Question

Member BCD is supported in the position shown using the 1/2 × 1/4 in2 rod AC and 1/4 in diameter pins at B, C&D. The ultimate bearing stress for the pin at B is 5ksi, the ultimate shear stress for the pin at B is 7 ksi and the ultimate normal stress for the link AC is 12ksi. Assuming that the pins at A and C are strong and a factor of safety of 3 is desired, determine

(a) The allowable force P that can be applied

(b) Explain, with a logical reason, why the allowable value is a maximum/minimum?

A in. 6 in. in. in. B С D in. 4 in. * 4 in. w Side view at B P.

Answer 1

Cut the bar AC at and consider FBD of the lower section. O= Tan (9 56.31 ° MB = 0 (sum of moment of forces about pivot point B = 0) -P.(4 + 4) + NAC sin (O).4 = 0 NAC NAC = 2.404.P E Fx = 0) (sum of forces in x direction = 0) HB - NAC COS(O) = 0 B с HB - 2.404.P.cos(O) = 0 → HB F HB 1.333P VB 0) Fy = 0 (sum of forces in y direction VB+ NAC sin (O) - P = 0 4" 4" + X VB + 2.404.P.sin(O) - P = 0 VB = -P 2 The resultant force acting on pin B, RB HB' + VB V (1.333p) + (-p)? RB = 1.666P To find maximum allowable force P if normal stress in AC not to exceed its maximum. Consider BAR AC: Axial force, NAC = 2.404.P 11 Area of bar AC, Area AC 24 = = 0.125 in? Ultimate normal stress, OU = 12 ksi Allowable stress, OU 12 Omax 4 ksi Factor of safety 3 Thus, Axial force in AC Area AC = Omax 2.404.P 4 0.125 P = 0.208kip To find maximum allowable force P if bearing stress stress in pin B is not to exceed its maximum.

Ultimate bearing stress for pin B, OB = 5ksi OB 5 Allowable bearing stress, OBmax = 1.667 ksi Factor of safety 3 Diameter of Pin, D in 4 a) Bearing between pin and Bar BCD Bearing Area, Perimeter of the pin Areabi · Thickness of bar BCD 2 TT-0.25 Areabi 0.131in? 2 Resultant at B, RB Thus, = ОВтах Aredbi 1.666P = : 1.667 0.131 P = 0.131kip (2) b) Bearing between pin and support Bearing Area, Area B2 Perimeter of the pin · Thickness of support.2 2 AredB2 = (72)2-(02) 92-u. 2 = 0.196in Resultant at B, RB Thus, = OBmax Areab2 1.666P = 1.667 0.196 P = 0.196kip (3) To find maximum allowable force P if shear stress in pin B is not to exceed its maximum. Ultimate shear stress, Tu = 7ksi Tu 7 Allowable shear stress, Tmax 2.333 ksi Factor of safety 3 T: Area of pin, 77.D 4 Areas 0.049 in? 4

a) Shear between pin and Bar BCD No. of shear surfaces, No = 2 Thus, Resultant at B, RB (Areab) (N) Tmax 1.666P 0.049.2 2.333 P = 0.137kip (4) b) Shear between pin and support No. of shear surfaces, Ns = 2 The number of shear surfaces is same as previous case. Hence P will be same. The allowable force P will be minmum of (1), (2), (3) and (4). P for (2) is minimum of all. Hence allowable force Pmax = 0.131kip The allowabe force P is a maximum. The relation between stress and force is given by Force Stress Area Area remaining constant if there is an increase in the Force, then the stress in the element will increase. The forces we found were by equating to maximum stress. Hence an increase in force will result in a stress more than the maximum for that element. In this case it will result in an increase in bearing stress in the pin B.

