Engineering Mechanics

Engineering Mechanics Continued (5) Mohammed Ameen, Ph.D Professor of Civil Engineering

B Section Forces in Beams Beams are thin prismatic members that are loaded transversely. Shear Force, Aial Force and Bending Moment Consider the beam shown below: y w() P Consider a cut at a distance of from the left end of the beam. At the cut surface, we have a single force and a single couple as the resultant of the distributed forces.

y w() P R A V y M z H Resolve the force into V y and H The couple is M z H is the aial force (AF), V y is the shear force (SF), and M z is the bending moment (BM). The other part of the beam looks like this: w() M z P H V y R B

Sign Convention: A force component at a section is positive if the area vector of the crosssection and the force component both have senses either in the positive or in the negative directions of the reference aes. + V y M z H y H M z + V y

For 3D beams there are Two shear forces (V y and V z ), Two bending moments (M y and M z ), One aial force (H), and One torsional moment (M y ). y M y V y H V z M z M z

Eample: Draw the shear force and bending moment diagrams of the simply supported beam shown below. y 6 kn m 4 m 0 < < m: V y 4 kn; M z 4 kn m H 0 4 kn V M H > > 6 m: V kn; M 4 6 ( ) m 6 kn V M + 1 kn m H 4 kn

y 6 kn Loading Diagram m 4 m 4 kn V kn + SFD M 8 kn m + BMD SFD Shear Force Diagram BMD Bending Moment Diagram

y W a + b l Wb/l a b Wb/l Wa/l + V SFD M Wab/l + BDM Draw the SFD and the BMD for a s.s. beam subjected to a u.d.l. w kn/m l

0 < < l: V wl/ + w wl/ w V M H M wl/. w./ w / (l ) wl/ w kn/m l wl/ wl/ V Loading diagram + wl/ SFD + M wl /8 BMD

y 6 kn 10 kn 7.333 8.667 m m m 7.333 1.333 V (kn) 8.333 + + 14.667 M (kn m) 17.333 4 < < 6 M V 8.667

Wl W W l W Wl V M Eercise: Draw SFD and BMD of: w kn/m l

wl / wl w kn/m l wl wl / V M

Differential Relations for Equilibrium There eists relations between the loading, the SF and the BM at any section. Consider the equilibrium of an infinitesimal strip of a beam as shown below: y w() d Note that the loading is assumed to be positive if it acts along the positive direction.

w()d M V+dV V d a M+dM ΣF y 0: V + (V + dv) + w() d 0 dv d w ΣM a 0: V d M w() d d/ + (M + dm) 0 dm d V

w kn/m l Obtain the SF and BM distributions in the s.s beam shown above subjected to a udl. wl/ y w kn/m l wl/ At, w() w; dv d ; V w + A At 0, V wl/; A wl/; V() w( l/) dm d w V w( l/) M() w / wl / + B; At 0, M 0; B 0;

wl/ w kn/m l wl/ wl/ V Loading diagram + wl/ SFD + M wl /8 BMD Note: Maimum BM at point of zero SF Maimum SF at point of zero load

Eercise 1: A s.s beam is loaded with a distributed load given by w( ) Asin π l Obtain the SF and BM distributions. y l A 0 Eercise : A s.s beam is loaded with a distributed load given by w() w 0 /l. Obtain the SF and BM distributions. y w 0 l

Draw the SFD and BMD of the following beams: 8 kn kn/m m m m m 4 kn 8 kn m 1 kn/m m m m m 4 kn 8 knm 1 kn/m 3 m 3 m 3 m 3 m

y 5 kn m m 3 m 1 kn V 3 kn m M + kn m

8 kn/m 5.33 10.67 m m m m 5.33 4.67 + V 6.67 4 10.67 +.67 m M 37.33 Point of contrafleure

6 kn 1 knm 4 kn/m 3 m 3 m 3 m 11 19 3 m 1 5 + + 6 7 V (kn) 18 18 15 M (knm) 1.5

Chains and Cables Relatively fleible chains and cables are often used to support loads Eamples: Suspension bridges Transmission lines Cable is assumed to be perfectly fleible and inetensible Fleibility assumption means that at the centre of any cross-section of the cable only a tensile force is transmitted and no moments can be present this force must be tangential to the cable profile Inetensibility assumptions denotes that the length of the cable remains a constant.

