UNIT-I STRESS, STRAIN 1. A Member A B C D is subjected to loading as shown in fig determine the total elongation. Take E= 2 x10 5 N/mm 2 Young s modulus E= 2 x10 5 N/mm 2 Area1=900mm 2 Area2=400mm 2 Area3=625mm 2 Length1=1500mm Length2=1000mm Length3=750mm To find Total elongation Consider tensile force +ve [increasing length] Compressive force ve [decreasing length] Step1 Consider part AB: To calculate the total load acts on suction AB (P 1 ) Left side load = right side load Step2 Consider part BC: To calculate the total load acts on suction BC (P 2 ) 1
Step3 Consider part CD: To calculate the total load acts on suction CD (P 3 ) Step 4 To calculate the change in length = + + 2. A reinforced concrete column 300 x 300 mm has 4 reinforcing steel bars of 25mm diameter in each corner. Find the safe axial load on the column when concrete subjected to a stress of 5N/mm 2 what is the corresponding stress in steel take E s /E s =18 Area of the column =300x300 Area of the steel bar= Stress on concrete =5N/mm 2 E s /E s =18 To find Load (P) Stress in steel Step: 1 To calculate total load on steel = 2
Step: 2 To calculate total load on concrete = Step: 3 To calculate total load on the beam = + Step: 4 Total Stress in steel = 3. A rod of made brass, copper and aluminum, as shown in fig is held between two rigid supports at A and D calculate the stresses developed in each material when the temperature of the system is raised by 40 o C. Take E s =2.1x10 5 N/mm 2. E c =1.1x10 5 N/mm 2. E a =0.7x10 5 N/mm 2 = 12x10-6 / 0 C; = 17x10-6 / 0 C.; = 21x10-6 / 0 C; 3
Area of the steel A s =100 mm 2 Length of the steel L s =75mm Area of the copper A c =200 mm 2 Length of the copper L c =150mm Area of the Aluminium A a =400 mm 2 Length of the Aluminium L a =200mm Rise in temperature T=40 0 C Young s modulus of steel E s =2.1x10 5 N/mm 2 Young s modulus of copper E c =1.1x10 5 N/mm 2 Young s modulus of aluminium E a =0.7x10 5 N/mm 2 Co- Efficient of linear expansions of steel = 12x10-6 / 0 C Co- Efficient of linear expansions of copper = 17x10-6 / 0 C. Co- Efficient of linear expansions of Aluminium = 21x10-6 / 0 C To Find 1. Stress in steel 2. Stress in copper 3. Stress in aluminium Step: 1 To calculate elongation of steel = Step: 2 To calculate elongation of copper = Step: 3 To calculate elongation of aluminum = 4
Step: 4 To calculate total elongation = + + Step: 5 Total force exerted in the bar P= = = Step: 6 To calculate total elongation = + + 5
4. Find the Young s modulus and poisons ratio of a metallic bar of length 300mm breadth 40mm depth 40mm when the bar is subjected to an axial load of 40KN decrease in length is 0.75 and increase in breadth in 0.03mm also find the modulus of rigidity of bar. Length L=300mm Breadth b =40mm Depth t = 40mm Axial load P= 40kN Decrease in length δ L =0.75mm Increase in breadth δ b =0.03mm To find Young s modulus E Poisson s ratio 1/m Modulus of rigidity G Step: 1 To calculate the passion ratio Passion ratio 1/m= = Lateral strain (e t ) = Lateral strain (e t ) = 6
Step: 2 To calculate the young s modulus Tensile stress σ= load /area E= tensile stress /tensile (or) longitudinal strain Step: 3 To calculate the modulus of rigidity Young s modulus (E) = 2 1 + UNIT II SHEAR AND BENDING OF BEAMS 5. A cantilever 1.5m long is loaded with uniformly distributed load of 2KN/m run over the length of 1.25 m from the free end. It also carries a point load of 3kNat a distance of 0.25m from the free end. Draw the shear force and bending moment diagram for the cantilever beam. 7
To draw: SFD and BMD. Step: 1 SF calculation draw: SFD and BMD. Step: 2 BM calculations 8
6. A simply supported beam is loaded as shown in fig Q. 12 (b). Draw the shear force and bending moment diagrams. (AU-APR-2005) Refer figure Q. 12 To draw: SFD and BMD. Draw: SFD and BMD. Step: 1 SF calculation 9
Step: 2 BM calculations 7. A beam 8 meters long rest on two supports one at the right end and the other two meters from the left end. The beam carries uniformly distributed load of 15kN/m over its entire length and point load of 80kN at a point 1.5m from the right end. Draw the SFD and BMD and find magnitude and location of maximum bending moment. Locate also the point of contra flexure 10
To draw: SFD and BMD. Step: 1 SF calculation draw: SFD and BMD. Step: 2 BM calculations 11