Answer 2

Cut the bar AC at and consider FBD of the lower section. O= Tan (9 56.31 ° MB = 0 (sum of moment of forces about pivot point B = 0) -P.(4 + 4) + NAC sin (O).4 = 0 NAC NAC = 2.404.P E Fx = 0) (sum of forces in x direction = 0) HB - NAC COS(O) = 0 B с HB - 2.404.P.cos(O) = 0 → HB F HB 1.333P VB 0) Fy = 0 (sum of forces in y direction VB+ NAC sin (O) - P = 0 4" 4" + X VB + 2.404.P.sin(O) - P = 0 VB = -P 2 The resultant force acting on pin B, RB HB' + VB V (1.333p) + (-p)? RB = 1.666P To find maximum allowable force P if normal stress in AC not to exceed its maximum. Consider BAR AC: Axial force, NAC = 2.404.P 11 Area of bar AC, Area AC 24 = = 0.125 in? Ultimate normal stress, OU = 12 ksi Allowable stress, OU 12 Omax 4 ksi Factor of safety 3 Thus, Axial force in AC Area AC = Omax 2.404.P 4 0.125 P = 0.208kip To find maximum allowable force P if bearing stress stress in pin B is not to exceed its maximum.

Ultimate bearing stress for pin B, OB = 5ksi OB 5 Allowable bearing stress, OBmax = 1.667 ksi Factor of safety 3 Diameter of Pin, D in 4 a) Bearing between pin and Bar BCD Bearing Area, Perimeter of the pin Areabi · Thickness of bar BCD 2 TT-0.25 Areabi 0.131in? 2 Resultant at B, RB Thus, = ОВтах Aredbi 1.666P = : 1.667 0.131 P = 0.131kip (2) b) Bearing between pin and support Bearing Area, Area B2 Perimeter of the pin · Thickness of support.2 2 AredB2 = (72)2-(02) 92-u. 2 = 0.196in Resultant at B, RB Thus, = OBmax Areab2 1.666P = 1.667 0.196 P = 0.196kip (3) To find maximum allowable force P if shear stress in pin B is not to exceed its maximum. Ultimate shear stress, Tu = 7ksi Tu 7 Allowable shear stress, Tmax 2.333 ksi Factor of safety 3 T: Area of pin, 77.D 4 Areas 0.049 in? 4

a) Shear between pin and Bar BCD No. of shear surfaces, No = 2 Thus, Resultant at B, RB (Areab) (N) Tmax 1.666P 0.049.2 2.333 P = 0.137kip (4) b) Shear between pin and support No. of shear surfaces, Ns = 2 The number of shear surfaces is same as previous case. Hence P will be same. The allowable force P will be minmum of (1), (2), (3) and (4). P for (2) is minimum of all. Hence allowable force Pmax = 0.131kip The allowabe force P is a maximum. The relation between stress and force is given by Force Stress Area Area remaining constant if there is an increase in the Force, then the stress in the element will increase. The forces we found were by equating to maximum stress. Hence an increase in force will result in a stress more than the maximum for that element. In this case it will result in an increase in bearing stress in the pin B.

Answer 3

Cut the bar AC at and consider FBD of the lower section. O= Tan (9 56.31 ° MB = 0 (sum of moment of forces about pivot point B = 0) -P.(4 + 4) + NAC sin (O).4 = 0 NAC NAC = 2.404.P E Fx = 0) (sum of forces in x direction = 0) HB - NAC COS(O) = 0 B с HB - 2.404.P.cos(O) = 0 → HB F HB 1.333P VB 0) Fy = 0 (sum of forces in y direction VB+ NAC sin (O) - P = 0 4" 4" + X VB + 2.404.P.sin(O) - P = 0 VB = -P 2 The resultant force acting on pin B, RB HB' + VB V (1.333p) + (-p)? RB = 1.666P To find maximum allowable force P if normal stress in AC not to exceed its maximum. Consider BAR AC: Axial force, NAC = 2.404.P 11 Area of bar AC, Area AC 24 = = 0.125 in? Ultimate normal stress, OU = 12 ksi Allowable stress, OU 12 Omax 4 ksi Factor of safety 3 Thus, Axial force in AC Area AC = Omax 2.404.P 4 0.125 P = 0.208kip To find maximum allowable force P if bearing stress stress in pin B is not to exceed its maximum.

