ENGINEERING MECHANICS AND STRENGTH OF MATERIALS QUESTIONS 1 Three forces, 20N, 30N and 40N are in equilibrium find the largest angle they make with each other.
CHOICES a. 104.48 degrees b. 105.58 degrees c. 106.69 degrees d. 107.96 degrees
2 Two forces of 20 units and 30 units act at right angle. What is the magnitude of the resultant force?
a. 36 b. 42 c. 40 d. 44
3 A load of 100lb is hung from a middle of a rope, which is stretched between two rigid walls 30ft apart. Due to the load,the rope sags 4ft in the middle. Determine the tension in the rope.
a. 165 lbs. b. 173 lbs c. 194 lbs d. 149 lbs
4 A boat moving at 12kph is crossing a river 500m wide a.19.47ºdownstream in which a current is flowing at 4kph. In what direction b.19.47ºupstream should the boat head if it is to reach a point on the c.18.43ºdownstream other side of the river directly opposite its starting point d.18.43ºupstream 5 A 100 kg weight rest on a 30º inclined plane. Neglecting friction, how much pull must one exert to bring the weight up the plane?
a. 88.67 kg b. 100kg c. 70.71 kg d. 50 kg
6 A man can exert a maximum pull of 1000N but wishes a. 10 times nearer to lift a new stone door for his cave weighing 20,000N. If he uses a lever, how much closer must the fulcrum b. 20 times farther c. 10 times farther be to the stone than to his hand? d. 20 times nearer
ANSWER
c. 104.48 degrees
DISCUSSION By cosine law 40²= 20² + 30² – 2(20)(30) CosѲ Ѳ= 104.48 degrees √∑Ѳ
a. 36
R = √{(20)² + (30)²} R = 36 units
c.194 lbs.
tan Ѳ = 15/4 Ѳ = 75.068 degrees ∑Fv = 0 2TCosѲ = 100 T = 194 lbs.
b.19.47ºupstream
sinѲ=4/12 Ѳ= 19.47ºupstream
d. 50 kg
d. 20 times nearer
∑F inclined = 0 P = WsinѲ = 100sin(30º) = 50 kg
P
∑Mс = 0 20(X2) = 1(X1) X1/X2 = 20 X1 = 20X2 Thus, the fulcrum must be placed 20 times nearer
7 What is the moment of inertia of a cylinder of radius 5m and mass of 5 kg?
a. 120 kg-m² b. 80 kg-m² c. 62.5 kg-m² d. 72.5 kg-m²
8 What is the magnitude of the resultant force of the two a. 332.5N b. 323.5N forces 200N at 20º and 400N at 144º? c. 313.5N d. 233.5N 9 A block weighing 500kN rest on a ramp inclined at 25º a. 121kN b. 265kN with the horizontal. The force tending to move the block down the ramp is______. c. 211kN d. 450kN 10 A 200 kg crate impends to slide down a ramp inclined a. 612.38N at an angle of 19.29º with the horizontal. What is the b. 628.38N frictional resistance? c. 648.15N d. 654.12N 11 Assume the three force vectors intersect at a single point. F1=4i + 2j +5k, F2=-2i + 7j -3k, F3=2i - j + 6k. What is the magnitude of the resultant force vector, R?
12 Assume the three force vectors intersect at a single point. F1=i + 3j +4k, F2=2i + 7j -k, F3=-i + 4j + 2k. What is the magnitude of the resultant force vector, R?
13 A certain cable is suspended between two supports at the same elevation at 500ft apart, the load is 500 lbs per horizontal foot including the weight of the cable. The sag of the cable is 30ft. Calculate the total length of the cable
a. 14 b. 12 c. 13 d. 15
a. 15 b. 13.23 c. 14.73 d. 16.16
a. 503.76ft. b. 502.76ft. c. 504.76ft. d. 501.76ft.
