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Examination: 2017 SUMMER
Que.No Marks
Q 3 d )

Question:

#### Compare welded joints with screwed joints. (Any six points)

Consideration in design of key: 1) Power to be transmitted. 2) Tightness of fit 3) Stability of connection 4) Cost 5) Crushing failure of key: 6) shearing failure of key 7) Material of key ,shaft should be same but key should be weaker than shaft .

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Q 4a)(iv)

Question:

#### Define following terms w.r.t. bolts: 1) Major diameter 2) Minor diameter 3) Pitch 4) Lead

Definition w.r.t. bolts 1) Major dia.- dia. Of imaginary cylinder parallel with the crest of the thread ,it is the distance from crest to crest largest dia. of an external or internal thread 2) Minor dia.-dia. Of imaginary cylinder which just touches the roots of an external thread or smallest dia.of an external or internal screw thread 3) Pitch-distance from a point on one thread to the corresponding point on the next thread. 4) lead- distance between two corresponding points on the same helix

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Q 6 c )

Question:

#### A bracket as shown in Figure No. 1 is fixed to a vertical steel column by means of five standard bolts.

Horizontal component of 45 KN, WH= 45Sin 600 =45x 0.866=38971N and vertical component of 45 KN, Wv,= 45xcos600 =45x0.5=22500N Direct tensile load in each bolt,Wt1= WH /5=38971/5=7794.20N Turning moment due to WH about G TH = WHx 25=38971x25=974275N (anticlockwise) direct shear load on each bolt =Ws=Wv/5 =22500/5=4500N Turning moment due to Wv about edge of the bracket, Tv= Wvx175=22500x175=3937500N-mm( clockwise( clockwise) Net turning moment =3937500-974275=2963225N---------(I) total moment of the load on the bolts @ th tilting edge = 2w x(L1) 2 + 2w x(L2) 2 =2xwx(50)2 + 2xwx(150)2 = 50000 w N-mm-----(II) from equations (I) and(II) 2963225N=50000 w N- w= 592.645 N max. tensile load on each of the upper bolt, Wt2= wL2 =592.645x150=88896.75 N tensile load on each of the upper bolt, Wt = Wt1+ Wt2 =7794.20+ 88896.75=96690.95N equivalent tensile load =Wte=1/2(Wt+ √‾(Wt)2 + 4(Ws)2 =1/2 ( 96690.95+97108.91)=96899.93 N Tensile load on each bolt = ∏/4(dc)2 x 6t =0.7854x(dc)2 x 70 dc = 41.98 mm from coarse series the standard core dia. Is 49.0177 mm and corresponding size of the bolt is M56 thickness of the arm of the bracket cross sectional area of the arm A = bXt =100x t

section modulus of the arm, Z = 1/6 t (b)2 = 1/6 xtx(100)2 =1666.67 xt direct tensile stress 6t1 = WH/A = 38971/100t =389.71/t bending stress 6t2 = MH/Z = 208/t bending stress 6t3 = Mv /Z = 2632.49/t net tensile stress, 6t1 +6t2 + 6t3 = 3230.20/t max. tensile stress , 6t max. 6t/2+ ½ √‾(6t)2 + 4(Ʈ)2 =70 t = 46.36 mm

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Q 6 e )

Question:

#### State the strength equation of double parallel fillet weld and single transverse fillet weld with neat sketches

Strength equation of double parallel fillet weld= throat area x allowable shear stress P= 2x 0.707x Sw x lwx Ʈ =1.414 x Sw x lwx Ʈ (1mark) Strength equation of single transverse fillet weld

P =throat area x allowable tensile stress P= 0.707x Sw x lw x σ -----------------------------------------------------------------------------------------------------
Examination: 2017 WINTER
Que.No Marks
Q 1 k )

Question:

#### Give two examples, where screwed joints are preferred over welded joints.

i) Cylinder head of the engine. (ii) Machine foundation. (iii) Assembly of fan, couplings. Any two examples (iv) Connect two bogies of the train with the turn buckle. (v) Structural bridges, pressure vessels, fly press (vi) Assembly of crank shaft and connecting rod.

