Question and answers

4

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

2

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.

8

Parallel and transverse Weld  

4

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

4

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

4

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.

4

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.

8

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

4

 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

6

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……………………