A Coplanar Cable with Loading as a Function of (Parabolic cable): Consider a cable suspended between two rigid supports A and B under the action of a loading w() given per unit length measured in the horizontal direction. The cable and the load are assumed to lie in one and the same plane. y l w h

Consider an element of the cable as shown below: T + T w T F 0: θ s T cos θ + (T + T) cos (θ + θ) 0 lim 0 or ( T + T ) cos( θ + θ ) T d ( T cos θ ) 0 d T cos θ H a constant, θ + θ cosθ where H is the horizontal component of the cable tension anywhere along the cable A 0

ΣF y 0: T sin θ + (T + T) sin (θ + θ) w av 0 or d ( T sinθ ) w d Integrating the above, we get B A, we get That is, T sin θ w () d + C 1 sinθ 1 cosθ H dy 1 tanθ d H w( ) d integrating the above, we obtain B + C y gives the equation of the deflection curve. C 1 and C are found from the boundary conditions. w( ) d + C [ ] w( ) d d + C1 C 1 y + H 1 1

Eample: Neglect the self weight of the cable and determine the shape, length and the maimum force in the cable. y w h l We have 1 y w d + C1 + H w + C1 + C H At 0, y 0 C 0 At 0, dy/d 0 C 1 0 Therefore, the deflection curve is y w H C

At l/, y h H y wl 8h 4h l h Parabola w H Now, tension in the cable is l 4 T ma corresponds to maimum θ. T H/ cos θ dy w tan θ ; d H θ ma tan 1 wl H T ma H cos tan 1 wl H which can be simplified as T ma wl 1+ l 4h

Length of the cable: Alternatively, + + + / 0 1/ / 0 / 0 / 0 8 1 1 l l l l d l h d dy d dy d ds L l h h l l h l 4 sinh 8 4 1 1 1/ + + Alternatively, This series converges for cables with small slopes (<45 o ); use first few terms only. + + + +... 5 3 3 8 1...] 8 1 1 [1 4 / 0 4 l h l h l d d dy d dy L l

Eample: The left side of a cable is mounted at an elevation of 7 m below the right side. The sag, measured from left support, is 7 m. Find the maimum tension if the cable has uniform loading in the vertical direction of 1500 N/m. 7 m y 7 m 1500 N/m 33 m a Choose the origin as shown above. 1 y w d + C1 + C H At 0, y 0 and y 0; hence, y w H

At a, y 7 m; hence 7 wa ; H a 14H w At 33 a, y 14; hence ( 33 a) w 14 ; H And, a 13.669 m H 0018.878 N T ma occurs at 33 a. T H/cos θ dy w w(33 a) tan θ ; θ d H H 0 55.38 T H cosθ ma 35,35.61 N

A Coplanar Cable with Loading being the self weight of the cable (Catenary): Now, the loading is a function of s, the position along the cable. We can replace in the earlier equations by s. T + T θ + θ T θ w s d ds ( T cos θ ) 0 T cos θ H a constant A d ds ( T sinθ ) w( s) T sin θ w (s) ds + C 1 B

B A, we get dy 1 tanθ d H w( s) ds + C 1 C We cannot integrate the above directly to obtain y. We have, ds d + dy dy ds d ; dy d ds d 1 From [C] we have ds d 1/ 1 1 w( s) ds + C H 1 ds 1 1 + w( s) ds + C d H 1 1/

Separating the variables and integrating, we obtain ds + C 1/ 1 1 + w( s) ds + C1 H Eample: Determine the shape of the cable shown below when loaded by self weight only. y s h l Let w be the weight per unit length of the cable. dy d 1 ws w( s) ds + C1 + C a 1 H H

With the coordinates as shown in figure, at s 0, dy/d 0. Hence C 1 0. Now, ds + 1/ ws 1 + H C Integrating, we obtain H w ws sinh C H 1 sinh + At 0, s 0. Hence C 0. s H w sinh w H From Eq [a], we get dy d sinh w H

H w y cosh + C 3 w H At 0, y 0. Hence C 3 H/w. H w y cosh 1 w H This curve is called a catenary (Latin catena means chain). At l /, y h. h H wl cosh 1 w H This equation can be solved using numerical methods to obtain H. Thus, T ma can be determined.

