ENGINEERING MECHANICS ELEMENTS OF CIVIL ENGINEERING AND ENGINEERING MECHANICS SLIDES OF ALL CHAPTERS

RESOLUTION F F Sin θ θ F cos θ F cos 0 F Sin 0

o θ Resolution of a FORCE A F sin θ F θ B Sin θ =Opp / Hyp = Fy / F F y = F Sin θ F cos θ A F Cos θ = Adj / Hyp = Fx / F F x = F Cos θ Hyp Opp F y θ O Adj F x B --------------------------------- Reverse Combination = √ ( F sin θ ) ² + ( F cos θ )² = √ F² ( sin² θ + cos² θ ) = F ---------------------------

75 N 40 N 15 N 35 N 50 N 25 N 40 N 20 N 50 N 30 N 45 N 55 N ∑ F x = 40 + 20 + 50 – 25 – 50 – 35 = 0 ∑ F y = 15 + 40 + 75 - 30 – 45 – 55 = 0

+, + - , + - , - + , - Sign Conventions I st Quadrant 2 nd Quadrant F2 sin θ I st Quadrant F1 sin θ 2 nd Quadrant F2 cos F1 cos θ θ F4 cos F3 cos θ θ 3 rd Quadrant F3 sin θ 4 th Quadrant F4 sin θ + , - - , -

- 100 cos 60 - 200 cos 65 - 200 sin 65 - 150 sin 25 Y-axis 250 N 100 N 45 100 cos 60 250 cos 45 X-axis 150 cos 25 200 cos 65 25 65 25 150 N 200 sin 65 150 sin 25 200 N - 100 cos 60 - 200 cos 65 ∑ F x = + 250 cos 45 + 150 cos 25 = 142.84 N - 200 sin 65 - 150 sin 25 ∑ F y = + 250 sin 45 + 100 sin 60 = 18.73 N

∑ F x = 142.84 N ∑ F y = 18.73 N ∑ F y R ∑ F y α Magnitude of the Resultant, R = √ ∑ F x ² + ∑ F y ² = √ 142.84 ² + 18.73² R = 144.06 N ∑ F x Direction of Resultant, tan α = ∑ F y / ∑ F x = 18.73 / 142.84 = 0.1311 α = 7º 28’ 13’’

COMPOSITION α P R Q θ O ---------------------------------------- R = √ p2 + q2 + 2 p q cos θ Tan α = l q sin θ / p + q cos θ l

Q α P tan α = Q / P R = √ P² + Q² Q R 90 MAGNITUDE OF RESULTANT DIRECTION OF RESULTANT ------------------ tan α = Q / P R = √ P² + Q²

1.2 STATICS Statics Deals With the Equilibrium of Bodies, Those That Are Either at Rest or Move With a Constant Velocity. Dynamics Is Concerned With the Accelerated Motion of Bodies and Will Be Dealt in the Higher Semester.

EQUILIBRIUM OF A PARTICLE CONCLUDED For equilibrium:  Fx = 0 and  F y = 0. Note: Considering Newton’s first law of motion, equilibrium can mean that the particle is either at rest or moving in a straight line at constant speed.

PRINCIPLE OF TRANSMISSIVITY OF FORCES Principle of transmissivity states that the condition of rest or motion of a rigid body is unaffected if a force, F acting on a point A is moved to act at a new point, B provided that the point B lies on the same line of action of that force. F F A B

MOMENT OF A FORCE ABOUT A POINT The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis.

Resolution of a Force into a Force at B and a couple A Force, F acting at point A on a rigid body can be resolved to the same force acting on another point B and in the same direction as the original force plus a couple M equal to r x F i.e. moment of F about B i.e Force in (a) equal to that in (b) equal to that in C M = 25 kN m F = 5 kN 5 kN 5 kN B A B 5 kN 5 m (a) (c) (b)

d d Force replaced by force and couple F A B = B A B = M= F x d F F F Moment of the pair of forces = F x d F

Varignon’s Theorm ( Moment Theorm) Summation of moments of all forces about one moment centre is equal to the moment due to RESULTANT alone about the same moment centre. Summation Mc = Rx Perpendicular Distance from C

Sign Conventions Clockwise Moments -- Positive + Counter Clockwise -- Negative -

For equilibrium of coplanar non- concurrent force system:  Fx = 0 ------ 1  F y = 0. ------ 2  M c = 0 ------ 3 Equations 1 and 2 are called force equations Equation 3 is called Moment equation

LAMI’S THEORM ץ β α P Q P = K sin α Q = K sin β R = K sin ץ Q P P Q R ------ ------ = = ------ Sin α Sin β Sin ץ R R

