SOLUTIONS D-02 APPLIED MECHANICS (June 2003)
Q1. a. A The friction force is absent hence the total reaction is normal to the surface at the point of contact.
b. C For equilibrium of a body subjected to two forces, the forces must be equal in magnitude, opposite in direction and must have the same line of action.
c. B For a plane area, with the x,y axes in the plane and z axis normal to the plane, the polar moment of inertia Izz = Ixx + Iyy.
d. B Let d be the depth of the well and t the time splash is heard. Then the relation t =√(2d/9.81) + d/350 = 4 is satisfied for d = 70.77 m.
e. A The normal acceleration = V2/R = (72×1000/3600)2/200 = 2 m/s2.
f. B The shear stress = Force/Area resisting shear = F/(πDt).
g. B The modulus of elasticity is defined as the ratio of direct stress and direct strain within elastic limit.
h. B The B.M. at any section at a distance x from the load P at the free end is Px. The maximum B.M. would be at the fixed end xmax = L.
Q.2.
Consider the I-section to be divided in three parts 1, 2 and 3 as shown in Fig.2. Let xi, yi be the coordinates of the centroids of the ith area Ai. The centroid of the whole section is at C(xC, yC).
As the section is symmetrical about the y axis its centroid would lie on the y axis, i.e. xC = 0.
yC = ∑Aiyi/∑Ai
= (150×2.5 + 75×12.5 + 100×22.5)/(150+75+100)
= 10.96 cm.
Q.3.
Consider the F.B.Ds. of the
joints E, D and C shown in Fig.3. The F.B.Ds are shown assuming tensile forces
in the various members.
Considering the equilibrium of the joint E:
∑Fx = - FCEcos45 = 0 → FCE = 0, ∑Fy = 0 → - FDE - FCEsin45 = 0 → FDE = 0.
Considering the equilibrium of the joint D:
∑Fy = FDE - FBDsin45 – 1 = 0 → FBD = -√2 kN, i.e. √2 kN(C).
∑Fx = - FCD – FBDcos45 = 0 → FCD = 1 kN, i.e. 1 kN(T).
Considering the equilibrium of the joint C:
∑Fx = FCD + FCEcos45 – FACcos45 = 0 → FAC = √2 kN, i.e. √2 kN(T).
∑Fy = - FBC + FCEsin45 - FACsin45 = 0 → FBC = - 1 kN, i.e. 1 kN(C).
Q.4.
A developed screw thread of
pitch p, mean diameter d is shown in Fig. 4. The helix angle with
the horizontal θ = tan-1 p/πd = tan-1
1.25/10π = 2.280.
The F.B.D. of the weight W while being lifted is also shown.
The friction angle φ = tan-1μ = tan-10.2 = 11.310.
The force F at the end of the lever of length R is equivalent to a horizontal force P at the screw thread,
P = F×R/(d/2) = 50F/(10/2) =10F.
Case(i)
Consider the equilibrium of the load at impending slip upward as shown. The friction force f = μN.
∑FV = Ncosθ – f sinθ – W → N = W/(cosθ + μsinθ).
∑FH = P – Nsinθ – f cosθ, → P = N(sinθ +μcosθ) = W(sinθ +μcosθ)/(cosθ + μsinθ).
→ P = Wtan(θ + φ) → F = Wtan(θ + φ)/10 = 50tan(2.28 + 11.31)/10 = 1.209 kN.
Case(ii)
For impending motion down, the direction of friction force is reversed and hence,
→ P = Wtan(θ - φ) → F = Wtan(θ - φ)/10 = 50tan(2.28 - 11.31)/10 = - 0.795 kN.
Efficiency = effort without friction/effort with friction, i.e.
η = tanθ/tan(θ + φ) = tan2.28/tan(2.28+11.31) = 0.165 = 16.5%.
As a force in the reverse direction is required to lower the load, the jack would be self locking.
Q.5.
The F.B.Ds. of the blocks are shown in Fig.5. The block on the horizontal surface is subjected to its weight M1g, normal reaction N, friction force μN and the tension T. The other block is subjected to its weight M2g and the tension T. The magnitude of acceleration of both the blocks is the same, i.e. a. The equations of motion for the blocks of mass M1 and M2 are,
N = M1g. (1)
T – μN = M1a (2)
M2g – T = M2a (3)
From equations (1) to (3)
a = (M2 - μ M1) g/( M1 + M2) = (5 – 0.25×10)×9.81/(10 + 5) = 1.635 m/s2.
T = M2( g – a) = 5(9.81 – 1.635) = 40.875 N.
Q.6.
Let at be the uniform tangential acceleration of the car. If it starts from rest, the speed V after time t would be, V = att.
Then, at = V60/t = (18×1000/3600)/60 = 1/12 = 0.083 m/s2.
The speed after 30 s would be V30 = att = (1/12)×30 = 2.5 m/s.
The normal acceleration an = V302/R = 2.52/250 = 0.025 m/s2.
Q.7.
As the cylinder rolls without slip on the horizontal plane AB, the instantaneous centre of rotation is at the point of contact O of the cylinder with the horizontal plane (see Fig.7). The velocity of any point on the cylinder is proportional to the distance from O and normal to the line joining the point to O.
