Free Body Diagram: Solution: The maximum load which can be safely supported by EACH of the support members is: ANS: A =0.217 in 2

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1 Problem 10.9 The angle β of the system in Problem 10.8 is 60. The bars are made of a material that will safely support a tensile normal stress of 8 ksi. Based on this criterion, if you want to design the system so that it will support a force F = 3 kip, what is the minimum necessary value of the cross-sectional area A? The maximum load which can be safely supported by EACH of the support members is: F MAX = σ MAX (A) = (8000 lb/in 2 )(A) Summing vertical forces on the FBD: ΣF y =0= 3000 lb + 2(8000 lb/in 2 )(A)(sin 60 ) A =0.217 in 2 Problem Suppose that the horizontal distance between the supports of the system in Problem 10.8 and the load F are specified, and the prismatic bars are made of a material that will safely a tensile normal stress σ 0. You want to choose the angle β and the cross-sectional area A of the bars so that the total volume of material used is a minimum. What are β and A? L = d/(cos β) [1] Summing vertical forces on the FBD to find the force supported by the two prismatic bars: ΣF y =0= F +2 [σ 0 A(sin β)] where σ 0 is the maximum allowable average normal stress. From the above equation, the cross-sectional area of one of the prismatic bars may be expressed as: F A = [2] 2σ 0 (sin β) Recall: sin 2β = 2 sin β cos β Using Equations [1] and [2], the volume of a single bar is: Fd V = AL = 2σ 0 sin β cos β = Fd σ 0 sin 2β We see that the volume will be minimum when sin2β is maximum, which is when sin 2β =1, so: β =5 [3] Using this value for β in Equation [2] to find A: A = F 2σ 0 (0.707) A =0.707 F σ 0

2 Problem The cross-sectional area of each bar is 60 mm 2.IfF =kn, what are the normal stresses on planes perpendicular to the axes of the bars? Summing horizontal forces on the FBD: ΣF x =0= F AB (cos 60 )+F AC (cos 5 ) F AB =1.1F AC [1] Summing vertical forces on the FBD: ΣF y =0= 0, 000 N + F AB (sin 60 )+F AC (sin 5 ) [2] Substituting Equation [1] into Equation [2]: 0, 000 N = [1.1F AC ](sin 60 )+F AC (sin 5 ) F AC =20, 708 lb Therefore: F AB =29, 281 lb The normal stresses in the two supporting members are: σ AB = F AB /A AB = (29, 281 lb)/( m 2 ) σ AB = 88 MPa σ AC = F AC /A AC = (20, 708 lb)/( m 2 ) σ AC = 35 MPa

3 Problem What tensile force would have to be exerted on the right end of the bar in Problem to increase its length to 9.02 in.? What is the bar s diameter after this load is applied? The strain in the bar will be: ε = L L L = 9.02 in 9.00 in 9.00 in The stress required to produce this strain is: = σ = Eε = ( lb/in 2 )( ) = 66, 667 lb/in 2 The load required to produce the stress is: ( π(0.75 in) P = σa = (66, 667 lb/in 2 2 ) ) P =29.5kip The radial strain in the deformed bar is: ε LAT = εν =( )( 0.32)= The diameter of the deformed bar is: D = D(1 ε LAT )=(0.75 in)( ) D =0.795 in Problem A prismatic bar is 300 mm long and has a circular cross section with 20-mm diameter. Its modulus of elasticity is 120 Gpa and its Poisson s ratio is Axial forces P are applied to the ends of the bar which cause its diameter to decrease to mm. (a) What is the length of the loaded bar? (b) What is the value of P? We can use Poisson s ratio to determine the extensional strain in the material. (19.98 mm 20 mm)/(20 mm) 0.33 = ε (a) The length of the loaded bar is: L = L(1 + ε) = (300 mm)( ) ε = L = mm (b) The value of P is determined using the definition of the modulus of elasticity. P = εea =(0.0079)( N/m 2 )(π)(0.01 m) 2 P = kn