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Q 4 a )

Question:

#### Fig 2 show a C.I. bracket to carryof a shaft -----------------------------------------------------------------------------------------------------
Q 5 a )

Question:

#### Parallel and transverse Weld  -----------------------------------------------------------------------------------------------------
Examination: 2016 SUMMER
Que.No Marks
Q 3 e )

Question:

#### Describe ‘bolt of uniform strength’ with neat sketch

When an ordinary bolt of uniform diameter is subjected to shock load stress concentration across at the weakest part of the bolt i.e. threaded portion (as shown in figure a), it means that greater portion of energy will be absorbed at the region of threaded part and it may cause the failure of threaded portion There are two methods to achieve bolts of uniform strength i. Turn down shank diameter of bolt equal or lesser than the core diameter of thread (dc) as shown in figure (b) and it gives bolt of uniform strength. ii. In this method an axial hole is drilled to the head as far as threaded portion such that area of shank become equal to the root area of thread as shown in figure (c). Where,d1= Diameter of hole to be drill do= Nominal diameter dc= Core diameter -----------------------------------------------------------------------------------------------------
Q 4a)(iii)

Question:

#### State any four advantages and disadvantages of welded joints over riveted joints.

1. The welded structures are usually lighter than riveted structures. This is due to the reason, that in welding, gussets or other connecting components are not used. 2. The welded joints provide maximum efficiency (may be 100%) which is not possible in case of riveted joints. 3. Alterations and additions can be easily made in the existing structures. 4. As the welded structure is smooth in appearance, therefore it looks pleasing. 5. In welded connections, the tension members are not weakened as in the case of riveted joints. 6. A welded joint has a great strength. Often a welded joint has the strength of the parent metal itself. 7. Sometimes, the members are of such a shape (i.e. circular steel pipes) that they afford difficulty for riveting. But they can be easily welded. 8. The welding provides very rigid joints. This is in line with the modern trend of providing rigid frames. 9. It is possible to weld any part of a structure at any point. But riveting requires enough clearance. 10. The process of welding takes less time than the riveting. Disadvantages

1. Since there is an uneven heating and cooling during fabrication, therefore the members may get distorted or additional stresses may develop. 2. It requires a highly skilled labour and supervision. 3. Since no provision is kept for expansion and contraction in the frame, therefore there is a possibility of cracks developing in it. 4. The inspection of welding work is more difficult than riveting work

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Q 6 c )

Question:

#### Determine the size of bolt in the cylinder head of a steam engine. The engine cylinder has a bore of 400 mm and the maximum steam pressure to which the cylinder is subjected is 1.5 N/mm2 . Cylinder head is held on the cylinder by 16 number of bolts. The permissible tensile stress for the bolt material is 25 N/mm2 Bolt size will be M 30 or M32

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Examination: 2016 WINTER
Que.No Marks
Q 3 d )

Question:

#### Explain why bolts of uniform strength are preferred. Draw sketches of two different types of bolts of uniform strength

bolts of uniform strength: if a shank dia.is reduced to a core dia.as shown in fig. the stress become same through out the length of the bolt. Hence impact energy is distributed uniformly throughout the bolt length, thus relieving the threaded portion of high stress. The bolt in this way becomes stronger and lighter. This type of bolt is known as bolt of uniform strength.Another method of obtaining the bolt of uniform strength is shown in fig.in this method, instead of reducing the shank dia.an axial hole is drilled through the head down to

the threaded portion such that the cross sectional area of the shank becomes equal to the area of the threaded portion If bolts of uniform strength are not used a large portion of impact energy will be absorbed in the threaded portion and relatively a small portion of energy is absorbed by a shank. This uneven distribution of impact energy may lead to the fracture of the bolt in threaded portion .hence bolts of uniform strength are preferred.

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Q 4a)(iv)

Question:

#### State four disadvantages of screwed joints.

Four Disadvantages of screwed joints: 1) Screwed joints are weaker than welded joint 2) Screwed joints weakens( due to holes) the parts that are to be joined. 3) Stress concentration in the threaded portion of screw makes them weak. 4) Locking arrangement is required in case of vibrations

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Q 6 a )

Question:

#### Derive strength equation for parallel fillet weld subjected to tensile load.

Derivation of strength equation for parallel fillet weld subjected to tensile load The parallel fillet welded joints are designed for shear strength. Consider a double parallel fillet welded joint as shown in Fig. -----------------------------------------------------------------------------------------------------
Q 6 c )

Question:

#### A steam engine cylinder has effective diameter of 250 mm. It is subjected to maximum steam pressure of 1.5 MPa. The cylinder cover is fixed to the cylinder with the help of 12 bolts. The pitch circle diameter of bolts is 400 mm. Permissible tensile stress of the bolt material is 30 MPa. Determine nominal diameter of the bolts. -----------------------------------------------------------------------------------------------------
Examination: 2015 SUMMER
Que.No Marks
Q 6 a )