Let ω be angular velocity of the cylinder. Then,
ω = VC/OC = 20/1 = 20 rad/s.
The velocity of point E, VE = ωOE = 20√2 m/s and is normal to the line OE as shown, i.e. VE = 20i + 20j m/s.
The velocity of point F, VF = ωOF = 20×2 = 40 m/s and is normal to the line OF as shown. i.e. VF = 40i m/s.
Q.8.
The F.B.D. of the beam is shown in Fig.8. Considering the equilibrium of the beam,
∑MA = RD×12.5 – (10×5)×2.5 - 80×7.5 – (16×2.5)×13.75 = 0 → RD = 102 kN.
∑Fy = RA + RD - 10×5 – 80 – 16×2.5 = 0 → RA = 68 kN.
The expressions for the S.F. V(in kN) and B.M. M(in kNm) between various sections are,
0 ≤ x ≤ 5 m V = - 68 + 10x, M = 68x – 10x2/2,
5 ≤ x ≤ 7.5 m V = - 68 + 10×5, M = 68x - 10×5(x-2.5),
7.5 ≤ x ≤ 12.5m V = - 68 + 10×5 + 80, M = 68x - 10×5(x-2.5) - 80(x-7.5),
12.5 ≤ x ≤ 15m V = - 16(15 - x), M = - 16(15 - x)2/2,
The S.F. and B.M. diagrams
with values at important sections are also shown in Fig.8.
The maximum S.F. is
Vmax = - 68 kN,
at A (x = 0).
The maximum B.M. is Mmax = 260 kNm,
at C (x = 7.5 m).
7.5 ≤ x ≤ 12.5m,
M = - 62x + 725 = 0,
at x = 725/62 = 11.7 m
which is a point of contraflexure.
Q.9.
The steel bar is loaded as shown in Fig.9.
The cross-sectional area of the bar A = πd2/4 = π×252/4 = 490.87 mm2.
The axial force F in various portions of the bar OD would be,
Portion OB (0 ≤ x ≤ 50cm):
POB = 40 kN,
Hence, stress σOB = POB/A = 40×103/(490.87×10-6) = 81.5×106 Pa = 81.5 MPa.
Portion BC (50 cm ≤ x ≤ 90cm):
PBC = 40 – 20 = 20 kN,
Hence, stress σBC = PBC/A = 20×103/(490.87×10-6) = 40.7×106 Pa. = 40.7 MPa.
Portion CD (90 ≤ x ≤ 110cm):
PCD = 30 kN,
Hence stress σCD = PCD/A = 30×103/(490.87×10-6) = 61.1×106 Pa. = 61.1 MPa.
The total elongation δ = δOB + δBC + δCD = (σOBOB + σOBBC + σOBCD)/E
= (81.5×106×0.5 + 40.7×106×0.4 + 61.1×106×0.2)/(210×109) = 0.33×10-3 m.
= 0.33 mm.
Q.10.
Let a circular shaft of radius r and length l be subjected to a torque T as shown in Fig.10. The following basic assumptions are made while deriving the torsion formula:
The material is linearly elastic following Hooke’s law.
A plane section normal to the axis of the shaft remains plane and normal during deformation.
A radial line in the section remains radial during deformation.
Consider a radial plane OAA1O1. After deformation the plane occupies the position OA'A1O1, where θ is the angle of twist over the length l.
Hence, the shear strain between circumferential and axial elements at radius ρ is γ = ρθ/l.
As the material follows Hooke’s law, the shear stress τ = Gγ = Gρθ/l.
Both the shear strain γ and shear stress τ are proportional to the radial distance ρ.
The maximum shear stress τmax = Grθ/l is at the outer fibers. (1)
The shear stress distribution over the cross-section is equivalent to the applied torque T
(2)
where, J = ∫AρdA = πr4/2
is the polar moment of inertia of the cross-sectional area A.
Combining the equations (1) and (2),
.
Q.11(i).
Complementary shear stresses:
Consider an infinitesimal element of material at O, subjected to shear stresses τxy on the planes OB, AC and τyx on the planes OA, BC. Considering the moment equilibrium of the element, ∑MO = 0 → τxy = τyx.
The pair of shear stresses like τxy, τyx are called complementary shear stresses. They act at mutually perpendicular planes at a point, both directed towards or away from the common edge and are equal in magnitude.
Q.11(ii).
Tensile stress strain curve for a ductile material:
The tensile engineering
stress σ and the engineering strain ε curve for a
ductile material like mild steel is shown in Fig.11(ii).
The engineering stress = axial load/original area of cross-section, i.e. σ = P/A0.
The engineering strain = change in length/original length, i.e. ε = (L – L0)/L0.
The important points on the curve are:
The proportional Limit P is the point upto which the stress is directly proportional to strain, i.e. follows Hooke’s law.
The elastic limit E is the point upto which only elastic deformation takes place and the material returns to its original undeformed state on unloading. Generally, P and E are so close that they are indistinguishable.