c. 62.5 kg-m²
I = (1/2)mr² =(1/2)(5)(5) ² I =62.5 kg-m² Where I = mass moment of inertia
a. 332.5N
By cosine Law: R ² = (200) ² + (400) ² 2(200)(400)Cos(36º+20º) R = 332.5N
c.211kN
c. 648.15N
∑F inclined = 0 P = WsinѲ = 500sin(25º) = 211kN
P
∑F inclined = 0 P = WsinѲ = 200(9.81)sin(19.29º) P = 648.15N
b.12
R=F1+F2+F3 =(4i + 2j +5k) + (-2i + 7j -3k) + (2i - j + 6k) IRI=√{(Ai) ²+(Aj) ²+(Ak) ²} IRI = √{(4) ²+(8) ²+(8) ²} IRI = 12 units
a. 15
R=F1+F2+F3 =(i + 3j 45k) + (2i + 7j -k) + (-i + 4j + 2k) IRI=√{(Ai) ²+(Aj) ²+(Ak) ²} IRI = √{(2) ²+(14) ²+(5) ²} IRI = 15 units
a. 504.76ft.
Let S=length of the cable S=L+(8d²/3L) - (32d^4/5L^3) =500+{8(30)²/3(500)} - {32(30)^4/5(500)^3} S = 504.76ft.
14 An iron column of annular cross-section has an outer diameter of 200mm and is subjected to a force of 74kN. Find the thickness of the wall if the allowable compressive stress is 10Mpa.
a. 12.75 mm b. 12.57mm c. 17.75mm d. 15.75mm
15 A force of 10N is applied to one end of a 10 inches diameter circular rod. Calculate the stress.
a. 0.20kPa b. 0.15kPa c. 0.05kPa d. 0.10kPa
16 A steel tie rod on must be made to withstand a pull of a. 0.75 5000lbs. Find the diameter of a rod assuming a factor b. 0.71 of safety of 5 and ultimate stress of 64000 psi. c. 0.84 d. 0.79 17 Determine the outside diameter of a hollow steel tube that will carry a tensile load of 500kN at a stress of 140Mpa. Assume the wall thickness to be one-tenth of the outside diameter.
18 If the ultimate shear strength of a steel plate is 42000 psi, what force is necessary to punch a 0.75 inch diameter hole in a 0.625 inch thick plate?
19 What force is required to punch a 0.5 inch hole on a 3/8 thick plate if the ultimate shearing strength of the plate is 42,000psi.
a. 12.75mm
a. 0.20kPa
b. 0.71
a. 24,940lbs. b. 26,620lbs. c. 24,960lbs. d. 24,740lbs.
d = 10in*(1ft/12in)*(1m/3.28) = 0.254m σ = P/A = P/(πd²/4) = 10/(π(.254)²/4) σ = 197.35 Pa
σ=kP/A, k=factor of safety A=kP/σ=(5(5000))/64000=0.3906in² A=π/4 d² d=0.71in.
d. 112.4 mm
σ= P/A; A=P/σ A= 500,000/140x〖10〗^6 = 0.00357 m² A=π/4(D²)-π/4(D-2t)² Note: t = 0.1D A = π/4(D)²-π/4[D-2(0.1D)]² A = 0.2827 D² 0.00357 = 0.2827 D² D = 0.1124 m. D=112.4 m.
c. 61,850 lbs.
σ = P/A P = σA = σ (πdt) =42,000 (π)(0.75)(0.625) P = 61,850 lbs.
a. 111.3 mm b. 109.7mm c. 113.7mm d. 112.4mm
a. 63,000 lbs. b. 68,080 lbs. c. 61,850 lbs. d. 66,800 lbs.
σ = P/A A = 75,000/10e6; A=.0075m² A=(πD²/4) – (πd²/4) 0.0075= π(0.2)²/4 – πd²/4 D=0.1745m. Solving for t: D=d+2t; 0.2 = 0.1745+2t t = 12.75mm
d. 24,740lbs.
σ = P/A P = σA = σ (πdt) = 42,000 (π)(0.5)(0.375) P = 24,740 lbs.
