By Joseph Lipka
In contrast to another reproductions of vintage texts (1) we haven't used OCR(Optical personality Recognition), as this ends up in undesirable caliber books with brought typos. (2) In books the place there are photos resembling snap shots, maps, sketches and so forth we've endeavoured to maintain the standard of those photos, in order that they signify competently the unique artefact. even if sometimes there is convinced imperfections with those previous texts, we believe they should be made to be had for destiny generations to take pleasure in.
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Example text
Determine the total elongation of the assembly. For steel, E = 200 GPa and for brass, E = 90 GPa. Ans. 24. A high-performance jet aircraft cruises at three times the speed of sound at an altitude of 25 000 m. It has a long, slender titanium body reinforced by titanium ribs. The length of the aircraft is 30 m and the coefficient of thermal –6 expansion of the titanium is 10 ×10 /°C. Determine the increase of overall length of the aircraft at cruise altitude over its length on the ground if the temperature while cruising is 500°C above ground temperature.
2-2 (b) Shearing stresses acting on an element. 3 Shear Stresses Shear Strain The change of angle at the corner of an originally rectangular element is defined as the shear strain. It must be expressed in radians and is denoted by g, as shown. Shear Modulus The ratio of the shear stress t to the shear strain g is called the shear modulus and is denoted by G. 2) G is also known as the modulus of rigidity. , N/m2, since the shear strain is dimensionless. Stress-strain diagrams for various materials may be drawn for shearing loads, just as they were drawn for normal loads in Chapter 1.
3-16 SOLUTION: (a) We have sx = 90 MPa, s y = 30 MPa, and txy = 40 MPa. The maximum normal stress, is given by Eq. 7) to be σ + σ y 2 2 σ max = x + [ 12 (σ x − σ y )] + (τ xy ) 2 = 9 0 − 30 2 90 + 30 2 + + 40 = 110 MPa 2 2 The minimum normal stress is given by Eq. 8) to be σ + σy 90 + 30 2 2 σ min = x − 3 02 + 402 = 10 MPa − [ 12 (σ x − σ y )] + (τ xy ) = 2 2 The directions of the principal planes on which these stresses occur are given by tan 2θ p = 1 2 τ xy = (σ x − σ y ) 1 2 − 40 4 =− 3 (90 − 30) ∴ θ p = − 26 .