Loading beyond the elastic limit also causes permanent deformation and the deformation continues with very little increment in the load. The upper yield point YU and the lower yield point YL correspond to upper and lower points on the kink (if present) in the stress strain curve beyond the elastic limit. Generally the lower yield point is more reliable and is taken as the yield strength σY of the material. In the absence of a well defined yield point, the yield strength is taken as the stress which causes 0.2% permanent strain.
The ultimate point U corresponds to the maximum load the material can withstand. The stress σU is called the ultimate strength of the material. The necking in the specimen initiates at the ultimate point.
The fracture point F corresponds to the fracture of the material. The stress σF is called the fracture stress. As the original area of cross-section is used in stress calculation, the fracture stress is less than the ultimate stress.
Q.11(iii).
Law of polygon of forces:
The resultant of a number of coplaner forces can be found graphically using the law of polygon of forces. The law can be stated as follows:
If several coplaner forces
are acting at a point such that they are represented in magnitude and direction
by the sides of a polygon taken in the same order, their resultant is
represented in magnitude and direction by the closing side of the polygon in
the reverse direction.
If the polygon is closed, i.e. no closing side is required, the forces would be in equilibrium.
For example, consider coplaner forces F1, F2, F3 and F4 acting at a point O as shown in Fig.11(iii). The forces are represented in magnitude and direction by the sides ab, bc, cd and de, respectively. The closing side ae of the polygon abcde represents the resultant R of the forces. If the points a, e are coincident, the resultant R = 0, and the forces would be equilibrium.
Q.11(iv).
General plane motion of a rigid body can be considered as the sum of a plane translation and a rotation about an axis perpendicular to the plane motion:
Consider
a body undergoing plane motion in the xy plane from position 1 at time t
to position 2 at time t + Δt as shown in Fig.11(iv). This
general plane motion can be thought of as translation and rotation as follows:
(a) Select a point A1 in the body and translate the whole body in the xy plane with displacement ΔRA such that the point A1 occupies its final destination A2. The body is in position 1*.
(b) Rotate the body about the z axis through the point A2 by an amount Δθ to obtain the final position 2.
If instead of A1, some other point B1 is chosen for translation, the translation ΔRB ≠ ΔRA, but the rotation Δθ would remain the same with the z axis passing through the final position B2 of the point B1.
SOLUTIONS D-02 APPLIED MECHANICS (December 2003)
Q1. a. A If the forces are concurrent, the moment about any point would also be zero which may not be so in other cases.
b. D At impending slip the frictional force is equal to limiting friction.
c. A The efficiency η = M.A./V.R. = (1000/90)/15 = 0.74 > 0.5.
d. C The altitude of the equilateral triangle which is also a median is (√3/2)a. One third of it is a/(2√3).
e. C In a compound lever, the simple leverages are all multiplied.
f. C The horizontal range of a projectile with projection angle θ is (u2/g)sin2θ which is maximum for θ = 450.
g. D The work done = torque×angle of rotation = 50×4π = 628 Nm.
h. C The point of contraflexure occurs in beams with two or more spans.
Q.2.
Let M1 be the mass of the shot and V its absolute velocity at an angle θ to the horizontal. The relative velocity of the shot Vr with respect to the gun would be along the gun barrel at an angle α to the horizontal. Let Vg be the recoil velocity of the gun in the opposite direction.
The equations relating the absolute and relative velocities of the shot and the recoil velocity of the gun are
Vrsinα = Vsinθ. (1)
Vrcosα = Vcosθ + Vg. (2)
The conservation of momentum in the horizontal direction yields,
M1Vcosθ - mM1Vg = 0. (3)
From equations (1) to (3)
tanα = mtanθ/(m+1).
Q.3.
Assume that the block of weight W is given a virtual displacement δW down along the plane. The corresponding upward virtual displacement δP of the effort P would be,
δP = δW/2. (1)
As the system is in equilibrium, according to the principle of virtual work, the virtual work done by all the forces must be zero. The forces which contribute to the virtual work are the effort P and the component of the weight of the block along the plane Wsin20.
P× δP – Wsin20×δW = 0. (2)
From equations (1) and (2),
P = 2Wsin20.
Q.4.
Let V be the velocity of the stone required to hit the bird and t the time at which it would be hit. Then,
25 = Vcos30×t → t = 25/Vcos30 (1)
10 = Vsin30×t – (½)gt2 (2)
Substituting for time t from equation (1) in equation (2),
10 = 25tan30 – 4.905(25/Vcos30)2
→ V = (25/cos30)/√((25tan30 -10)/4.905) = 30.4 m/s.
Q.5.
The F.B.D. of the safe for
possible slide down the plank of its own is shown in Fig.5. The safe is
subjected to its 4000 N weight, the normal reaction N of the inclined
plane and the limiting friction force μN. The inclination of the
plank to the horizontal is α = tan-11.2/2.4 = 26.5650.
The forces normal to the plank must be in equilibrium,
N = 4000cos26.565 = 3577.7 N.
The limiting friction μN = 0.3×3577.7 =1073.3 N.
The component of the weight of the safe parallel to the plank is 4000sin3577.7 = 1788.9 N
As the weight component parallel to the plank is more than the limiting friction force, the safe can slide down the plank of its own.
Q.6.
The F.B.Ds. of the joints F, D and E are shown in Fig.6. The forces are assumed tensile in the members and are assigned positive sign. All the inclined members are at 450 to the horizontal.