Eample: A large balloon has a buoyant force of 100 lb. It is held by a 150-ft cable whose weight is 0.5 lb/ft. What is the height h of the balloon above the ground when a steady wind causes it to assume the position shown? What is the maimum tension in the cable? y s h 30 o dy d 1 ws + C (a) 1 H

At s 0, dy/d tan 30 1/ 3; Hence C 1 1/ 3 ds + 1/ ws 1 1 + + H 3 C H w ws H 1 3 1 sinh + + C (b) s H w ( C) w sinh H 1 3 (c) From Eq. (a), we have dy ( C w ) sinh d H (d)

Integrating the above, we have H ( C ) w y cosh + w H C 3 (e) At s 150, from Eq. (a) we get dy d ws 1 0.5 150 + H 3 H 100 tanθ H + V 100 lb 1 3 T (f) V T sin θ 100 lb θ H s h 30 o

At 0, s 0, we get from Eq. (b) H w 1 1 0 sinh + C C H w 3 sinh 1 1 3 From Eq. (f), we can write H 3 H 100 0.5 100 5 3 5; (g) 75 1 tan θ θ 5 3 3 T 0 ma + ; ma 66. 587 H ma cosθ ma 108.97 lb From Eq. (g), C 5 1 3 0.5 sinh 1 3

Introduction Friction Forces Friction is the force distribution at the contact surface between two bodies that impedes sliding motion of one body relative to the other. This force distribution is tangential to the contact surface and acts opposite to the direction of any possible movement. Frictional forces are associated with energy dissipation They are at times undesirable, and at times beneficial! Coulomb friction occurs between bodies with dry contact. Lubrication problems are completely different, and come under the purview of fluid mechanics.

Major cause of dry friction is associated with microscopic roughness of the surface of contact. Interlocking microscopic protuberances oppose the relative motion between the surfaces. During sliding, some of these protuberances are sheared off and some are melted which eplains the high rate of wear during dry friction. A smooth surface of contact can only support a normal force; a rough one in addition supports a tangential force. P

W P N Body on a smooth surface W P Body on a rough surface S N Laws of Coulomb Friction When we slide a block of wood over a surface, we eert continuously increasing load which is completely resisted by friction until the object starts moving, usually with a lurch.

P condition of impending motion or impending slippage dynamic static t Idealised plot of applied force P Coulomb in 1781 presented certain conclusions after carrying out eperiments on blocks which tend to move or are moving without rotation. These conclusions are applicable at the condition of impending slippage or once slippage has begun. These are known as Coulomb s laws of friction.

The laws of Coulomb friction are: 1. The total force of friction is independent of area of contact.. For low relative velocities between sliding objects, the frictional force is independent of velocity. However, the sliding frictional force is less than the frictional force corresponding to impending slippage. 3. The total frictional force is proportional to the normal force transmitted across the surface of contact. Static Coefficient of Friction Steel on cast iron 0.4 Copper on steel 0.36 Hard steel on hard steel 0.4 Mild steel on mild steel 0.57 Rope on wood 0.7 Wood on wood 0.0 0.75

Thus, the frictional force f is proportional to the normal reaction. That is P W f N f µ N f µ N N Note: Direction of f is such that it opposes the motion. Eample 1: A heavy bo weighing 1 kn rests on a floor. The static coefficient of friction between the surfaces is 0.35. What is the largest force P, and what is the highest position h for applying this force that will not allow the bo either to slip or to tip? P h 1 kn 1 m

Consider the free body diagram shown below corresponding to the condition of impending tipping. ΣF y 0: N 1 kn ΣF 0: P 0.35N 0.35 kn ΣM A 0: 1 kn Ph 1 0.5; h 0.5/0.35 1.4 m P h < h ma to avoid tipping. h 1 m N 0.35 N Eample : Will P hold block A in equilibrium, move block A up or is it too small to prevent A from coming down? W A 1000 N W B 400 N P 50 N µ 0. for all contact surfaces. A B 0 o P

N A 1000 N 0. N 0 0 f.b.d. 1 N 1 0. N 1 N 1 0 0 0. N 1 B f.b.d. 400 N P 0. N 3 N 3 Find N 1 and N from f.b.d 1. f.b.d corresponds to B moving to left. Find P. If B moves to the right, determine P 1 (reverse directions of P and the frictional forces).

