EXERCISES OF CHAPTER 1 + 2 + 3 Exercise 1: Two solid cylindrical rods AB and BC are welded together at B and loaded as shown (Fig. 1). Knowing that the everage normal stress must not exceed 150 MPa in either rod, determine the smallest allowable values of the the diameters d 1 and d 2 . Exercise 2: The uniform beam is supported by two rods AB and CD that have cross-sectional areas of 10 mm 2 and 15 mm 2 , respectively (Fig. 2). Determine the position d of the distributed load so that the average normal stress in each rod is the same 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: Fig. 1 Fig.2 125kN 125kN 60kN 0.9m 1.2m 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 sectional area of 400 mm 2 and the contact area at C is 650 mm 2 . Solution Internal loading. The free body diagram for member AC is shown in Fig. (4b). There are three unknowns, namely, F AB , F C , and x. The equilibrium of AC will give: ( ) ( ) (1) + (2) y AB C A C F 0 F F 3000N 0 M 0 - 3000N (x) F 200mm 0 + ↑ = ⇒ + − = = ⇒ + = ∑ ∑ N Average Normal Stress. A necessary third equation can be written that requires the tensile stress in the bar AB and the compressive stress at C to be equivalent: i.e., ( ) 3 CAB AB C 2 2 F F F 1.625F 400 mm 650 mm σ = = → = Substituting (3) into Eq. 1, solving for F AB then solving for F C , we obtain: AB C F 1143 N; F 1857 N= = The position of the applied load is determined from Eq. 2.: x = 124 mm Note that 0 < x < 200 mm, as required. Exercise 5: The steel column is used to support the symmetric loads from the two floors of a building (Fig.5). Determine the loads P 1 and P 2 if A moves downward 3 mm and B moves downward 2 mm when the loads are applied. The column has a cross-sectional area of 645 mm 2 . E st = 200 Gpa. Exercise 6: Fig.4 Fig.5 3,6m 3,6m 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? Exercise 7: In the figure 6 of the precedent exercise, suppose that link BC is 6 mm thick and is made of a steel with a 450-MPa ultimate strength in tension. What should be its width w if the structure shown is beeing designed to support a 20-kN load P with a factor of safety of 3? Exercise 8: Both portions of the rod ABC are made of an aluminum for which E = 70 Gpa (Fig.7). Knowing that the magnitude of P is 4 kN. Determine: (a) the value of Q so that the deflection at A is zero; (b) the corresponding deflection of B; Fig.7 Exercise 9: A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of brass (E b = 105 GPa, α b = 20.9 x 10 -6 / 0 C) and portion BC is made of aluminum (E a = 72 GPa, α a = 23.9 x 10 -6 / 0 C) (Fig.8). Knowing that the rod is initially unstressed, determine (a) the normal stresses induced in portions AB and BC by a temperature rise of 42 0 C; (b) the corresponding deflection of point B. Fig.6 Fig.8 Exercise 10: A axial centric force of magnitude P = 450 kN is applied to the composite block shown by means of a rigid end plate (Fig.9). Knowing that h = 10 mm, determine the normal stress in (a) the brass core, (b) the aluminum plate Exercise 11: The rigid bar is supported by the two short white spruce wooden posts and a spring. If each of the posts has an unloaded length of 1 m and a cross-sectional area of 600 mm 2 , and the spring has a stiffness of k = 2 MN/m and an unstretched length of 1.02 m, determine the vertical displacement of A and B after the load is applied to the bar Exercise 12 The rigid bar shown in the figure is fixed to the top of the three posts made of steel and aluminum. The posts each have a length of 250 mm when no load is applied to the bar and the temperature is T 1 = 20 0 C. Determine the force supported by each post if the bar is subjected to a uniform distributed load of 150 kN/m and the temperature is raised to T 2 = 80 0 C. The diameter of each post and its material properties are listed in the figure. Solution Fig.9 Fig.10 40mm 40mm 60mm 150kN/m 300mm 300mm Steel Steel E st = 200 Gpa α st = 12(10 -6 )/ 0 C Al. Aluminum E Al = 70 Gpa α st = 23(10 -6 )/ 0 C Fig. 11 Equilibrium: ( ) 1) ( 3 y st Al F 0; 2N N - 90 10 0+ ↑ = → + = ∑ Compatibility: Due to load, geometry, and material symmetry, the top of each post is displaced by an equal amount. Hence: ( ) ( ) st Al 2+ ↓ δ = δ The final position of the top of each post is equal to its displacement caused by temperature, plus its displacement caused by the internal axial force. ( ) ( ) ( ) (3) st st st T N + ↓ δ = − δ + δ ( ) ( ) ( ) (4) Al Al Al T N + ↓ δ = − δ + δ Introducing into (2) gives: ( ) ( ) ( ) ( ) st st Al Al T N T N − δ + δ = − δ + δ (5) Using Eqs. (2) – (5) , we get: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )                 st -6 0 0 0 2 9 2 Al -6 0 0 0 2 9 2 N 0.250m - 12 10 / C 80 C - 20 C 0.250m + p 0.02m 200 10 N/m N 0.250m = - 23 10 / C 80 C - 20 C 0.250m + p 0.03m 70 10 N/m or ( ) 3 st Al N = 1.270N -165.9 10 (6) Solving (1) and (6) simultaneously yields: N st = - 14.6 kNN Al = 119 kN The negative value of N st indicate that this force acts opposite to that shown in figure. In other words, the steel posts are in tension and the aluminum post is in compression. 90 kN N st N st N Al (δ st ) T (δ st ) N (δ Al ) T (δ Al ) N Final position Initial position