Considering the equilibrium of joint F:
∑Fx= - FDFcos45 – FEFcos45 = 0, ∑Fy= FDFsin45 – FEFsin45 -15 = 0
→ FDF = 15/√2 kN, i.e 10.6 kN(T) and FEF = - 15/√2 kN, i.e. 10.6 kN(C).
Considering the equilibrium of joint D:
∑Fx= FDFcos45- FADcos45 – FDEcos45 = 0, ∑Fy= FADsin45– FDEsin45 – FDFsin45 = 0.
→ FAD = 15/√2 kN, i.e. 10.6 kN(T) and FDB = 0.
Considering the equilibrium of joint E:
∑Fx= FEFcos45 – FBEcos45 - FCEcos45 = 0, ∑Fy= FEFsin45 + FBEsin45 – FCE sin45 = 0.
→ FCE = - 15/√2 kN, i.e. 10.6 kN(C) and FBE = 0.
Q.7.
A vehicle of mass m is negotiating a level curve of radius R with uniform speed V as shown in Fig.7. The C.G. of the vehicle C is at a height h above the ground and the distance between the inner and outer wheels is b.
The normal acceleration of C, an = V2/R.
At impending overturning, the normal reaction N* and friction force f* on the inner wheels are zero.
Hence for vertical equilibrium,
∑FV = N –mg = 0 → N = mg. (1)
The equations of motion in the radial direction is,
∑Fn = man → f = m V2/R. (2)
The moment equilibrium about C yields,
∑MC = 0 → N×b/2 - f ×h ≥ 0. (3)
Substituting for N and f from equations (1) and (2) into inequality (3),
V ≤ √(bgR/2h). Hence, Vmax = √(bgR/2h).
Q. 8.
The F.B.D. of the structure is shown in Fig.8. The vertical reactions RA and RC act at the roller supports A and C, respectively.
Considering the equilibrium of the beam,
∑MA = 120×20 - RC×60 + (4×30)×75 = 0.
→ RC = 190 kN.
∑Fy = RA -120 + RC – 4×30 = 0.
→ RA = 50 kN
The expressions for S.F. V in kN and B.M. M in kNm at various sections are
0 ≤ x ≤ 20m:
V = -50, M = 50x.
20 m ≤ x ≤ 60m:
V = -50+120= 70, M = 50x – 120(x – 20).
60 m ≤ x ≤ 90m:
V = - 4(90 – x), M = - 4(90 – x)2/2.
The S.F. and B.M. diagrams are also shown in Fig.8.
The maximum S.F. Vmax = -120 kN occurs at the right support, x = 60m.
The maximum B.M. Mmax = -1800 kNm also occurs at the right support x = 60 m.
For 20 m ≤ x ≤ 60m, M = 50x – 120(x – 20) = 0 at x = 34.3 m which is a point of contraflexure.
Q.9.
As the material is the same, the allowable shear stress τo is the same for the shafts.
From the torsion formula for shafts, the torque Tsolid = τ0J/rmax.
For a solid shaft of diameter d: The polar moment of inertia Jsolid = πd4/32.
Tsolid = τ0J/rmax = τ0(πd4/32)/(d/2) = τ0πd3/16. (1)
For a hollow shaft of internal diameter di and external diameter do:
The polar moment of inertia, Jhollow = π(do4 - di4)/32.
Thollow = τ0J/rmax = τ0(π(do4 - di4)/32)/(do/2) = 16τ0(π(do4 - di4)/do) (2)
For the shafts of the same length and weight, the cross-sectional area must be equal, i.e. do2 - di2 = d2 → do/d = √(1 + di2/d2) (3)
From equations (1) to (3),
Thollow/Tsolid
= (do4 - di4)/(dod3) = (do2 + di2)/(dod) = do/d + di2/(dod) > 1. (proved).
Q.10.
Consider a uniform bar of original cross-sectional area A and length L0 subjected to equal and opposite axial forces P at each end. The final deformed length of the bar is Lf. Then,
Stress σ is defined as the force per unit area, i.e. σ = P/A. This stress is known as the normal stress and may be tensile or compressive.
Strain ε is defined as the change in length per unit length, i.e. ε = (Lf – L0)/L0. This strain is known as the normal strain and may be tensile or compressive.
Young’s modulus of elasticity E is defined as the ratio of stress to strain within elastic limit, i.e. E = σ/ε, σ < σelastic.
The F.B.D. of the rigid
horizontal bar is shown in Fig.10.
From the equilibrium equations of the bar,
∑FV = FA + FB - 50000 = 0.
∑MB = FA×60 - 50000×40 = 0.
→ FA = 100000/3 N, FB = 50000/3 N.
Let AB be the cross-sectional area and σB the stresss in the rod B. Then, AB = FB/ σB = (50000/3)/50 = 333.3 mm2.
As the rods are of the same length and their lengths remain equal after deformation, the strains in the rods must be equal, εA = εB = σB/EB = 50/90000 = 1/1800.
Hence, the stress in rod A, σA = EAεA = 200000/1800 = 1000/9 N/mm2.