Ultimate bearing stress for pin B, OB = 5ksi OB 5 Allowable bearing stress, OBmax = 1.667 ksi Factor of safety 3 Diameter of Pin, D in 4 a) Bearing between pin and Bar BCD Bearing Area, Perimeter of the pin Areabi · Thickness of bar BCD 2 TT-0.25 Areabi 0.131in? 2 Resultant at B, RB Thus, = ОВтах Aredbi 1.666P = : 1.667 0.131 P = 0.131kip (2) b) Bearing between pin and support Bearing Area, Area B2 Perimeter of the pin · Thickness of support.2 2 AredB2 = (72)2-(02) 92-u. 2 = 0.196in Resultant at B, RB Thus, = OBmax Areab2 1.666P = 1.667 0.196 P = 0.196kip (3) To find maximum allowable force P if shear stress in pin B is not to exceed its maximum. Ultimate shear stress, Tu = 7ksi Tu 7 Allowable shear stress, Tmax 2.333 ksi Factor of safety 3 T: Area of pin, 77.D 4 Areas 0.049 in? 4

Answer 4

Cut the bar AC at and consider FBD of the lower section. O= Tan (9 56.31 ° MB = 0 (sum of moment of forces about pivot point B = 0) -P.(4 + 4) + NAC sin (O).4 = 0 NAC NAC = 2.404.P E Fx = 0) (sum of forces in x direction = 0) HB - NAC COS(O) = 0 B с HB - 2.404.P.cos(O) = 0 → HB F HB 1.333P VB 0) Fy = 0 (sum of forces in y direction VB+ NAC sin (O) - P = 0 4" 4" + X VB + 2.404.P.sin(O) - P = 0 VB = -P 2 The resultant force acting on pin B, RB HB' + VB V (1.333p) + (-p)? RB = 1.666P To find maximum allowable force P if normal stress in AC not to exceed its maximum. Consider BAR AC: Axial force, NAC = 2.404.P 11 Area of bar AC, Area AC 24 = = 0.125 in? Ultimate normal stress, OU = 12 ksi Allowable stress, OU 12 Omax 4 ksi Factor of safety 3 Thus, Axial force in AC Area AC = Omax 2.404.P 4 0.125 P = 0.208kip To find maximum allowable force P if bearing stress stress in pin B is not to exceed its maximum.

Ultimate bearing stress for pin B, OB = 5ksi OB 5 Allowable bearing stress, OBmax = 1.667 ksi Factor of safety 3 Diameter of Pin, D in 4 a) Bearing between pin and Bar BCD Bearing Area, Perimeter of the pin Areabi · Thickness of bar BCD 2 TT-0.25 Areabi 0.131in? 2 Resultant at B, RB Thus, = ОВтах Aredbi 1.666P = : 1.667 0.131 P = 0.131kip (2) b) Bearing between pin and support Bearing Area, Area B2 Perimeter of the pin · Thickness of support.2 2 AredB2 = (72)2-(02) 92-u. 2 = 0.196in Resultant at B, RB Thus, = OBmax Areab2 1.666P = 1.667 0.196 P = 0.196kip (3) To find maximum allowable force P if shear stress in pin B is not to exceed its maximum. Ultimate shear stress, Tu = 7ksi Tu 7 Allowable shear stress, Tmax 2.333 ksi Factor of safety 3 T: Area of pin, 77.D 4 Areas 0.049 in? 4

a) Shear between pin and Bar BCD No. of shear surfaces, No = 2 Thus, Resultant at B, RB (Areab) (N) Tmax 1.666P 0.049.2 2.333 P = 0.137kip (4) b) Shear between pin and support No. of shear surfaces, Ns = 2 The number of shear surfaces is same as previous case. Hence P will be same. The allowable force P will be minmum of (1), (2), (3) and (4). P for (2) is minimum of all. Hence allowable force Pmax = 0.131kip The allowabe force P is a maximum. The relation between stress and force is given by Force Stress Area Area remaining constant if there is an increase in the Force, then the stress in the element will increase. The forces we found were by equating to maximum stress. Hence an increase in force will result in a stress more than the maximum for that element. In this case it will result in an increase in bearing stress in the pin B.