Lami’s Theorm Statement When three concurrent forces are acting on a body simultaneously, be in equilibrium then any force is directly proportional to the SINE of the angle subtended by other two forces. P Q R ------ ------ = ------ Sin α Sin β Sin ץ

Find Resultant completely

∑ F x = 5000 KN 120 KN ∑ F y = +120 – 1120 - 420 = - 1420 KN 5000 KN 1120 KN Magnitude of Resultant R = √ ∑ F x ² + ∑ F y ² ------------------------- 420 KN ------------------------- R = √ 5000 ² + 1420 ² R = 5197.7 kN 5000 1420 tan α = 5000 α 1420 R α = 15º 51´ 16 ´´ ( Fourth Quadrant )

α Applying Varignon’s Theorem Taking Moments about ‘ o ‘ ┴ d ∑ M o = R x ┴ d R 5197.7 kN + 5000 x 4 + 1120 x 2 – 120 x 4 + 420 x 5 = 5197.7 x ┴ d 23860 ┴ d = = 4.59 m 5197.7

15 sin 60 40 sin 60 20 kN 15 cos 60 10 sin 30 40 cos 60 60 10 cos 30

y 40 sin 60 20 15 sin 60 10 sin 30 x 40 cos 60 10 cos 30 15 cos 60 ∑ F x = + 10 cos 30 + 15 cos 60 – 40 cos 60 = - 3.84 kN ∑ F y = 10 sin 30 + 15 sin 60 + 20 + 40 sin 60 = 72.63 kN

R α ∑ F x = - 3.84 kN ∑ F y = 72.63 kN --------------------------- = √ 5289.86 α R = 72.73 kN 72.63 tan α = - 3.84 α = 86 º 58 ´ 24 ´´ Resultant 72.73 kN acts in second Quadrant at angle of 86 º 58 ´ 24 ´´

α R 20 kN 25 kN 50 kN 35 kN ------------------ ∑ F x = 25 – 20 = 5 kN = √ 7250 ∑ F y = - 50 – 35 α R = 85.15 kN = - 85 kN 85 tan α = 5 α = 86 º 38 ´ 1 ´´ R Resultant 85.15 kN acts in Fourth Quadrant at angle of 86 º 38 ´ 1 ´´

X X √ √ α ∑ M o = R x ┴ d R = 85.15 kN Applying Varignon’s Theorem ┴ d Taking Moments about ‘ o ‘ ┴ d α ∑ M o = R x ┴ d R = 85.15 kN X X + 25 x 3 + 35 x 4 = 85.15 x ┴ d ┴ d = 2.53 m √ √

1500 sin 60 1805 Sin 33.67 1500 cos 60 1805 cos 33.67 2240 cos 63.40 2240 sin 63.40

1500 sin 60 1805 Sin 33.67 1500 cos 60 1805 cos 33.67 2240 cos 63.40 2240 sin 63.40

y 1500 sin 60 1500 cos 60 1805 cos 33.67 2240 cos 63.40 x 1805 Sin 33.67 2240 sin 63.40 ∑ F x = +2240 cos 63.40 – 1805 cos 33.67 – 1500 cos 60 = - 1249.22 N ∑ F y = 1500 sin 60 – 1805 sin 33.67 – 2240 sin 63.40 = - 1704.58 N

α R 1249.22 1704.58 ---------------------------------- = √ 4466143.6 R = 2113.3 N 1704.58 tan α = 1249.22 α = 53 º 45 ´ 49´´ Resultant 2113.3 N acts in Third Quadrant at angle of 53 º 45 ´ 49´´

α 1500 sin 60 1805 Sin 33.67 1500 cos 60 1805 cos 33.67 √ ┴ d 2240 cos 63.40 √ √ √ √ X 2240 sin 63.40 - 1500 cos 60x 3 – 1500 sin 60 x 2 – 1805 cos 33.67 x 3 + 1805 sin 33.67 x 3 + 2240 sin 63.4 x 4 = - 2113.3 x ┴ d

α ∑ M o = R x ┴ d R = 2113.3 N Applying Varignon’s Theorem Taking Moments about ‘ o ‘ ∑ M o = R x ┴ d ┴ d R = 2113.3 N 0.79m - 1500 cos 60x 3 – 1500 sin 60 x 2 – 1805 cos 33.67 x 3 + 1805 sin 33.67 x 3 + 2240 sin 63.4 x 4 = - 2113.3 x ┴ d 1659.07 Negative sign indicates that the RESULTANT lies below point ‘o’ at same distance 0.79 m ┴ d = - 2113.3 ┴ d = - 0.79 m 36 36 36