20 What is the stress in a thin-walled spherical shell of diameter D and a wall thickness t when subjected to internal pressure?
a. S = D/ρt b. S = 4D/ρt c. S = ρD/4t d. S = ρD/t
21 A single bolt is used to lap joint two steel bars together. Tensile force on the bar is 20,000N. Determine the diameter of the bolt required if the allowable shearing stress is 70Mpa.
a. 17 mm b. 18 mm c. 19 mm d. 20 mm
22 A cylindrical water tank is 8m in diameter and 12m high. If the tank is to be completely filled, determine a. 11.77 mm the minimum thickness of the tank plating if the stress b. 13.18 mm is limited to 40MPa. c. 10.25 mm d. 12.6 mm
23 A water reservoir of 24m high and 12m in diameter is to be completely filled with water. Find the minimum a. 24.5 mm thickness of the reservoir plating if the stress is limited b. 28 mm to 50MPa. c. 21 mm d. 26mm
24 A solid shaft 48.2cm long is used for a transmission of mechanical power at a rate of 37kW running at 1760 a. 30 mm rpm. The stress is 8.13MPa. Calculate the diameter. b. 35 mm c. 40 mm d. 50 mm
25 A cylindrical tank with 10 inches inside diameter a. 11.44 mm contains oxygen gas at 2,500 psi. Calculate the b. 11.34 mm required thickness in mm under a stress of 28,000 psi. c. 10.60 mm d. 10.30 mm
c. S = ρD/4t
c. 19 mm
a. 11.77 mm
σL = pD/4t (formula)
σ = P/A = P/(π/4 d²) , d = diameter of the bolt 70 x 〖10〗^6 = 20000/(π/4 d²) d = 0.019 m. d = 19mm. σL = pD/2t Note : The biggest pressure occures at the bottom of the tank P = ωh = 9810 N/m² (12m) = 117.720 Pa t = "pD" /("2" ρt ) = (117.720(8))/(2(40 x 〖10〗^6))= 0.01177 m. t = 11.77 mm.
b. 28 mm
σT = pD/2t Note : The biggest pressure occurs at the bottom of the tank P = ωh = 9810 N/m^3(24m) = 235,440Pa t = "pD" /("2S" ) = (235,440(12))/(2(50x 〖10〗^6))= 0.028m. t = 28 mm.
d. 50 mm
P =2πfT/60 37000= 2π(1760)(T) T = 200.75 N-m T=16T/πd^3 8.13x10^6=16(200.75)/ πd^3 d=0.050 m. d=50 mm.
b. 11.34 mm
σT = pD/2t 28,800=2500(10)/2t t=0.4464in(2.4cm/1in)(10mm/1cm) t=11.34mm
26 What is the modulus of elasticity if the stress is 44,000 a. 41.905e6 b. 42.300e6 psi and unit strain of 0.00105? c. 41.202e6 d. 43.101e6 27 A 2 inch solid shaft is driven by a 36-inch gear and transmits power at 120rpm. If the allowable shearing stress is 12ksi, what horsepower can be transmitted?
a. 41.905e6
E= σ/ ∑=44,000/0.00105 =41.905x10^6Pa
T=16T/ πd^3 12,000=16T/π(2)^3 T=18,849.55lb-in P=2πfT/33,000(12) =2π(120)(18,849.55)/33,000(12) =35.89Hp
a. 29.89 b. 35.89 c. 38.89 d. 34.89
b. 35.89
28 A hollow shaft has an inner diameter of 0.035 m and a. 4500 an outer diameter of 0.06m. Compute for the torque in b. 4100 N-m, if the stress is not to exceed 120MPa. c. 4300 d. 4150
a. 4500
T=16TD/ π(D^4-d^4) 120x10^6=16T(0.06)/ π〖(0.06)^4-(0.035)^4〗 T=4,500N-m
d. 380
P= 2πfT/60 750,000=2π(1500)T/60 T=4,774.648N-m T=16T/ πd^3 =16(4,774.648)/ π(0.04)^3 =380MPa
d. 1.125e-6m^4
J=π/32(D^4-d^4) = π/32〖(0.06)^4-(0.035)^4〗 J=1.125x10^-6 m^4
c. 96.88kW
T=16T/ πd^3 59=16T/ π(55)^3 T=1,927,391.637 N-mm(1m/1000mm) =1,927,39N-m P=2πfT/60=2π(480)(1927.39)/60 =96.8kW
d. 7.46cm
&=PL/AE -----1 σS=P/A -------2 Substitute (2) in (1) &= σL/E=(172x10^6)(30)/69,116x10^6) =0.0746m &=7.46cm
29 Compute the nominal shear stress at the surface in MPa for 40mm diameter shaft that transmits 750kW at a. 218 b. 312 1500rpm. Axial and bending loads are assumed c. 232 negligible d. 380
30 A hollow shaft has an inner diameter of 0.035 m and an outer diameter of 0.06m. Determine the polar moment of inertia of the hollow shaft.
