JNTU
JNTU B.Tech II STRENGTH OF MATERIALS I Previous Question Paper (March - 2021 - R15)
Answer any five questions. All questions carry equal marks.
Main
Derive the relationship between bulk modulus and young’s modulus.
Answer
Consider a cube of unit volume subjected to a hydrostatic stress σ acting on all three faces. The volumetric strain εv is given by the sum of linear strains in three mutually perpendicular directions: εv = εx + εy + εz.
Using Hooke's Law, the linear strain in any direction is ε = (σ/E) - (μσ/E) - (μσ/E) = (σ/E)(1 - 2μ). Since the stress is uniform in all directions, εx = εy = εz.
By definition, the Bulk Modulus K is the ratio of direct stress to volumetric strain:
Simplifying this expression yields the relationship:
Three bars made of copper, zinc and aluminium are of equal length and have cross sections of 500, 700, and 1000 sq.mm respectively. They are rigidly connected at their ends. If this compound member is subjected to a longitudinal pull of 400kN, estimate the proportion of the load carried on each rod and the induced stresses. Take the value of E for copper = 1.3×105 N/mm2, for zinc = 1×105 N/mm2 and for aluminium = 0.8×105 N/mm2.
Answer
Since the bars are rigidly connected at their ends, they undergo the same deformation (ΔL), meaning the strain ε is equal for all three bars. Let Pc, Pz, Pa be the loads and Ac, Az, Aa be the areas for copper, zinc, and aluminium respectively.
The total load is P = Pc + Pz + Pa = 400,000 N. Since P = σA = (Eε)A, the total load is:
Calculating the axial rigidities (EA):
- EcAc = 1.3×105 × 500 = 6.5×107 N
- EzAz = 1.0×105 × 700 = 7.0×107 N
- EaAa = 0.8×105 × 1000 = 8.0×107 N
Sum of rigidities = 21.5×107 N. The common strain ε = 400,000 / 21.5×107 = 0.0018605.
Load distribution:
- Pc = 6.5×107 × 0.0018605 = 120,932.5 N
- Pz = 7.0×107 × 0.0018605 = 130,235 N
- Pa = 8.0×107 × 0.0018605 = 148,840 N
Induced stresses (σ = P/A):
- σc = 120,932.5 / 500 = 241.87 N/mm2
- σz = 130,235 / 700 = 186.05 N/mm2
- σa = 148,840 / 1000 = 148.84 N/mm2
A tension bar is made of 2 parts. The length of 1st part is 300 cm and area is 20cm2 while the second part is of length 210 cm and area 30 cm2. An axial load of 120 kN is gradually applied. Find the total strain energy produced in the bar and compare its value with that obtained in a uniform bar of same length and having same volume under same load. Take E = 2.1×105 N/mm2.
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A simply supported beam of length 10m carries the uniformly distributed load and two point loads as shown in Figure 1. Draw the S.F and B.M diagrams for the beam and also calculate the maximum bending moment.
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Draw S.F.D and B.M.D for the Over hanging beam shown in figure 2.
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A simply supported beam of span 5m carries a udl of 8 kN/m over the entire span. If the maximum allowable stress due to bending is restricted to 150 N/mm2, determine the cross sectional dimensions if the section is:
Rectangular with depth twice the breadth
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Solid circular section.
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A beam of I section has overall dimensions of 400×200 mm. The thickness of flange and web are 10 mm. If this section is used as a simply supported beam for a span of 8m carrying a udl of 10kN/m over entire span, draw the shear stress distribution at:
At starting of beam
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At one-fourth span of beam
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At center of the beam.
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Explain about Maximum principal stress theory, maximum principal strain theory and shear strain energy theory.
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A beam of length 6 m is simply supported at its ends and carries two point loads of 60 kN and 40 kN at a distance of 1.5 m and 3.5 m respectively from the left support. Solve for:
Deflection under each load
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Maximum deflection
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Slope at the left support.
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