F B D Lami’s thoerem T1 T2 1000 T1 = 500 N = = Sin 90 Sin 150 Sin 120

TCB TCA TCB = 7.76 N TCA = 10.98 N T CA T CB F B D 75 150 135 15 N Lami’s theorem TCB TCA 15 TCB = 7.76 N = = Sin 75 Sin 150 Sin 135 TCA = 10.98 N

Lami’s thoerem T1 T2 5 T1 = 2.89 N = = Sin 150 Sin 90 Sin 120 T2 = 5.77 N

R 5 kN S 10 kN 7 kN

R S ∑ F y = 0 ∑ F x = 0 R - 10 – 7 sin 45 - 0.058 sin 30 =0 30º 5 kN 45º S 10 kN 7 kN ∑ F y = 0 ∑ F x = 0 R - 10 – 7 sin 45 - 0.058 sin 30 =0 - S cos 30 + 5 – 7 cos 45 = 0 R =14.98 kN S = 0.058 kN

60 30 105º 120º 135º At Joint B ∑ Fy = 0 At Joint D T3 sin 60 – 183.01 sin 30 – 200 = 0 Lami’s thoerem T3 = 336.6 N T1 T2 250 = = ∑ Fx = 0 Sin 120 Sin 135 Sin 105 336.6 cos 60 + 183.01 cos 30 – T4 = 0 T1 = 224.14 N T4 = 326.79 T2 = 183.01 N

250 125 mm 125 mm α 110 Cos α = 110 250 α = 63º 53 ‘ 46’’

R B R A RA = 49 N R B R A RB = 111.36 N Consider F B D of cylinder o1 116º 6’ 14’’ R A 63º 53 ‘ 46’’ 90 153º 53’ 46’’ 100 N Lami’s theorem RA = 49 N R B R A 100 = = Sin 153º53’46’’ Sin 90 Sin 116º6’14’’ RB = 111.36 N

R C R D R B R D – 111.36 cos 63º53’46’’ = 0 R c = 200 N R D = 49 N Consider F B D of cylinder o2 R C R D 63º 53 ‘ 46’’ R B 111.36 N 100 N ∑ Fy = 0 ∑ Fx = 0 R D – 111.36 cos 63º53’46’’ = 0 R c – 100 – 111.36 sin 63º53’46’’ = 0 R c = 200 N R D = 49 N

Q. Find Reactions at all points of contacts, force P required to keep system in equilibrium and force in connecting member A B 60 54

60

Lami’s theorem T AB 4000 R 1 R1 = 5464.1 N T AB = - 4898.98 N 240º 45º Consider F B D of cylinder A W A T AB 4000 R 1 = = 30 Sin 45 Sin 75 Sin 240 15 60 T AB R1 R1 = 5464.1 N 60 T AB = - 4898.98 N 240º 45º Negative sign indicates TAB must act in opposite direction. i. e. Acts in Second quadrant 30 15 R1 W A = 4000 N T AB 75º

Lami’s theorem T AB 4000 R 1 R1 = 5464.1 N T AB = 4898.98 N T AB 135 Consider F B D of cylinder A TAB in opposite direction Lami’s theorem W A T AB 4000 R 1 = = 30 Sin 135 Sin 105 Sin 120 15 60 T AB R1 R1 = 5464.1 N 60 W A = 4000 N T AB = 4898.98 N 135 60º Here TAB is positive and acts in Second Quadrant with same Magnitude 15º 120 R1 T AB 105 57

60 58 58

2000 N 60º P 15º T AB 15 30º 15º 45º R 2 P 45º T AB 15º 30º Consider F B D of cylinder B 45º 45º R 2 2000 N

P T AB 15º 30º 45º 45º R 2 60 2000 N Consider F B D of cylinder B

R 2 T AB= 4898.98 N P Consider F B D of cylinder B TAB cos 15 15º 30º R2 sin 45 R 2 P cos 30 R2 cos 45 TAB cos 15 45º 4898.98 sin 15 15º 30º P sin 30 T AB= 4898.98 N 2000 N P Resolving forces vertically Resolving forces Horizontally ∑ F y = 0 ∑ F x = 0 TAB cos 15 – R2 cos 45 – P cos 30 = 0 R2 sin 45- P sin 30 -4898.98 sin 15 – 2000 = 0 4898.98 cos 15 – R2 cos 45 – P cos 30 = 0 R2 cos 45 + P cos 30 = 4732.05 R2 sin 45- P sin 30 = 3267.95

R2 cos 45 + P cos 30 = 4732.05 ------Eq 1 x sin 30 R2 sin 45 - P sin 30 = 3267.95 -----Eq 2 x cos 30 R2 cos 45x + P cos 30 x = 4732.05 x sin 30 sin 30 sin 30 R2 sin 45 - P sin 30 = 3267.95 x cos 30 x cos 30 x cos 30 R1 = 5464.1 N R2 x 0.966 = 5196.15 R2 = 5379.04 N T AB = 4898.98 N P = 1072.13 N R2 = 5379.04 N P = 1072.13 N