31 What power would a spindle 55mm in diameter transmit at 480rpm. Stress allowed for short shaft is 59N/mm^2
32 A 30m long aluminum bar is subjected to a tensile stress of 172MPa. Find the elongation if E=69,116MPa?
a. 1.512e-6m^4 b. 1.215e-6m^4 c. 1.152e-6m^4 d. 1.125e-6m^4
a. 42.12 kW b. 50.61kW c. 96.88kW d. 39.21kW
a. 0.764m b. 0.007m c. 6.270mm d. 7.46cm
Substitute (2) in (1) &= σL/E=(172x10^6)(30)/69,116x10^6) =0.0746m &=7.46cm 33 A steel wire is 4.0 m long and 2mm in diameter. How much is it elongated by asuspended body of mass 20kg? Young's modulus for steel is 196,000MPa.
a. 1.123mm b. 1.385mm c. 1.374mm d. 1.274mm
d. 1.274mm
&=PL/AE Where: P= weight of the body P=mg=(20x9.81)=196.2N A= π/4(d²)= π/4(0.002)²=3.1416x10^-6m² Substitute &=196.2(4)/(3.1416x10^-6)(196,000x10^6) =1.274x10^-3 =1,274mm
d. 106.48Gpa
&=PL/AE 0.0186=350(10)/(π/4)(0.0015)²E E=1.0648x10^11 Pa E=106.48 GPa
&T = 1-0.99970=0.0003cm. &T= ѲL(t2-t1) 0.0003=1.2x10^-5(1)(t2-30º) T2=55ºC
34 A copper rolled wire 10 m long and 1.5mm diameter when supporting a weight of of 350N elongates 18.6mm. Compute the value of the Young's modulus of this wire.
a. 200GPa b. 180.32GPa c. 148.9GPa d. 106.48GPa
35 A cylinder of diameter 1.0cm at 30ºC is to be slid into a hole on a steel plate. The hole has a diameter of 0.99970 cm at 30ºC. To what temperature the plate must be heated? Coefficient of linear expansion for steel is 1.2e-5cm/ºC.
a. 62ºC b. 65ºC c. 48ºC d. 55ºC
d. 55ºC
36 An iron steam pipe is 200ft long at 0ºC. What will its increase in length when heated to 100ºC? Coefficient of linear expansion 10e-6ft/ºC.
a. 0.18 ft b. 0.12 ft c. 0.28ft d. 0.20ft
d. 0.20ft
37 What is the acceleration of a body that increases in velocity from 20m/s to 40m/s in 3 seconds?
a. 5.0m/s ² b. 6.67m/s ² c. 7.00m/s ² d. 8.00m/s ²
b. 6.67m/s ²
38 How far does an automobile move while its speed increases uniformly from 15kph in 20 seconds? a. 185m b. 167m c. 200m d. 172m
b. 167m
&T= ѲL(t2-t1) = 10x10^-6(200)(100-30º) &T=0.20 ft. V =Vo+ at 40=20+a(3) a=6.67 m/s² Vo=15km/hr(100m/1km)(1hr/3600)=4.167m /sec V=45(1000/3600)=12.5 m/sec. V=Vo+at 12.5=4.167+a(20) a=0.41665m/s² S=Vot +1/2(at²)=4.167(20)+1/2(0.4166650)(20)² S=167 m.
39 From a speed of 75 kph, a car decelerates at a rate of 500m/min² along a straight path. How far in meters will a.790.293m it travel in 45 seconds? b. 791.357m c. 793.238m d. 796.875m
40 A train starting at initial velocity of 30 kph travels a distance of 21 km In 18 minutes. Determine the acceleration of the train at this instant.
