Bài tập sức bền vật liệu (Giáo trình tiếng Anh)

Tài liệu tham khảo về các bài tập sức bền vật liệu của chương 1, 2 và 3. Tài liệu bằng tiếng Anh.
Nội dung trích xuất từ tài liệu:
Bài tập sức bền vật liệu (Giáo trình tiếng Anh) EXERCISESOFCHAPTER1+2+3 Exercise1: Two solid cylindrical rods AB and 125kN BC are welded together at B and loaded as shown (Fig. 1).60kN Knowing that the everage normal 125kN stress must not exceed 150 MPa 0.9m 1.2m in either rod, determine the smallest allowable values of the Fig. 1 the diameters d1 and d2. Exercise2: The uniform beam is supported by two rods AB and CD that have cross- sectional areas of 10 mm2 and 15 mm2, respectively (Fig. 2). Determine the position d of the distributed load so that the average normal stress in each rod is the same Fig.2 Exercise 3: Fig.3a Fig.3b The stress-strain diagram for a polyester resin is given in the figure 3b. If the rigid beam is supported by a strut AB and post CD, both made from this material, and subjected to a load of P = 80 kN (Fig.3a), determine the angle of tiltof the beam when the load is applied. The diameter of the strur is 40 mm and the diameter of the post is 80 mm. Exercise 4: Member AC is subjected to a vertical force of 3 kN. Determine the position x of this force so that the compressive stress at C is equal to the tensile stress in the tie rod AB (Fig.4a). The rod has a cross F Fig.4 sectional area of 400 mm2 and the contact area at C is 650 mm2.SolutionInternalloading. The free body diagram for member AC is shownin Fig. (4b). There are three unknowns, namely, FAB, FC, and x. Theequilibrium of AC will give:+ ↑ ∑F = 0 ⇒ y F B + F − 3000N = 0 A C (1) ∑∑+ M A = 0 ⇒ ( 3000N ) (x) + F ( 200m m ) = 0 C (2)AverageNormalStress.A necessary third equation can be writtenthat requires the tensile stress in the bar AB and the compressivestress at C to be equivalent: i.e., FB F σ= A 2 = C C 625F B ( 3) → F = 1. A 400 m m 650 m m 2Substituting (3) into Eq. 1, solving for FAB then solving for FC, weobtain: F B = 1143 N; A F = 1857 N CThe position of the applied load is determined from Eq. 2.: x=124mmNote that 0 < x < 200 mm, as required.Exercise5: The steel column is used to support the symmetric loads from the two floors of a building (Fig.5). 3,6 Determine the loads P1 and P2 if A m moves downward 3 mm and B Fig.5 moves downward 2 mm when the loads are applied. The column has a cross-sectional area of 645 mm2. Est 3,6m = 200 Gpa.Exercise6: Link BC is 6 mm thick, has a width w = 25 mm, and is made of a steel of 480-MPa ultimate strength in tension (Fig.6). What was the safety factor used if the structure shown was designed to support a 16-kN load P? Fig.6Exercise7: In the figure 6 of the precedent exercise, suppose that link BCis 6 mm thick and is made of a steel wi ...