The cross-sectional are of the rod A, AA = FA/σA = (100000/3)/(1000/9) = 300 mm2.
Q.11.
Let RA and RB be the support reactions. As the loading is symmetrical about the centre,
RA = RB = P/2.
The expression for the B.M. M between A and D is, M = RAx = Px/2, where x is measured from A along the beam.
Using the moment curvature relation, the deflection v is obtained as follows.
EId2v/dx2 = - M = - Px/2
Integrating, EIdv/dx = - Px2/2 + C1
Due to symmetry, the slope dv/dx = 0 at x = l/2, C1 = Pl2/8.
Integrating once more,
EIv = - Px3/6 + C1x + C2 = - Px3/6 + Pl2x/8 + C2
As the deflection v = 0 at the support x = 0, C2 = 0.
Hence, the deflection at D (x = l/2), vD = Pl3/(24EI).
SOLUTIONS D-02 APPLIED MECHANICS (June 2004)
Q.1. a. D If a two force body is in equilibrium, the forces must be equal, opposite and collinear.
b. C 
c. A The limiting force of friction F = μR.
d. C The force required to move the load W down is Wtan(α - φ). For α < φ, the load would not move down for zero force.
e. B The moment of inertia of a semicircular section about its diameter I = (1/2)(πR4/4) = 32π cm4.
f. B V = dS/dt = 14t +10. At t = 0, V = 10 m/s.
g. C The point of contra-flexure is a point where the beam curvature changes sign and hence bending moment changes sign.
h. C Torsional rigidity of a shaft is given by GJ.
Q.2.
The F.B.D. of the wheel when it is just about to roll over the block is shown in Fig.2. Just when the wheel begins to roll over, there is no force from the ground on the wheel at B. The wheel is subjected to its own weight of 1000 N, the force P and the reaction R of the block.
As the wheel is in equilibrium under three forces, W, P and R only, the forces are concurrent and pass through the point D.
From the geometry of the figure,
cosECA = CE/CA = 15/30 =1/2 → 2θ = 600.
angleCDA = (1/2)angleECA = θ = 300.
From Lame’s theorem,
P/sinθ = W/sin(90-θ) → P = W tanθ = 1000tan30 = 1000/√3 = 577.4 N.
Q.3.
Consider the T section to consist of two rectangular parts 1 and 2 as shown in Fig.3.
Let C(xC, yC) be the centroid of the T section. Let Ai be the area and xi, yi the cooordiantes of the centroid of the ith part.
The y coordinate of
the centroid is obtained as
yC = ∑Aiyi/∑Ai
= [(6×4)×3 + (2×8)×7]/( 6×4 + 2×8)
= 184/40 = 4.6 cm.
The moment of inertia ICxx of the T section about an axis parallel to the x axis through C would be
ICxx = ∑[(bihi3)/12 + (bihi)(yC – yi)2]
= 4×63/12 + 4×6(4.6 – 3)2
+ 8×23/12 + 8×2(4.6 – 7)2
= 231 cm4.
Q.4.
The F.B.Ds. of the joints E, D and C are shown in Fig.4. The forces are assumed tensile in the members and are assigned positive sign.
Consider the equilibrium of joint E:
∑Fx = - FCEcos45 = 0 → FCE = 0, ∑Fy = FDE - FCEsin45 = 0 → FDE = 0.
Consider the equilibrium of joint D:
∑Fy= - FBDsin45 + FDE = 0 → FBD = 0, ∑Fx = -FCD – FBDcos45 + 2 = 0 → FCD = 2 kN(T).
Consider the equilibrium of joint C:
∑Fx = FCD + FCEcos45 – FACcos45 = 0 → FAC = 2√2 kN(T).
∑Fy= FCEsin45 - FACsin45 – FBC = 0 → FBC = -2 kN, i.e. 2 kN(C).
Q.5.
The F.B.D. of the ladder is shown in Fig.5. The man is at D (AD = d) for the ladder to start slipping. The ladder is subjected to its own weight W at C, the weight of the man W/2 at D, the normal reaction NA and the limiting friction force NA/2 from the floor at A and the normal reaction NB and limiting friction force NB/3 from the wall at B.
From the force equlibrium equations,
∑Fx = NB – NA/2 = 0 and ∑Fy = NA + NB/3 – W – W/2 = 0.
NA = 9W/7 and NB = 9W/14.
From moment equilibrium about A,
∑MA = – NB×7/√2–(NB/3)×7/√2+W×3.5/√2+(W/2)d/√2 = 0.
Substituting for NB in the moment equation, d = 5m.
Q.6.
Let the effort P be given a virtual displacement δP downward, then the virtual displacement of the load W would be δW = δP/2 upward.
If the system is in equilibrium, the sum of the virtual work must be zero, i.e.
PδP – WδW = 0. → P = W/2 = 500 N.
Q.7.
Let at be the uniform tangential acceleration of the car. If it starts from rest, the speed V after time t would be,
V = att.
→ at = V60/60= (18×1000/3600)/60 = 1/12 = 0.083 m/s2.
The speed after 30 s would be V30 = at×30 = (1/12)×30 = 2.5 m/s.
The normal acceleration an = V302/R = 2.52/250 = 0.025 m/s2.