Question:

#### State the strength equations of double parallel fillet weld and single transverse fillet weld with neat sketches. Let t = Throat thickness (BD), s = Leg or size of weld, = Thickness of plate, and l = Length of weld, From Fig. 10.7, from above we find that the throat thickness, t = s × sin 45° = 0.707 s ∴*Minimum area of the weld or throat area, A = Throat thickness ×Length of weld= t × l = 0.707 s × l…….. If σt is the allowable tensile stress for the weld metal, then the tensile strength of the joint for single fillet weld, P = Throat area × Allowable tensile stress = 0.707 s × l × σt…….. and tensile strength of the joint for double fillet weld, P = 2 × 0.707 s × l × σt = 1.414 s × l × σt……..If τ is the allowable shear stress for the weld metal, then the shear strength of the joint for single parallel fillet weld, P = Throat area × Allowable shear stress = 0.707 s × l × τ and shear strength of the joint for double parallel fillet weld, P = 2 × 0.707 × s × l × τ = 1.414 s × l × τ …… The strength of the joint is given by the sum of strengths of single transverse and double parallel fillet welds. Mathematically, P = 0.707s × l1 × σt + 1.414 s × l2 × τ

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Examination: 2015 WINTER
Que.No Marks
Q 3 d )

Question:

#### Explain with neat sketch the bolts of uniform strength. In an ordinary bolt shown in Fig. (a), the effect of the impulsive loads applied axially is concentrated on the weakest part of the bolt i.e. the cross-sectional area at the root of the threads. In other words, the stress in the threaded part of the bolt will be higher than that in the shank. Hence a great portion of the energy will be absorbed at the region of the threaded part which may fracture the threaded portion because of its small length. If the shank of the bolt is turned down to a diameter equal or even slightly less than the core diameter of the thread (d) as shown in Fig. (b), then shank of the bolt will undergo a higher stress. This means that a shank will absorb a large portion of the energy, thus relieving the material at the sections near the thread. The bolt, in this way, becomes stronger and lighter and it increases the shock absorbing capacity of the bolt because of an increased modulus of resilience. This gives us bolts of uniform strength. The resilience of a bolt may also be increased by increasing its length. A second alternative method of obtaining the bolts of uniform strength is shown in Fig. (c). An axial hole is drilled through the head as far as the thread portion such that the area of the shank becomes equal to the root area of the thread

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Q 4a)(iv)

Question:

#### A cylinder head of steam engine is held in position by M20 bolts. The effective diameter of cylinder is 350 mm and the steam pressure is 0.75 N/mm2. If the bolts are not initially stressed, find the number of bolts required. Take working stress for bolt material as 20 N/mm2. -----------------------------------------------------------------------------------------------------
Q 4b)(ii)

Question:

#### Explain the following types of stresses: (1) Transverse shear stress (2) Compressive stress (3) Torsional shear stress

When a section is subjected to two equal and opposite forces acting tangentially across the section such that it tends to shear off across the section. The stress produces is called as transverse shear stress. …………………… From figure Mathematically transverse shear stress is represented as, τ = / Where, F = Tangential force applied A = Area of cross section = ( /4 ) 2 d = Diameter of rivet. ……………………ii )Compressive Stress: When a body is subjected to two equal & opposite axial pushes ,then the internal resistances set up in the material is called as compressive stress.……………………It is denoted by σc σc =P/A Where ,P: Axial compressive force ,A : Cross Sectional Area.…………… iii) Torsional stress: When a machine component is under the action of two equal and opposite couples i.e. twisting moment or torque, then component is said to be torsional and the stresses set up due to torsion are called as torsional shear stress. ……………………Consider a component of circular cross-section. ‘d’ in diameter, subjected to torque T, Torsional shear stress is given by, basic torsion equation /J = / =Gθ/L τ = . / J Where, r = distance of outer fibre from neutral axis = d/2 J = Polar moment of inertia of cross- section = ( /64)4……………………

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Q 6 c )

Question:

#### A wall bracket is fixed to the wall by means of three bolts, one bolt at a distance of 25 mm from the lower edge and remaining two bolts at a distance of 175 mm from the lower bolts. It supports a load of 7.5 kN at a distance of 250 mm from the wall. The bolts are made from plain carbon steel 45C8 with tensile yield strength of 380 N/mm2. If factor of safety is 2.5, estimate the size of the bolts. Sketch the arrangement  