Find reactions at all points of contacts RA , RB, RC , and RD Both cylinders weigh 200 N each R B R D R B R A R C

Consider F B D of cylinder 1 R B R A 200 N 30º 60º R A R B

Consider F B D of cylinder 1 30º 60º R A R B 66

Lami’s theorem R A R A 200 RA = 173.2 N RB = 100 N R B RB Consider F B D of cylinder 1 R A Lami’s theorem 90º R B R A RB 200 = = 150º 120º Sin 90 Sin 120 Sin 150 200 N RA = 173.2 N RB = 100 N 67 67

R D R D R C R C R D– R C cos 60 = 150 Rc = 330.94 N Consider F B D of cylinder 2 200 N 200 N R B 173.2 N R B R D R D 30 60 R C R C Resolving forces Horizontally Resolving forces vertically ∑ F x = 0 X or Y First ∑ F y = 0 R D – 173.2 cos 30 – R C cos 60 = 0 Rc sin 60- 173.2 sin 30 – 200 = 0 R D– R C cos 60 = 150 Rc = 330.94 N R D = 330.94 cos 60 + 150 R D = 315.47 N

R2 45 R1 Lami’s Theorem W1 Lami’s Theorem R1 W1 R2 W2 Resolution R2 R3 R4 Resolution R4 45 R3 Resolution 45 R2 W2

Sequence of equilibrium of cylinders to be considered are--- Lami’s Theorem Lami’s Theorem Resolution Lami’s Theorem Resolution Resolution Sequence of equilibrium of cylinders to be considered are--- First---- Cylinder A * Lami’s Theorm Second-- Cylinder C * Lami’s Theorem Third ----- Cylinder B * Resolution Sequence of equilibrium of cylinders to be considered are--- First---- Cylinder C * Lami’s Theorm Second-- Cylinder A or B * Resolution Third ----- Cylinder B or A * Resolution

80 N B 80 N 100 mm 100 sin 60 100 N 60º 100 mm C 120 cos 30 100 cos 60 A 30º 100 sin 60 120 N 120 sin 30 80 N 120 cos 30 100 cos 60 80 N 120 sin 30

∑ F x = +80 – 100 cos 60 - 120 cos 30 = - 73.9 N ∑ F y = 100 sin 60 – 80 – 120 sin 30 = - 53.39 N 100 sin 60 80 N Magnitude of Resultant R = √ ∑ F x ² + ∑ F y ² 120 cos 30 100 cos 60 80 N 120 sin 30 R = √ ( - 73.9 ) ² + ( - 53.39 ) ² R = 91.17 N - 53.39 tan α = - 73.9 α = 35º 50´ 48 ´´ ( Third Quadrant )

X X X √ √ √ α ∑ M A = R x ┴ d R = 91.17 N 80 N B 80 N α = 35º 50´ 48 ´´ Applying Varignon’s Theorem Taking Moments about A 100 mm 100 sin 60 100 N ∑ M A = R x ┴ d 60º 100 mm R = 91.17 N C 120 cos 30 100 cos 60 A 30º ┴ d 120 N 120 sin 30 X R = 91.17 N +120 sin 30 x 100 + 80 x 100 + 80 x 50 = + 91.17 x ┴ d X X 18000 ┴ d = √ 91.17 √ √ ┴ d = 197.4 mm

2kN 1.2m 2 3 1m θ1 1 θ2 Each Box measures 1 m x 1.2 m 4 5 θ3 2 kN 2 cos 39º 48’ 20’’ = 1.54 kN 2 sin 39º 48’ 20’’ = 1.28 kN 6 15 kN 5 kN tan θ1 = 1 / 1.2 5 kN 5 cos 78º 13’ 54’’ = 1.02 kN 5 sin 78º 13’ 54’’ = 4.89 kN θ1 = 39º 48’ 20’’ tan θ2 = 3 / 4.8 θ2 = 78º 13’ 54’’ tan θ2 = 1 / 2.4 15 kN 15 cos 22º 37’ 11’’ = 13.85 kN 15 sin 22º 37’ 11’’ = 5.77 kN θ3 = 22º 37’ 11’’

. . CENTROID Y 20 mm X2 Y2 X1 50 mm Y1 25 mm X 30 mm 30 1/3x20=6.67 15 50 mm Y1 25 25 mm 25 X 30 mm

. . CENTROID Y 20 mm X2 Y2 X1 50 mm Y1 25 mm X 30 mm 30 1/3x20=6.67 15 50 mm Y1 25 25 mm 25 X 30 mm