(b) 4 5 3 5 C 3N 1.1.4 ddition of Forces Consider three forces P, Q and S acting on a bod as shown in Fig 1.3. S P Q Fig 1.3 (a) Their resultant R is defined b R P + Q + S Resolving the forces into rectangular components, we have R + R P + P + Q + Q + S + S (P + Q + S ) + (P + Q + S ) This implies that
R P + Q + S R P + Q + S or R F R f Thus the scalar components R and R of the resultant R of several forces acting on a z z Fig. 1.4 particle are obtained b adding the corresponding scalar components of the given forces algebraicall. 1.2 Three-Dimensional Force Sstem F F i + F j + F k F co sθ i + F co sθ j + F co sθ k z F λ co sθ i + co sθ j + co sθz k F li + m i + n k 2 λ co sθ + co sθ + co sθ 1 λ 1 z Eample 1.3 wooden plank is held b cable B, C and DB as shown
B 27cm C 8cm 11cm 16cm Fig. E1.3(a) If the tension is 84N in cable b and 1.2kN in cable C, determine the magnitude and direction of the resultant of the force eerted b cables B and C in state. Solution z FC FB Fig. E1.3(b) If i,j and k are unit vector along, and z respectivel B 16i + 8 j + 11k, B 21cm C 16i + 8 j 16k, C 24cm λ B u n it v ec to r a lo n g B F F λ F B B B B 8 4 2 1 Substitute B. B B F ( 6 4 i + 3 2 j + 4 4 k ) N B S im ilarl, F ( 8 i + 4 j 8 k ) N C R F + F ( 1 4 4i + 7 2 j 3 2 k ) N B 2 2 2 R R + R + R z 1 6 5N R 1 4 4 c o s θ, θ 1 5. 8 λ 1 6 5 S im ilarl, θ 6 4. 1, θ 1 2. 6 C z
1.3 Moments Moment is a tendenc to rotate or twist a bod about an ais which does not intersect the line of action of the force. The magnitude F Perpendicular Distance from its line of action to point of Moment arm. Y Moment arm for P B P Moment arm for B X Fig. 1.4 Most moments bends causing bending while some moments twist causing torque or torsional moment. It is sometimes easier to work with the components of the force rather than with the force itself. Y Y P5kN P17.1kN 25 2m P47 d 3m 1' Fig. 1.5(a) Moment arm with respect to d M p d X Fig. 1.5(b) X Counter Clockwise Clockwise lso, M P ( ) + P ( ) Credit for this observation is given to Pierre Varignon (1654-1722), a French Mathematician. Varignon s theorem states that for coplanar forces, the moment of a force about an point is equal to the sum of the moments of the components of the forces about the same point. lso referred to as the Principle of Moments. In the Fig1.5b M + P ( 2) P ( 3) 17kNm M 17kNm C lockwise or determine the using triangle KLM
P K M 3 tan 2 8. 24m 2m L 2 d 3m K 2 K O 6. 24m d 6. 24 sin 2 2. 14m Fig. 1.5c M M P d 17kN m Moment of P is independent of the location of the force along its line of action e.g. if P acts at point L, then using Varignon s theorem M r P (O n ) - P (6.24) -17kNm. This technique of moving force until it passes there, the derived moment centre can be a valuable shortcut for man moment calculations. 1.4 Moment of a Force about a Point M o r F rf sin θ z O r F Fig. 1.6 r (F 1 + F 2...) r F 1 + r F 2... Varignon s theorem M o i j k z F F F z
i k + j Eample 1.4 rectangle plate is supported b brackets and B and b a wire CD as shown in Fig. E 1.4(b). If the tension in the wire is 2N, determine the moment about of the same force etended b the wire on point C. 24mm 24mm 8mm 8mm B 3mm C Fig. E1.4(a) Solution 2N r C C Fig. E1.4(b) M eerted b wire on point C is M r F r C C (. 3i +. 8k ) m F λ C 2 N directed alon g C D C D ( i j k C D C D. 3 / +. 24. 32 ) m. 5 F F λ ( 12i + 96 j 128k ) N C D M r F ( 7. 68i + 28. 8 j + 28. 8 k ) N m C
i j k. 3. 6 12 9 128 1.5 Equilibrium of Concurrent Forces Equilibrium is a state of balance created b opposing forces which act on a bod in such a manner that their combined net effect is zero. Newton s first law states that: When a bod is at rest (or is moving with a constant velocit, with zero acceleration) in a straight line, the resultant of the force sstem acting on the bod is equal to zero. For coplanar sstem F F 1.6 Couple: Concept of a pair of equal and opposite parallel forces, If produces onl rotation, no translation, hence pure Moment. Magnitude Product of one of the forces and the perpendicular distance between the forces and its sense in either clockwise or anticlockwise. Equilibrium Couple consists of forces with identical magnitude but opposite sense to balance a couple. n understanding of what a couple is enables us to work more closel at what happens inside a bending member (beam). Consider beam B carring a point load P as shown in Fig. 1.6(a). If the beam is cut at section X- X, shear force V develops to balance reaction R at as shown in Fig. 1.6(b) X P X (a) B V R (b) Fig. 1.6 R and V constitute a couple of magnitude V (or R) which can onl be balanced b an opposing couple. That is, the beam portion might be in force (translational) equilibrium ( F), but it is not in moment (rotational) equilibrium. The opposing couple can onl be provided b the other beam part upon the cut face as shown in Fig. 1.6(c).
R V Fig. 1.6(c) Q Q Because the moment arm is limited b the beam depth the forces Q (Tensile/Compressive) developed will be quite large. For moment equilibrium ( M), Q times must equal V. This introduces the third equation of static equilibrium which joins the previous two equation as follows: F F M