4 Problem Bar AB has cross-sectional area A = 100 mm 2 and modulus of elasticity E = 102 Gpa. The distance H = 00 mm. If a 200-kN downward force is applied to bar CD at D, through what angle in degrees does bar CD rotate? (You can neglect the deformation of bar CD.) Strategy: Because the bar s change in length is small, you can assume that the downward displacement v of point B is vertical, and that the angle (in radians) through which bar CD rotates is v/h. We can determine the angle of rotation by finding the vertical displacement at point B. Find the axial load in member AB by summing moments about point C: ΣM c =0= (200 kn)(0.6 m)+f AB (sin 60 )(0. m) F AB = 36. kn(c) The strain in member AB will be: ε = P/A E = (36,00 N)/( m) N/m 2 ε =0.03 The original length of member AB is: L = (300 mm)/(sin 60 ) = 36.3 mm The new length of member AB is: L = L(1 + ε) = (36.3 mm)(1 0.03) = 33.5 mm The original height of point B is h = 300 mm. The deformed height of point B is: h = (L ) 2 [300 mm/tan 60 ] 2 = (33.5 mm) 2 (173.2 mm) 2 = 286.2mm The change in vertical height at point B is: v = h h = 300 mm mm=13.8 mm The angle, in radians, through which the bar CD rotates is: θ =13.8 mm/00 mm = radians = 1.98 clockwise

5 Problem Bar AB in Problem is made of a material that will safely support a normal stress (in tension or compression) of 5 GPa. Based on this criterion, through what angle in degrees can bar CD safely be rotated relative to the position shown. Maximum allowable strain in the material is: ε = σ E = N/m N/m 2 =0.09 Maximum allowable change in length for H is: H = Lε =(0. m)(.09) = m In the diagram, maximum allowable distance d is: d =( H)(cos 30 )=( m)(cos 30 ) d = m The angle through which bar AB may rotate is: θ = d = ± m L AB (0.3/ sin 60 = ±0.09 rad ) θ = ±2.8 Problem 10.3 If an upward force is applied at H that causes bar GH to rotate 0.02 degrees in the counterclockwise direction, what are the axial strains in bars AB, DC, and EF? (You can neglect the deformation of bar GH.) The vertical displacement at point H is: θ =(0.02 /180 )( radians) = radians The vertical displacement of points B, D and F are: V B = ( radians)(00 mm) = 0.1 mm V D = ( radians)(800 mm) = 0.28 mm V F = ( radians)(1200 mm) = 0.2 mm The strains in each of the vertical bars is: 0.1 mm ε AB = 00 mm = ε 0.28 mm CD = 00 mm = ε 0.2 mm EF = 00 mm =

6 Problem The bar has cross-sectional area A and modulus of elasticity E. The left end of the bar is fixed. There is initially a gap b between the right end of the bar and the rigid wall (Figure 1). The bar is stretched until it comes into contact with the rigid wall and is welded to it (Figure 2). Notice that this problem is statically indeterminate because the axial force in the bar after it is welded to the wall cannot be determined from statics alone. (a) What is the compatibility condition in this problem? (b) What is the axial force in the bar after it is welded to the wall? The compatibility condition requires that the bar s change in length must be limited to the amount of the gap, b. Using the relationship δ = PL/AE to find the axial load: P = δae L = bae L Problem The bar has cross-sectional area A and modulus of elasticity E. If an axial force F directed toward the right is applied at C, what is the normal stress in the part of the bar to the left of C? (Strategy: Draw the free-body diagram of the entire bar and write the equilibrium equation. Then apply the compatibility condition that the increase in length of the part of the bar to the left of C must equal the decrease in length of the part to the right of C.) Summing horizontal forces on the FBD: [1] ΣF x =0=F R L R R The compatibility condition requires that the change in length of the left portion of the bar must equal the change in length for the right portion of the bar. R L L L = R RL R A L E L A R E R Since the denominators of the above equation are identical, we need only consider the numerators. [2] R R =(L L /L R )(R L )=[(L/3)/(2L/3)](R L )=R L /2 Substituting Equation [2] into Equation [1]: F = R L + R R = R L +(R L /2)=3R L /2 or R L =(2/3)F The stress in the left-hand portion of the bar is: σ =2F/3A σ = P A = R L A

