Solution: The moment of inertia for the cross-section is: ANS: ANS: Problem 15.6 The material of the beam in Problem

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1 Problem 15.4 The beam consists of material with modulus of elasticity E 14x10 6 psi and is subjected to couples M 150, 000 in lb at its ends. (a) What is the resulting radius of curvature of the neutral axis? (b) Determine the maximum tensile stress due to bending? The moment of inertia for the cross-section is: πr4 4 π(in)4.57in 4 4 (a) Using Equation (15-10) to determine the magnitude of the radius of curvature: 1 ρ M E 150, 000 in lb ( lb/in )(.57 in 4 ) in 1 ρ in 97.7 ft Using Equation (15-) to determine the applied moment: σ MAX My MAX σ MAX 3.9ksi (150, 000 in lb)(in).57 in 4 Problem 15.5 The material of the beam in Problem 15.4 will safely support a tensile or compressive stress of 30, 000 psi. Based on this criterion, what is the largest couple M to which the beam can be subjected? Using Equation (15-) to determine the applied moment: σ MAX My MAX M 188, 550 in lb M σ MAX (30, 000 lb/in )(.57 in 4 ) y MAX in Problem 15.6 The material of the beam in Problem 15.4 will safely support a tensile or compressive stress of 30, 000 psi. f the beam has a hollow circular cross-section, with -in. outer radius and 1-in. inner radius, what is the largest couple M to which the beam can be subjected? The moment of inertia for the cross-section is: π 4 (r4 o ri 4 ) π [ ( in) 4 (1 in) 4] in 4 4 Using Equation (15-) to determine the applied moment: σ MAX My MAX M 176, 700 in lb M σ MAX (30, 000 lb/in )(11.78 in 4 ) y MAX in

2 Problem 15.7 Suppose that the beam in Example 15-1 is made of a brittle material that will safely support a tensile stress of 0 MPa or a compressive stress of 50 MPa. What is the largest couple M to which the beam can be subjected? From the solution to Example 15-1, we know that: 1.85x10 6 m 4 and y m from the top of the cross-section. Using the maximum tensile stress in Equation (15-) to determine the allowable moment: σ MAX My MAX M σ MAX (0 106 N/m )( m 4 ) y MAX 0.08 m m M 1138 N m Using the maximum compressive stress in Equation (15-) to determine the allowable moment: σ MAX My MAX M σ MAX ( N/m )( m 4 ) y MAX m M 1947 N m We realize that the bar will fail if either of the calculated moments is exceeded, so the maximum allowable moment must be the smaller moment. M 1138 N m Problem 15.8 What is the maximum tensile stress due to bending in the beam in Example 15-, and where does it occur? Summing moments about point B to determine the reaction at point A: ΣM B [ 1 (w 0)(L) ]( L 3 ) A y(l) A y w 0 L/6 B y w 0 L/3 The bending moment is maximum where the shear force equals zero. Summing vertical forces on an arbitrary length of the left-hand portion of the beam: ΣF y 0 w 0L 6 [ ] 1 w 0 L x (x) x 0.577L at the bottom of the cross-section Summing moments on the free-body diagram at x 0.577L: ( ) [ w0 L 1 ( w0 ) ]( ) 0.577L M (0.577L)+ (0.577L)(0.577L) 6 L 3 M w 0 L ( 0.064) Using Equation (15-) to determine the bending stress: (σ T ) MAX My (σ T ) MAX 0.384w 0 L h w 0L (h/) (h(h) 3 /)