Q.8.
The stress σ = P/A = P/(πd2/4) = 100×103/(πd2/4) = 100×106.
Hence, d = √(4/1000π) = 0.03568 m = 3.568 cm.
The total elongation δ = (P/E)[L1/A1 + L2/A2 + L3/A3]
= 100×103/(290×109)[0.1/(π×0.035682/4) + 0.15/(π×0.12/4) + 0.15/(π×0.082/4)]
= 0.0514×10-3 m = 0.0514 mm.
Q.9.
The F.B.D. of the girder is
reproduced in Fig.9. The support reactions are determined by considering its
equilibrium,
∑MA=9RB–180×4.5-30×6-40×7.5= 0. → RB = 143.33 kN.
∑Fx = RA + RB - 20×9 – 30 -40 = 0. → RA = 106.67 kN.
The S.F. V and the B.M. M are,
0 ≤ x ≤ 6m:
V = - 106.67 + 20x.
M = 106.67x – 20x2/2.
6 m ≤ x ≤ 7.5m:
V = - 106.67 + 20x + 30.
M = 106.67x – 20x2/2 + 30(x - 6).
7.5 m ≤ x ≤ 9m:
V = 143.33 – 20(9 - x).
M = 143.33(9 – x) – 20(9 - x)2/2.
The S.F. and B.M. diagrams are also shown in Fig.9.
The maximum S.F. Vmax = 143.33 kN at the right support, i.e. x = 9 m.
The S.F. changes sign at x = 5.33 m.
The maximum B.M. is at x = 5.33 m
Mmax=284.4 kNm.
Q.10.
The F.B.D. of the girder is shown in Fig.10. RA and RB are the support reactions. From the equilibrium equations,
∑MA = RB×L –P×a = 0. → RB = Pa/L.
∑Fy = RA + RB - P = 0. → RA = P(L – a)/L.
The expression for the B.M. at any section x using singularity functions is:
M = - RAx + P < x - a> = - P(L – a)x/L + P < x - a>.
If v is the deflection of the elastic line of the beam,
EI d2v/dx2 = M = - P(L – a)x/L + P < x - a>.
Integrating twice,
EI dv/dx = - P(L – a)x2/2L + P < x - a>2/2 + C1
EI v = - P(L – a)x3/6L + P < x - a>3/6 + C1x + C2
At x = 0, v = 0 → C2 = 0.
At x = L, v = 0 → C1 = Pa(L - a)(2L - a)/6L.
Hence, v = - P(L – a)x3/6L + P < x - a>3/6 + Pa(L - a)(2L - a)x/6L.
The deflection vP under the load P at x = a,
vP = Pa2(L – a)2/(3EIL)
= 120×103×4.52 (14 – 4.5)2/(3×210×109×16×10-4×14) = 1.55×10-2 m = 1.55 cm.
Q.11.
A shaft of diameter d is subjected to a torque T. The shear modulus of the material is G. The shear stress τ at any radius r and the angle of twist θ over a length L is given by the torsion formula for circular shafts as
τ/r = T/J = Gθ/L,
where the polar moment of inertia of the cross-section J = πd4/32.
(i) The maximum shear stress τmax would be in the outer fibers at rmax = d/2.
τmax= T(d/2)/J = 16T/(πd3) = 16×560/(π×0.033) = 105.63×106 N/m2 =105.63 MPa.
(ii) The angle of twist θ over a length L = 1m would be
θ = TL/GJ = 32TL/(Gπd4) = 32×560×1/(82×109×π×0.034) = 0.0859 rad. = 4.920.
(iii) The shear stress τ at r = 0.01m would be
τ = Tr/J = 32Tr/( πd4)= 32×560×0.01/(π×0.034) = 70.42×106 N/m2 = 70.42 MPa.
SOLUTIONS D-02 APPLIED MECHANICS (December 2004)
Q1. a. C Displacement magnitude = [(12×1/2)2+(8×1)2]1/2 = 10 km.
b. A As the body moves in the horizontal plane, the work done by weight is zero.
c. D The distance of the centroid from each leg is b/4 and hence from the corner is b/√8.
d. B The mechanical efficiency = output/input = 120×10/(20× 80) = ` 0.75. or 75%.
e. D Simple harmonic motion is always in a straight line.
f. C The kinetic energy = Iω2/2 = (mL2/3)ω2/2 = mL2ω2/6.
g. A The maximum shear strain at the outermost fibers γmax = θr/L.
h. B In pure bending, the bending moment is constant and so the curvature is constant.
Q.2a.
Resultant force in the x direction, Fx = 8 – 16cos120 = 8 – 16/2 = 0.
Resultant force in the y direction Fy = 16sin120 = 16√3/2 = 8√3 N
Acceleration the x direction ax = Fx/m = 0.
Acceleration the y direction ay = Fy/m = 8√3/2 = 4√3 = 6.93 m/s2.
Q.2b.
Let the resultant R
act downward at a point (xR, yR).
∑Fz = 0 → R = 3 + 6 + 4 + 7 = 20 N.
∑MOy = 0 → xRR = 1×3 + 1×6 + 4×4 + 2×7 = 39
→ xR = 39/20 = 1.95 m.