8. A timber beam of rectangular section is to support a load of 20 kn uniformly distributed over a span of 3.6 m, when the beam is simply supported. If the depth is twice the width of the section and the stress in timber is not to exceed 3.5 N/mm 2, find the dimensions of the cross section? UDL load=20kn/m Length (L) =3.6m Depth=twice the width Stress= 3.5 N/mm 2 To find Dimensions of the cross section Step: 1 To calculate the bending moment M= for SSB Step: 2 To calculate section modulus of the beam (Z) Bending equation = = = = y=d/2 for rectangular crass section 12
Step: 3 To calculate the dimensions of the beam Section modulus of the beam (Z) = for rectangular crass section 9. A I section beam 350mm x 200mm has a web thickness of 12.5mm and a flange thickness of 25mm. it carries a shearing force of 20 tonnes at a section. To calculate shear stress a crosses the section Breath b=350mm Depth d=200mm Flange thickness =25mm Web thickness =12.5mm Shear force (F) =20tonnes To find Shear stress of the beam 13
Step: 1 To calculate the center of gravity section from top face. = + + + + Step: 2 To cal calculate the total moment of inertia of the beam (I) = + + Step: 3 To calculate shear stress in upper flange with web (q) = 14
Step: 4 To calculate shear stress in web with upper flange (q) = Step:5 To calculate maximum shear stress of the beam (q max ) = 8 + 15
UNIT III TORSION 10. A shaft is subjected to a torque of 1.6kNm. Find the necessary diameter of the shaft, if the allowable shear stress is 60Mpa, the allowable twist is 1 o for every 20 diameters length of the shaft, take C=80Gpa. Torque (T)=1.6 knm Allowable shear stress ( )=60Mpa Angle of twist =1 o Length l= 20d To find Diameter of the shaft (d) Step:1 To calculate the diameter considering shear stress Torque (T) = 16
Step:2 To calculate the diameter considering angle of twist Torsion equation = = for solid shaft 11. Calculate the power that can be transmitted at a300rpm by a hollow steel shaft of 75mm external diameter and 50mm internal diameter when the permissible shear stress for the steel is 70N/mm 2 and the maximum torque is 1.3 times of the mean torque. Compare the strength of this hollow shaft with that of an solid shaft. The same material, weight and length of both the shafts are same. Speed N=300rpm External dia D=75mm Internal dia d=50mm Shear stress ( ) = 70N/mm 2 Torque T max =1.3T mean To find Power Strength of hollow shaft Strength of solid shaft Step: 1 To calculate maximum torque (T max ) (T h ) =(T max )= 17
Step :2 To calculate mean torque (T mean ) T max =1.3T mean Step: 3 To calculate power Power (P) = Step :4 Strength compression Area of the hollow shaft = area of the solid shaft = 4 4 To calculate solid shaft diameter Step : 5 Torque transmitted to the solid shaft (T s ) Torque (T) = 18
Step :6 = 12. Hollow steel shaft of 75mm outside diameter is transmitting a power of 300kW at 2000rpm. Find the thickness of the shaft if the maximum shear stress should not exceed 40N/mm 2 Speed N=2000rpm External dia D=75mm Power (P)= 300KW Shear stress ( ) = 40N/mm2 To find Thickness of the shaft Step :1 To calculate maximum torque (T) Power (P) = Step :2 (T)= 19
Step :3 To calculate thickness t= 13. A closed coil helical spring of circular crass section wire 18mm diameter is loaded by a force of 500N. the mean coil diameter of the spring is 125mm. the modulus of rigidity is 80 kn/mm 2 determine the maximum shear in the material of the spring. What number of coils must the spring have for its deflection to be 6mm and calculate the stiffness of the spring Wire diameter d=18mm Mean diameter D= 125mm Load P= 500N Modulus of rigidity C= 80 kn/mm 2 Deflection δ=6mm To find Shear stress (τ) Number of coils (n) Stiffness of the spring (K) Step: 1 To calculate the Shear stress (τ) Shear stress (τ) = 20
Step: 2 To calculate number of coils = 8 Step: 3 To calculate the stiffness of the spring = = 64 Step: 4 To calculate strain energy (U) = l=πdn 21
14. An open coiled helical spring made of 5mm diameter wire 16 coils 100mm inner diameter with the helix angle of 16 o calculate the deflection, maximum direct and shear stresses induced due to an axial load of 300N. take G= 90Gpa and E=200Gpa Wire diameter d=5mm Inner diameter Di= 125mm Load P= 300N Modulus of rigidity C= 90 Gpa Young s modulus E=200Gpa Number of coils (n) =16 Helix angle = 16 To find Maximum direct& Shear stress (τ) Deflection δ Stiffness of the spring (K) Step: 1 Deflection = + Step: 2 Calculate bending stress 22
= 32 Step: 3 Calculate shear stress = 16 Step: 4 Calculate maximum shear stress = 16 Step: 4 Calculate maximum principal stress (direct stress) = 16 + 1 23