3 Problem 15.9 The beam consists of material that will safely support a tensile or compressive stress of 350 MPa. Based on this criterion, determine the largest force F the beam will safely support if it has the cross section (a); if it has the cross section (b). (The two cross sections have approximately the same area.) The moment of inertia of the cross-section in case (a) is: a bh3 (0.033 m)(0.060 m) m 4 The moment of inertia for the cross-section in case (b) is: [ ] (0.050 m)(0.060 m)3 (0.00 m) (0.040 m) 4 b m 4 Summing moments about point B to find the reaction at point A: ΣM B 0F (0.6 m) A y(1.6 m) A y 0.375F Maximum bending moment occurs at the point where the concentrated load is applied, so we calculate maximum bending stresses 1.0 mto the right of point A. Maximum bending moment is: M MAX (0.375F )(1.0 m)0.375f N m For cross-section (a): N/m F kn For cross-section (b): N/m F 1.4 kn (0.375F )(0.03 m) m 4 (0.375F )(0.03 m) m 4 Problem f the beam in Problem 15.9 is subjected to a force F 6kN, what is the maximum tensile stress due to bending at the cross section midway between the beam s supports in cases (a) and (b)? The moment of inertia of the cross-section in case (a) is: a bh3 (0.033 m)(0.060 m) m 4 The moment of inertia for the cross-section in case (b) is: [ ] (0.050 m)(0.060 m)3 (0.00 m) (0.040 m) 4 b m 4 Summing moments about point B to find the reaction at point A: ΣM B 0(6, 000 N)(0.6m) A y(1.6m) A y 50 N The bending moment at the midpoint of the beam is: M (50 N)(0.8 m) 1800 N m n case (a), the maximum bending stress is: (σ a) MAX (σ a) MAX 8.8 MPa (1800 N m)(0.03 m) m 4 n case (b), the maximum bending stress is: (σ b ) MAX (σ b ) MAX 78.6 MPa (1800 N m)(0.03 m) m 4

4 Problem 15. The beam is subjected to a uniformly distributed load w lb/in. Determine the maximum tensile stress due to bending at x 0inif the beam has the cross section (a); if it has the cross (b). (The two cross-sections have approximately the same area.) The moments of inertia for the cross-sections in the two cases are: a (4.47 in)(4.47 in) in 4 b (6 in)(6 in)3 Equilbiruim to find A y + B y (4 in)(4 in) in 4 ΣM B 0A y(85) 5, 500(4.5) A y, 500 N Summing moments about the cut through the beam at x 0in: M (, 750 lb)(0 in) (6, 000 lb)(10 in) 195, 000 in lb We see that the maximum tensile occurs at the bottom of each crosssection. n case (a), the maximum tensile stress is: (σ T ) MAX (195, 000 in lb) (4.47 in/) 33.7 in 4 (σ T ) MAX 13.1 ksi at the bottom of the cross-section n case (b), the maximum tensile stress is: (σ T ) MAX (195, 000 in lb)(3in) in 4 (σ T ) MAX 6.75 ksi at the bottom of the cross-section

5 Problem The beam in Problem 15. consists of material that will safely support a tensile or compressive stress of 30ksi. Based on this criterion, determine the largest distributed load w 0 (in lb/in) the beam will safely support if it has the cross section (a); if it has the cross section (b). From the symmetry of the loading, we see that: A y B y (85w 0 /) lb The bending moment is maximum at the point on the beam where the shear stress is zero (the middle of the beam). ΣF y 0A y w 0 x 85w 0 w 0 x x 4.5 in The bending moment about a cut through the center of the beam is: [ ]( ) ( )( ) w0 (85in) 85 in 85 in 85 in M w w 0 [1] 4 To find the maximum allowable bending moment in case (a), the maximum allowable normal stress is used. 30, 000 lb/in M (4.47 in/) 33.7 in 4 M 446, 577 in lb [] Solving equations [1] and [] together: ANS w lb/in To find the maximum allowable bending moment in case 9b0, the maximum allowable normal stress is again used. 30, 000 lb/in M (3 in) M 866, 700 in lb [3] in4 Solving equations [1] and [3] together: w lb/in

6 Problem 15.3 A beam with the cross section is sublected to a shear force V 8kN. What is the shear stress at the neutral axis (y 0)? Using Equation (15-18) to determine the average shear stress at the neutral axis: τ AVG 3V A 3(8, 000 N) (0.04 m)(0.06 m) τ AVG 5 MPa Problem 15.4 n Problem 15.3, determine the average shear (a) at y 0.01 m; (b) at y 0.0 m. (a) Using Equation (15-17) to determine the average stress at y 0.01 m: τ AVG 6V bh 3 [ ( ) h ( ] y ) 6(8, 000 N) (0.04 m)(0.06 m) 3 [ ( ) ] 0.06 m (0.01 m) τ AVG 4.44 MPa (b) Using Equation (15-17) to determine the average stress at y 0.0 m: [ ( τ AVG 6V ) h bh 3 ( [ ( ] y ) ) ] 6(8, 000 N) 0.06 m (0.04 m)(0.06 m) 3 ( 0.0 m) τ AVG.78 MPa Problem 15.5 n Example 15-5, consider the cross section at x 3m. What is the average shear stress at y 0.05 m. Summing moments about point B to determine A y: MB 0[(6, 000 N/m)(8 m)] (4 m) A y(8 m) A y 4, 000 N Cut the FBD where x 3mand draw the FBD. Summing vertical forces to determine the shear force V : Fy 04, 000 N (6, 000 N/m)(3 m) V V 6, 000 N Using Equation (15-17) to determine the average stress at y 0.05 m: [ ( τ AVG 6V ) h bh 3 ( [ ( ] y ) ) ] 6(6, 000 N) 0.5 m (0.5 m)(0.5 m) 3 (0.05 m) τ AVG 1 kpa