7 Problem 10.0 The bar has a circular cross section and modulus of elasticity E =70GPa. Parts A and C are 0 mm in diameter and part B is 80 mm in diameter. If F 1 =60kNand F 2 =30kN, what is the normal stress in part B? We must first determine the reactions at the left and right walls. We allow the right-hand side of the bar to float. The displacement of the free right-hand side of the bar is: δ R = F 1L A F 2L B F 2L A A A E A A B E B A A E A (60,000 N)(0.2 m) δ R = ( π (0.0 m)2 )( N/m 2 ) ( (30,000 N)(0. m) (0.08 m)2 π )( N/m 2 ) ( (30,000 ) N)(0.2 m) (0.0 m)2 π ( N/m 2 ) δ R =0.031 m = 3.1 mm The reaction at the right wall must be sufficient to prevent ANY displacement. Its magnitude is: ( ) ( ) R m = R 0.2 m lb/in 2 ( ) 0. m 0.2 m + + (0.0 m)2 (0.08 m)2 (0.0 m)2 π π π R R = 6000 N The reaction at the left wall is: ΣF x =0=60, 000 N 6, 000 N 30, 000 N R L R L =2, 000 N The stress in section A is: σ B = P 2,000 N+60,000 N = A π (0.08 m)2 Note: The negative sign indicates a compressive stress. σ A = 7.16 MPa Problem 10.1 In Problem 10.0, if F 1 =60kN, what force F 2 will cause the normal stress in part C to be zero? For the normal stress in section C to be zero, we see that the displacement of the intersection of sections B and C must be zero. The equation for the displacement of the intersection of sections B and C is: (60, 000 N)(0.2 m) δ BC =0= ( π ((0.0 m) 2 /) N/m 2) F 2 (0.2 m) ( π ((0.0 m) 2 /) N/m 2) F 2 (0. m) π ((0.08 m) 2 /) ( N/m 2) F 2 =0, 000 N = 0 kn

8 Problem 10.2 The bar in Problem 10.0 consists of a material that will safely support a normal stress of 0 MPa. IfF 2 =20kN, what is the largest safe value of F 1? We see that section B has a cross-sectional area which is four times that of sections A and C. The cross-sectional areas are: A A = A C = π (0.02 m) 2 = m 2 A B = π (0.0 m) 2 = m 2 With both ends of the bar restrained, total displacement must be zero. R δ R =0= L (0.2 m) ( m 2 )( n/m 2 ) (F 1 R L )(0. m) ( m 2 )( N/m 2 ) (F 1 R L 20,000 N)(0.2 m) ( m 2 )( N/m 2 ) R L R L F R L F R L =0 R L =0.6F 1 8, 000 N Summing horizontal forces to find R R : ΣF x =0= R L +F 1 20, 000 N R R = 0.6F N+F 1 20, 000 N R R R R =0.F 1 12, 000 N The axial loads in sections A, B and C are: σ a = = 0.6f 1 8, F 1 =97.3 kn σ b = = 0.6f 1 8,000 F F 1 = 523 kn σ c = = 0.f 1 12, F 1 = 156 kn The smallest of these three values for F 1 is the highest allowable value. F 1 =97.3 kn Problem 10.3 Two aluminum bars (E AL = psi) are attached to a rigid support at the left and a cross-bar at the right. An iron bar (E FE = psi) is attached to the rigid support at the left and there is a gap b between the right end of the iron bar and the cross-bar. The cross-sectional area of each bar is A = 0.5 in 2 and L =10in. The iron bar is stretched until it contacts the cross-bar and welded to it. Afterward, the axial strain of the iron bar is measured and determined to be ε FE = What was the size of the gap b? The foreshortening of the two aluminum bars plus the lengthening of the steel bar must equal the gap, b. The lengthening of the steel bar is (approximately): δ FE =(0.002)(10 in) = 0.02 in Calculating the force in the steel bar: P FE = εea =(0.002)( lb/in 2 )(0.5in 2 )=28, 500 lb This same force is compressing the TWO aluminum bars. The aluminum bars are shortened by an amount of: δ AL = PL AE = (28, 500 lb)(10 in)) 2 ( 0.5 in 2) ( = in lb/in 2) The total original gap is: b = δ AL + δ FE = in in b =0.085 in