41 If a particle position is given by the expansion x(t) = 3.4t³ - 5.4t meters, what is the acceleration of the particle after t=5 seconds?
a. 0.0043m/s² b. 0.0206m/s² c. 0.0865m/s² d. 0.3820m/²
a. 1.02m/s² b. 102m/s² c. 3.4m/s² d. 18.1m/s²
42 The distance a body travels is a function of time and is a. 36 b. 54 given by x(t) = 18t+9t². Find its velocity at t=2. c. 24 d. 20
d. 796.875m
b. 0.0206m/s²
Vo=30km/hr(1hr/3600sec)(1000/1km)=8.33 3 m/sec S=Vot+1/2(at²) 21000=8.333〖18(60)〗-1/2(a)〖18 (60)〗² a=0.0206m/s².
b. 102m/s²
X=3.4t³-5.4t V=dx/dt=3(3.4)t²-5.4 V=10.2t²-5.4 a=dV/dt=20.4t=20.4(5) a=102 m/s²
b. 54
x= 18t + 9t² V=dx/dt=18 + 18t=18 + 18(2) V=54m/s
b. 19.8m/s
D=20t + 5/(t+1) V= dV/dt =20 + -5/(t+1)² =20-5/(4 + 1)² V=19.8 m/s
43 Determine the velocity of progress with the given equation: D = 20t + 5/(t+1) when t=4 seconds.
a. 18.6m/s b. 19.8m/s c. 21.2m/s d. 22.4m/s
44 A ball is dropped from a building 100m high. If the mass of the ball is 10gm, after what time will the ball strike the earth?
a. 4.52s b. 4.42s c. 5.61s d. 2.45s
a. 4.52s
a. 50m/s b. 28m/s c. 19.8m/s d. 30m/s
b. 28m/s
45 A ball is dropped from a roof of a building 40 meters tall will hit the ground with a velocity of:
Vo=75km/hr(1000m/1km)(1hr/60min)=1250 m/min S=Vot-1/2(at²) =1250(45/60)-1/2(500)(45/60)² S=796.875 m.
Note: Since the ball was dropped,initial velocity of the ball is zero. h= Vot + ½(gt²) 100=0+1/2(9.81)t² t²=4.52 seconds V²= Vo²+2gh V²=0+2(9.81)(40) V=28 m/s ΄
46 A ball is thrown vertically upward from the ground and a student gazing out of the window sees it moving upward pass him at 5m/s. The window is 10m above the ground. How high does the ball go above the ground?
a. 15.25m b. 14.87m c. 9.97m d. 11.30m
47 A ball is dropped from a height of 60 meters above the a. 2.1 s b. 3.5 s ground. How long does it take to hit the ground? c. 5.5 s d. 1.3 s 48 The muzzle velocity of a projectile 1500fps and the distance of the target is 10 miles. The angle of elevation of the gun must be:
a. 21º59΄ b. 22º41΄ c. 24º33΄ d. 25º18΄
49 The flywheel of a puncher is to be brought to a. 5 turns avomplete stop in 8 seconds from a speed of 60 revolutions per minute. Compute the number of turns b. 3 turns the flywheel will still make if its deceleration is uniform. c. 4 turns d. 6 turns
50 What is the speed of the synchronous earth satellite situated 4.5e7m from the earth?
a. 11,070.0kph b. 12,000.0kph c. 11,777.4kph d. 12,070.2kph
d. 11.30m
b. 3.5 s
V²= V1²-2gh 0=5² - 2(9.81)(h) V=28 m/s h=1.3m H= 10+h H=10+1.3 =11.3m h=Vot + ½ gt² 60=(0)t + ½(9.81)t² t=3.5 s
c. 24º33΄
R= Vo²sin2Ѳ/g 10(5280)=(1500)²sin2 Ѳ/32.2 Ѳ=24.54 º = 24 º (0.54)(60)’ Ѳ= 24 º32.4’
c. 4 turns
ωo = 60rev/min(1min/60s)=1 rev/s ω = ωo- Ѳt 0=1- Ѳ(8) Ѳ= 0.125 rev/s ω²= ωo²-2(alpha)( Ѳ) 0= 1² - 2(0.125) (Ѳ) Ѳ= 4 rev. or 4 turns
c. 11,777.4kph
V = rω ω = 1rev/24hr(2 πrad/rev)(hr/3600 s) = 7.27x10^-6 rad/s V = (4.5x10^7)(7.27x10^-5 rad/s) V = 3271.5m/s (1km/1000m) (3600 s/ 1hr) V=11.777.4 kph