7 Problem 15.8 Solve Problem 15.7 for the cross section at x 80in. Summing the moments about point B to determine the reaction at point A: MB 0 [(1500 lb/in)(60 in)] (30 in) A y(0 in) A y, 500 lb Draw the FBD at x 80in. Summing the vertical forces to determine the shear force: Fy 0, 500 lb( 1, 500 lb/in)(0 in)+v V 7, 500 lb (a) Using Equation (15-18) to determine the average shear stress at the neutral axis: τ 3V 3(7, 500 lb) A (1 in)(4 in) τ AVG, 8.5 lb/in (b) Using Equation (15-17) to determine the average stress at y 1.5 in: [ ( τ AVG 6V ) h bh 3 ( [ ( ] y ) ) ] 6(7, 500 lb) 4in (1 in)(4 in) 3 (1.5 in) τ AVG 1, lb/in

8 Problem 15.9 What is the maximum magnitude of the average shear stress in the beam in Problem 15.7, and where does it occur? Summing the moments about point B to determine the reaction at point A: MB 0 [(1500 lb/in)(60 in)] (30 in) A y(0 in) A y, 500 lb We see that the maximum shear stress exists at x 0 in. V 67, 500 lb (a) Using Equation (15-18) to determine the average shear stress at the neutral axis: τ 3V 3(67, 500 lb) A (1 in)(4 in) τ AVG 5, 3, 5lb/in 5.3 kip/in B y 67, 500 lb Problem By integrating the stress distribution given by Equation (15-17), confirm that the total force exerted on the rectangular cross section by the shear stress is equal to V. Starting with Equation (15-17) and integrating over the dimensions h/ to h/: τ AVG 6v h/ [ h ] bh 3 h/ (y ) da 6v h/ [ h ] bh 3 h/ (y ) bdy 6vb h/ [ h ] bh 3 h/ (y ) dy Doing the integration: τ AVG 6v [( h y h 3 (y ) 3 )] h/ 4 3 h/ τ AVG 6v [( h 3 ) ( h 3 8 h3 h 3 )] h τ AVG 6v [ h 3 ] h 3 4 h3 6v h 3 τ AVG V [ h 3 ] 6

9 Problem The beam whose cross section is shown consists of three planks of wood glued together. At a given axial position it is subjected to a shear force V 400 lb. What is the average shear stress at the neutral axis y 0? Finding the centroid of the entire cross section (measuring from the TOP): y ( in)(8 in)(4 in) + () [( in)(4 in)(7 in)] ( in)(8 in) + () [( in)(4 in)] 5.5 in Calculating y : Calculating A : y 5.5 in.75 in A ( in)(5.5nches) 11in Calculating Q: Q (.75 in)(11 in ) 30.5 in 3 Calculating the moment of inertia: [ ( in)(8 in) 3 [ (4 in)( in) + ( in)(8 in)(4 in 5.5 in) ]+ 3 Now calculating the average shear stress: τ AVG VQ b (, 400 lb)( 30.5 in3 ) ( in)(16.7 in 4 ) τ AVG 3.1 lb/in + (4 in)( in)(7 in 5.5 in) ] 16.7in 4 Problem n Problem 15.35, what are the magnitudes of the average shear stress acting on each glued joint? Finding the centroid of the entire cross section (measuring from the TOP): ( in)(8 in)(4 in) + () [( in)(4 in)(7 in)] y 5.5 in ( in)(8 in) + () [( in)(4 in)] Calculating the moment of inertia: [ ( in)(8 in) 3 [ (4 in)( in) + ( in)(8 in)(4 in 5.5 in) ]+ 3 Calculating y : Calculating A : Calculating Q: y 7in 5.5 in1.5 in A ( in)(4 in) 8 in Q y A (1.5 in)(8 in )in 3 Now calculating the average shear stress: τ AVG VQ b (, 400 lb)( in3 ) ( in)(16.7 in 4 ) τ AVG 88.5 lb/in + (4 in)( in)(7 in 5.5 in) ] 16.7in 4