By V.I. Feodosiev, Sergey A. Voronov, Sergey V. Yaresko
This ebook is a set of difficulties for complex scholars within the region of power of fabrics. It attracts the reader?s awareness additionally to difficulties which are usually ignored and solutions questions which are a ways past a coaching direction and require extra primary realizing. All difficulties are supplied with targeted strategies to allow the reader to both know about the problem-solving method or simply to envision his/her personal approach of answer. The examine and academic paintings of V.I. Feodosiev used to be performed within the Bauman Moscow kingdom technical collage the place he held the path on power of fabrics for fifty years. Deep perception into engineering difficulties, clearness of thoughts and magnificence of ideas observed by way of pedagogical expertise are the most gains of his type.
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Extra resources for Advanced Stress and Stability Analysis: Worked Examples (Foundations of Engineering Mechanics)
Fig. 83 Using the maximum shear stresses strength criterion analyze the calculated ¾ eq stress dependence on moment M under given pressure p. 89. A thin-walled spherical vessel of radius R = 0:5 m and of thickness h = 1 cm is loaded with internal pressure p1 = 32 M P a and external pressure p 2 = 30 M P a (Fig. 84). 3. Complex Stress State, Strength Criteria, Anisotropy. Part I. Problems 35 It is necessary to determine a vessel wall’s safety factor sy if the yield limit stress of its material is known to be ¾ y = 300 M P a: Let’s consider an element taken from the wall near the internal surface of the vessel (Fig.
The lower end of the tube is clamped in a solid foundation. Pressure p is fed into the tube (Fig. 112). Can this tube buckle under su¢ciently high pressure? 123. A liquid of speci…c weight ° runs through the tube pinned at both ends (Fig. 113). Fig. 113 Show that under some value of liquid velocity v the tube looses its stability like Euler’s column. 46 4. Stability Part I. Problems 124. Let us consider the following problem. A rod having roundings of radius R at its ends (Fig. 114) is compressed without friction between two solid slabs.
Stability Part I. Problems 47 The …rst (the least) root of equation (1) must be substituted here instead of ®l. Now let us …nd the least root ®l of equation (1) as a function of R=l. The most convenient way to ascertain this relationship is specifying values of ®l and then determining R=l from equation (1). Fig. 115 The calculation results are shown as a curve (Fig. 115). We reveal that ®l = ¼ in case of R=l = 0; and therefore according to (2) : ¼2 E J l2 The critical load is equal to the ordinary Euler force as one would expect.