N A 1000 N 0. N 0 0 0. N 1 N 1 N 1 0 0 0. N 1 B 400 N P 1 0. N 3 N 3

Eample: (a) An automobile is shown in the figure below on a roadway inclined at an angle θ with the horizontal. If the coefficient of static and dynamic friction between the tires and the road are 0.6 and 0.5 respectively, what is the maimum inclination θ ma that the car can climb at uniform speed? It has rearwheel drive and has a total loaded weight of 0 kn. The centre of gravity for this loaded condition has been shown in the diagram. cg 400 mm θ 1.4 m 1.7 m

We assume that the drive wheels do not spin (i.e., there is no slip between the wheels and the road surface). The maimum friction force possible is µ s times the normal force as shown below: y cg 400 mm µ s N 1 a N 1 θ 0 kn N ΣF 0: 0.6 N 1 0 sin θ ma ΣF y 0: N 1 + N 0 cos θ ma (i) (ii) ΣM a 0: 3.1 N 0 cos θ ma 1.4 0 sin θ ma 0.4 (iii)

Eliminate N 1 from (i); then eliminate N from (ii); put these in (iii) and simplify to get tan θ ma 0.3566. Hence, θ ma 19.68 0 If the drive wheels spin, we need to use µ d in the place of µ s. We will get a smaller value for θ ma in that case (θ ma 15.33 0 ). (b) Using the above data, compute the torque needed by the drive wheels to move the car at a uniform speed up an incline with θ 1 0. Wheel dia 450 mm. cg 400 mm f a N 1 1 0 0 kn N

In this case, the friction force f has to be determined from Newton s laws (and not by Coulomb s) as there is no impending slippage between the wheel and the road. ΣF 0: f 0 sin 1 0 Hence, f 4.158 kn. torque Now, looking at the free body diagram of the wheel alone, we get f N 1 torque f r 4.158.45 1.87 knm (c) What force is needed to tow the car up the incline when it is parked with brakes locked on all the four wheels with θ 1 0? What force is necessary to tow it down the incline?

F up cg 0.5 N 1 a N 1 1 0 0 kn 0.5 N N N 1 + N 0 cos 1 0 F up 0.5(N 1 + N ) + 0 sin 1 0 13.94 kn Similarly calculate F down. ( 5.63 kn) cg 0.5 N F 1 0 down a 0 kn 0.5 N 1 N 1 N

Eercise: A ladder 4 m long and 00 N weight is placed as shown. A man weighing 600 N climbs up the ladder. At what position will he induce slipping? Coefficient of static friction for all contacts is 0.. 30 0 0. N B 600 N N B N A 30 0 00 N 0. N A N B 0. N A N A + 0.N B 800 N N A 769.31 N N B 153.846 N

Belt Friction Consider the fleible belt shown in figure wrapped around a portion of a drum. The angle of wrap is β. Assume that the drum is stationary and the belt is just about to move (impending motion). Also, T 1 > T (and hence the belt moves clockwise with respect to the drum). ds θ dθ β T T 1

Consider an infinitesimal segment of the belt as a free body diagram. dθ/ dθ/ T t µ dn s dn T + dt n dθ ΣF t 0: dθ dθ T cos + ( T + dt )cos µ sdn 0 or dt µ s dn (i) ΣF n 0: T dθ dθ sin ( T + dt )sin + dn or T dθ dn (ii) 0

dt T µ dθ s Integrating the above, we obtain That is T 1 β T dt T 0 T 1 ln µ s β T µ dθ s Or T 1 T e µ s β Thus, the ratio of tensions depend only on the angle of wrap β. Note: (i) β should be in radians. (ii) Appropriate µ should be used.