9 Problem 10.5 From x =0to x = 100 mm, the bar s height is 20 mm. From x = 100 mm to x = 200 mm, its height varies linearly from 20 mm to 0 mm. From x = 200 mm to x = 300 mm, its height is 0 mm. The flat var s thickness is 20 mm. The modulus of elasticity of the material is E =70GPa. If the bar is subjected to tensile axial forces P =50kNat its ends, what is its change in length? Free Body Diagrams: The problem is solved by considering each of the three sections of the bar separately. Elongation of the left-hand section of the bar is: δ L = PL AE = (50, 000 N)(0.1 m) = mm (0.02 m)(0.02 m)( N/m Elongation of the right-hand section of the bar is: δ R = PL AE = (50, 000 N)(0.1 m) = mm (0.02 m)(0.0 m)( N/m Determining the elongation of the center section of the bar will require integration. The cross-sectional area at any point in the center section is: A C =(0.02 m)[0.02 m+((0.02 m)/(0.1 m)(x)]= x The stress at any point in the center section of the bar is: σ = P A = 50, x n/m2 Total elongation of the center section of the bar is: ( σ ) ( 50, 000 N/( x) m 2 ) dδ C = dx = E N/m 2 dx 0.1 δ C = dx = x x dx δ C = [ln ( (0.1)) ln ( (0))] δ C = mm Total change in length of the bar is: δ = δ L + δ C + δ R = mm mm mm δ =0.392 mm

10 Problem A cylindrical bar with 1 in diameter fits tightly into a circular hole in a 5 in thick plate. The modulus of elasticity of the marerial is E = psi. A 1000 lb tensile force is applied at the left end of the bar, causing it to begin slipping out of the hole. At the instant slipping begins, determine (a) the magnitude of the uniformly distributed axial force exerted on the bar by the plate; (b) the total change in the bar s length. The distributed load along the 5 inch section of the bar is: 1,000 lb q = = 200 lb/in 5in The elongation of the 10 inch section of the bar is: δ 10 = PL AE = (1, 000 lb)(10 in) π(0.5) in) 2 ( lb/in 2 = in ) The elongation of the 5 inch section of the bar is: δ 5 = εl = σ 5 E L = ( x) lb/π(0.5 in 2 ) lb/in 2 dx = in Total change in length of the bar is: δ = δ 10 + δ 5 = in in δ = in Problem The bar has a circular cross section with m diameter and its modulus of elasticity is E =86.6 GPa. The bar is fixed at both ends and is subjected to a distributed axial force q =75kN/m and an axial force F =15kN. What is its change in length? Cross-sectional area of the bar is: ( ) m 2 A = π = m 2 2 Summing horizontal forces on the FBD to find the magnitude of R: ΣF x =0= R 15, 000 N + (75, 000 N/m)(0.8 m) R =5, 000 N Total change in length of the bar is: 0.8 (5, 000 N 75, 000(x) N)dx δ = 0 ( m 2 )( m 2 ) δ =0.01 m =.1 mm

11 Problem The bar is fixed at A and B and is subjected to a uniformly distributed axial force. It has crosssectional area A and modulus of elasticity E. What are the reactions at A and B? We recognize that the sum of horizontal forces on the bar must be zero. ΣF x =0= R A R B + ql [1] Removing the fixed structure at the right-hand end of the bar, the distributed load will produce a change in length of: L qx δ = 0 AE dx = qx2 L 2AE = ql2 0 2AE R B = ql 2 Substituting this value for R B in Equation [1]: R A = R B + ql = (ql/2) + ql R A = ql/2 Problem What point of the bar in Problem undergoes the largest displacement, and what is the displacement? We regognize that the sum of horizontal forces on the bar must be zero. L qx Σ= 0 AE dx = qx2 L 2AE = ql2 0 2AE [1] The reaction at the right-hand support must be sufficient to produce a reduction in length of (ql 2 )/(2AE). ql 2 2AE = R BL AE R B = ql 2 Substituting this value for R B in Equation [1]: R A = RB + ql = (ql/2) + ql R A = ql/2 The expression for displacement of any point on the bar is: L (ql/2) qx δ = dx [2] 0 AE We know that maximum deflection occurs where the expression dδ/dx =0. Setting the derivative of Equation [2] equal to zero: (ql/2) qx =0 AE x = L/2 [3] Using the value of x from Equation [3] in Equation [2] and evaluating: L/2 [ (ql/2) qx (qlx/2) qx 2 ] L/2 /2 δ = dx = 0 AE AE 0 δ MAX = ql2 8AE