∑MOx = 0 → yRR = 1×3 + 3×6 + 0.25×4 + 4×7 = 50
→ yR
= 50/20 = 2.5 m.
Q.3a.
The F.B.Ds. of ball A and ball B are shown in Fig.3a.
Q.3b.
The F.B.Ds. of the joints C,
B, D and E assuming tensile forces in the members, are shown in Fig.3b.
Considering the equilibrium of joint C:
∑FV = - FBCsin45 – 2000 = 0
→ FBC = 2000√2 N = 2828.4 N(T).
∑FH = - FDC + FBCsin45 = 0
→ FDC = -2000 N = 2000 N(C).
Considering the equilibrium of joint B:
∑FH = - FBA+ FBCsin45 = 0
→ FBA = 2000 N = 2000 N(T).
∑FV =
- FBD - FBCcos45 = 0
→ FBD = - 2000 N = 2000 N(C).
Considering the equilibrium of joint D:
∑FV = FDAsin45 + FBD -2000 = 0
→ FDA = 4000√2 N = 5656.9 N(T).
∑FH = -FDE + FDC – FDAcos45 = 0
→ FDE = - 6000 N = 6000 N(C).
At support E, only normal reaction RE is possible. Therefore FAE = 0.
Q.4.
The I section is divided in
parts 1, 2, 3 as shown in Fig.4. Let xC, yC
be the coordinates of its centroid. Let Ai be the area and xi,
yi the coordinates of the centroid of the ith
part.
By symmetry → xC = 0.
yC = ∑Aixi/∑Ai
= (20×13 + 20×7 + 40×1)/(20 + 20 + 40) =.5.5 cm.
Let Ixx and Iyy be the second moments of area about the coordinate axes.Then,
Ixx = ∑Ixxi = ∑Aiyi2 + ∑(bihi3/12)
∑Aiyi2 = 20×132 + 20×72 + 40×12= 4400 cm4
∑bihi3/12=(10×23+ 2×103+ 20×23)/12= 186.67 cm4
→ Ixx = 4400 + 740/3 = 4586.67 cm4
Iyy = ∑Aixi2 + ∑(hibi3/12)
= 0 + (2×103 + 10×23 + 2×203)/12 = 1506.67 cm4.
Let ICxx and ICyy be second moments of area about the centroidal axes parallel to Ox, Oy.
ICxx = Ixx – (∑Ai) yC2 = 4586.67 - 80×5.52 = 2166.67 cm4
ICyy = Iyy – (∑Ai) xC2 = 1506.67 – 80×02 = 1506.67 cm4
The polar moment about the axis through centroid ICzz = ICxx + ICyy = 3673.3 cm4.
Q.5.
The F.B.D. of the wedge being
driven in the wood is shown in Fig.5. Considering the equilibrium at impending
motion,
∑Fy = μN1sin(θ/2)–N1cos(θ/2)+N2cos(θ/2) -μN2 sin(θ/2)= 0.
→ (N2 –N1)[cos(θ/2)-μ sin(θ/2)]= 0.→ N1=N2 =N.
Σ Fx = P - N1sin(θ/2) - μN1cos(θ/2) - N2sin(θ/2) - μN2 cos(θ/2) = 0.
→P = 2N[sin(θ/2) + μcos(θ/2)].
For the wedge, depending on the coefficient of friction μ, a force P1 may be required to keep the wedge in place without being squeezed out. The friction forces are reversed and the force P1 is obtained from the expression for P by changing the sign of μ.
P1 = 2N[sin(θ/2) - μcos(θ/2)].
The wedge is self locking if P1≤ 0, → μ ≥ tan(θ/2).
Q.6a.
Horizontal velocity of the plane = 200 km/h = 200×1000/3600 = 55.56 m/s
At the time of firing, the plane is vertically above the gun. Therefore, for the shell to hit the plane, the horizontal velocity of the shell must be the same as that of the plane.
Let θ be the inclination of the gun to the horizontal. Then,
300cosθ = 55.56 → θ = cos-1 = 55.56/300 = 79.30.
The shell hits the plane at time t and travels a vertical distance of 1000 m as a projectile.
300sin79.3 t – 9.81 t2/2 = 1000 → t2 – 60.1 t + 203.9 = 0 → t = 3.6s, 56.5s
The shell hits the plane at t = 3.6s.
The horizontal distance of the plane from the gun = 55.56×3.61 = 200.6 m.
Q.6b.
The centripetal acceleration ar = V2/R = ω2R = 22×0.5 = 2 m/s2.
The tangential acceleration at = (dV/dt) = αR = 3×0.5 = 1.5 m/s2.
The total acceleration = a = √(ar2 + at2) = √(22 + 1.52) = 2.5 m/s2.
Q.7a.
The F.B.D. of the 10 kg mass
and 50 kg drum are shown in Fig.7a.
The equation of motion for downward acceleration am of the mass is
10g - T = 10am (1)
The equation of motion for the clockwise angular acceleration α of the drum is
0.4T = Iα = 50×(0.3)2α (2)
The relation between angular acceleration of the drum and linear acceleration of the mass is
am = 0.4α (3) From equations (1) to (3) → α = 6.43 rad/s2 and T = 72.3 N.