y T 1 W A r T A F If the drum A is forced to the right, as long as the angle of wrap β remains unchanged, the ratio of T 1 /T for impending or actual constant speed slippage will remain unaffected. The torque developed by the belt on the drum as a result of friction is affected by the force F. The torque is obtained as torque T 1 r T r (T 1 T ) r

Eample: A drum having a radius of 400 mm as shown in the figure below needs a torque of 500 Nm to get it to start rotating. If µ s 0.5, determine the minimum aial force F needed to create sufficient tension to start the rotation of the drum. T 1 15 0 Torque r F T T 1 π 0.5 150 µ sβ 180 e e Torque 500 Nm (T 1 T ) r Hence get T 1 and T ; then F (T 1 + T ) cos 15 0 15 0 T

Eample: The seaman pulls with 100 N force and wants to stop the motor boat from moving away from the dock. How few wraps n of the rope must he make around the post if the motor boat develops a thrust of 3500 N. (µ s 0. between rope and the post.) T 1 T µ s β e T 1 3500 N; T 100 N Hence β 17.776 0.83 wraps.

Eample: What is the maimum weight W that can be supported by the system in the position shown? Pulley B cannot turn. Bar AC is fied to cylinder A, which weighs 500 N. µ s 0.3 for all contact surfaces. B 6 m m W C A

6 m T 500 N m 0.3 N 1 C A N 1 ΣF 0: 0.3 N N 1 N 0.3 N ΣF y 0: 0.3 N 1 + N + T 500 N (500 T)/1.09 ΣM a 0: 6 T (0.3 N 1 + 0.3 N ) 0.78 N T 53.78 N

F B W T 0.3π e W T Hence, W 136.731 N Eample: Find the force F needed to start the block of mass 100 kg moving to the right if µ s 0.35. 0 o 50 kg 100 kg F

The Square Screw Thread Consider the action of a nut on a screw with square screw thread. z r p α df dn r is the mean radius from the centre-line of the screw thread; p is the pitch (distance along screw between adjacent threads); L is the lead (distance the nut will advance in the direction of ais in one revolution). And for an n-threaded screw, L np

Forces are transmitted from screw to nut over several revolutions of the thread. Hence, we have a distribution of normal and friction forces. Due to the narrow width, we may assume that the distribution is confined to the centre-line, thus forming a line load distribution over several turns of the thread. The local slope is given by L tan α π r np π r Infinitesimal normal and friction forces dn and ds have been indicated in figure. All elements of the distributed forces have the same inclination with respect to the z-direction.

Hence, the force distribution could be replaced by a single normal force N and a single friction force f at the inclinations shown in figure below. Note also that elements of the distribution have the same moment arm about the centre-line. N P M ΣF z 0: N cos α µ s N sin α P f α ΣM z 0: M z rµ s N cos α + rn sin α

These equations are used to eliminate the force N and to get a relation between P and M z. Thus we get M z P r( µ s cosα + sinα ) cosα µ sinα Is the screw self-locking is an important question. That is, having raised the load, if the torque M z is released will the load P stay at the raised position or will it unwind under the action of the load? For this, go back to the equations of equilibrium. Set M z 0, and simultaneously reverse the direction of friction forces. Eliminating N, we get as the condition for impending unwinding of the screw as: s

which requires that That is, P r( µ s cosα + sinα) cosα + µ sinα s Thus, if the coefficient of friction µ s equals or eceeds tan α, the screw is of self-locking type; else it will unwind under the load. µ s cos α + sinα µ s tanα Eample: A single threaded jackscrew has a pitch of p mm and a mean diameter of 50 mm. What torque M is needed to lift a load P 5000 N up? If the torque is released, will the load stay in the raised position? µ s 0.5. 0 0

P tan α p π r π (5) F Hence, α 0.79 0 M z P r( µ s cosα + sinα) cosα µ sinα s 5000 5 (0.5cosα + sinα) cosα 0.5sinα 3946.4 Nmm It is self-locking as µ s > tan α.