UNIT IV DEFLECTION 15. A cantilever AB, 2m long is carrying a load of 20kN at free end and 30kN at a distance of 1 m from the free end. Find the slop and deflection at the free end. Take E= 200Gpa and I= 150 x 10 6 mm 4 E= 200Gpa I= 150 x 10 6 mm 4 To find Slop and deflection of the beam at the free end Double integration method Step: 1 To calculate Slope at the free end due W 1 = 2 24
Step: 2 To calculate Slope at the free end due W 2 = 2 Step: 3 To calculate total slope at the free end = 2 + 2 = + Step: 4 To calculate the deflection at the free end To calculate deflection at the free end due W 1 = 3 Step :5 To calculate deflection at the free end due W 2 = 3 25
Step: 6 To calculate total deflection at the free end = 3 + 3 = + Macaulay s method Moment area method 26
Conjugate beam method 16. A beam is simply supported as its ends over a span of 10m and carries two concentrated loads of 100kN and 60 kn at a distance of 2m and 5m respectively from the left support. Calculate (i) slope at left support (ii) slope and deflection at under the100 kn load. Assume EI=36x10 4 kn-m 2 EI=36x10 4 kn-m 2 To find (i) slope at left support (ii) slope and deflection at under the100 kn load 27
Step-1 Macaulay s method Step-2 Slope at the left support Step-3 Deflection at 100kN load 28
Step-4 Slope at 100 kn load 17. An I section joist 400mmx 200mmx 20mm and 6m long is used as strut with both ends fixed. What is Euler s crippling load for the column? Take E=200Gpa. Depth (d) = 400mm Breath (b) =200mm Thickness (t) = 20mm Length l= 6m Young s modulus (E) = 2x 10 5 N/mm 2 To find Critical load Step-1 To calculate the moment of inertia about x-x axis = 29
Step-2 To calculate the moment of inertia about y-y axis = Step-3 To calculate the crippling load (P) ( ) = 4 18. Find Euler s crippling load for a hollow cylindrical cast iron column of 20mm external diameter, 25mm thick and 6m long hinged at both ends. Compare the load with 30
crushing load calculated from Rankin s formula f c = 550N/mm 2 Rankin s constant =1/1600, E=1.2x 10 5 N/mm 2 Major dia D = 200mm Minor dia d=(200-25-25) Thickness (t) = 25mm Length l= 6m σ c = 550N/mm 2 Rankin s constant = Young s modulus (E) = 1.2x 10 5 N/mm 2 To find Critical load Step-1 To calculate the moment of inertia for circular section = 1 64 Step-2 Radius of gyration K= area (A) = 31
Step-3 To calculate the crippling load (P) ( ) = Critical load by using Rankin s formula Step-4 P= UNIT-5 COMPLEX STRESSES AND PLANE TRUSSES 19. A steel cylinder shell 3m long which is closed at its ends, had an internal diameter of 1.5m and a wall thickness of 20mm calculate the circumferential and longitudinal stress induced and also change in dimensions of the shell if it is subjected to an internal pressure of 1.0 N/mm 2 assume the modulus of elasticity and passion ratio for steel as 200 kn/mm 2 and 0.3 respectively Length of the shell L= 3m Diameter of the shell = 1.5m Pressure (P) = 1.0 N/mm 2 Thickness of shell t= 20mm Modulus of elasticity (E)=200 kn/mm 2 Passion ratio (1/m) = 0.3 To find 32
Circumferential stress Longitudinal stress Change in dimensions Step-1 To calculate circumferential stress = 2 Step-2 To calculate longitudinal stress = 4 Step-3 To calculate maximum shear stress = 2 Step-4 To calculate circumferential strain = 33
Change in dia = Step-5 To calculate longitudinal strain = Change in length = Step-6 To calculate volumetric strain = + Change in length = 34
20. A spherical shell of 2m diameter is made up of 10mm thick plates. Calculate the change in diameter and volume of the cylinder, when it s subjected to an internal pressure of 1.6 Mpa. Take E= 200Gpa and 1/m= 0.3 Diameter of the shell = 2m Pressure (P) = 1.6 N/mm 2 Thickness of shell t= 10mm Modulus of elasticity (E) =200 Gpa Passion ratio (1/m) = 0.3 To find Change in dimensions Step-1 Change in dia = 4 1 1 Step-2 Change in dia = 8 1 1 35
21. The state of stress at a certain point in a strained material is shown in fig calculate the principal stress inclination of the principal plane normal, shear and resultant stress on the plane MN Direct stress σ 1 =200N/mm 2 Direct stress σ 2 = 150N/mm 2 (compressive) Shear stress q= 100 N/mm 2 To find 1. Principal planes and principal stresses 2. Maximum shear stress and plane Step-1 To calculate principal plane 2 2 = Step-2 Major Principal Stress = + + 4 Step-3 MinorPrincipal Stress = + 4 36
Step-4 Maximum shear Stress = + 4 A steel cylinder shell 3m long which is closed at its ends, had an internal diameter of and wall thickness 20mm calculate the circumferential and longitudinal stress induced and also chane in dimensions of the shell if it is subjected to an internal pressure 1.0 N/mm 2 assume the modulus of elasticity and passion ratio for steel as 200kN/mm 2 and 0.3 respectively. 37