Q.7b.
The hammer falls freely. Hence the velocity just before impact with the plate,
V1 = √(2gh) = √(2×9.81×2.5) = 7 m/s
As the impact is perfectly plastic, both the hammer and pile move together after impact. Their common velocity V just after impact is obtained from momentum conservation.
(50 + 20)×v = 50×7 → V = 5 m/s.
Let R be the average resistance of the ground. Then from the work-energy principle
Work done against the resistance = Loss of Kinetic Energy + Loss of Potential Energy
→ 0.1×R = (50 + 20)×52/2 + 0.1× (50 + 20) ×9.81→ R = 9436.7 N.
Q.8a.
The tensile load extension
diagram is as shown in Fig.8a. The elongation is 0.8 mm at the elastic limit
load of 30 kN. Young’s modulus
E = (P/A)/(ΔL/L) for P ≤ 30 kN.
→ E = (30×103/100×10-6)/( 0.8/200)
= 75×109 Pa = 75 GPa.
For 0.2% proof stress σ0 the permanent strain must be .002, i.e. ΔL = 200×0.2/100 = 0.4 mm. This 0.4 mm permanent elongation is obtained by unloading from the 45 kN load.
Hence, σ0 = 45×103/100×10-6 Pa = 450 MPa.
Ultimate stress σu = Max. load /Area of section.
→σu =60×103/100×10-6 Pa = 600 MPa.
The permanent elongation after fracture = 3.6 mm. The permanent strain =3.6/200.
Hence, the % elongation = (3.6/200)×100 = 1.8%.
Q.8b.
The area A resisting the shear is A = πdt.
The punch force P = τ × A = τ πdt.
Q.9.
The relation between power P, torque T and rpm N is P = T×2πN/60.
→ T = 628×103×60/(2×3.14×200) = 30000 Nm.
Let do be the outer diameter and di the inner diameter of the shaft.
The polar moment of area of the shaft section is J = π(do4 - di4)/32.
The torsion formula for a hollow shaft is τmax/(d0/2) = T/J = Gθ/L.
From τmax/(d0/2) = Gθ/L,
→ d0 = 2τmaxL/Gθ = 2×80×106×4/[80×109×(3×3.14/180)] = 0.153 m = 153 mm.
From T/J = Gθ/L and J = π(do4 - di4)/32,
→ J = π(do4 - di4)/32 = TL/Gθ → di4 = do4 – (32/3.14)× TL/Gθ
→ di = [do4 – (32/3.14)×30000×4/[80×109×(3×3.14/180)]]1/4 = 0..126 m = 126 mm.
Q.10.
The F.B.D. of the beam is shown in Fig.10. From beam equilibrium,
∑MB = - 4RA + 6×1 + 1×(3×2) = 0.
→ RA = 3 kN.
∑Fy = RA + RB – 1 - 3×2 = 0
→ RB= 4 kN
The S.F. V ( kN) and B.M. M (kNm) are:
0 ≤ x ≤ 2 m
V = 1, M = -x.
2 m ≤ x ≤ 4 m
V = 1 – 3 = - 2, M = -x + 3(x – 2).
4 m ≤ x ≤ 6 m
V = 1 - 3 + 3( x – 4), M = -x + 3(x- 2) – 3(x – 4)2/2.
The S.F. and B.M. diagrams are also shown in Fig.10.
The maximum shear Vmax = 4 kN at the right support, i.e. x = 6 m.
V = 1 - 3 + 3( x – 4) = 0 at x = 4.67 m. Hence, Mmax = 2.67 kNm at x = 4.67 m.
The B.M. M = 1 - 3 + 3(x – 4) = 0 at x = 3 m which is a point of contraflexure.
Q.11a.
Let A and B be points on the elastic curve of a beam of flexural rigidity EI. The slopes of the tangents to the elastic curve at A and B are θA and θB, respectively. The tangential deviation tB/A is the displacement of point A from the tangent at B in a direction normal to the undeformed elastic curve. Let M be the bending moment at any section.
The first theorem: The rotation between the tangents at points A and B is equal to the area of the (M/EI) diagram between these points.
The second theorem: The tangential deviation of point A from the tangent at B, tA/B is equal to the first moment of the area of (M/EI) diagram about an axis normal to the undeformed elastic curve through A.
Q.11b.
The B.M. at any section x,
M = - Px.
The B.M. M and (M/EI) diagrams are as shown in Fig.11b.
As the tangent to the elastic curve at D is horizontal θD = 0, the slope at B,
θB= - area (M/EI) diagram B to D= -∑Ai.
A1 = (PL/2EI)(L/2)/2 = - PL2/8EI,
A2 = (PL/4EI)L/2 = - PL2/8EI,
A3 = (PL/4EI)(L/2) /2 = - PL2/16EI
→ θB = 5PL2/16EI.
Distances xi of the centroids of Ai from the vertical axis through B are
x1 = L/3, x2 = 3L/4, x3 = 5L/6
The deflection δB = tB/D = ∑Aixi.
→ δB = - (PL2/8EI)L/3 - (PL2/8EI)×3L/4 - (PL2/16EI)( 5L/6) = - 3PL3/16EI.