STRESS! Stress Evisdom! verage Normal Stress in an xially Loaded ar! verage Shear Stress! llowable Stress! Design of Simple onnections 1
Equilibrium of a Deformable ody ody Force w F R x w(s). D s y Support Reaction D y
Table 1 Supports for oplanar Structures Type of onnection Idealized Symbol Reaction Number of Unknowns (1) θ Light cable θ θ One unknown. The reaction is a force that acts in the direction of the cable or link. () Rollers and rockers F One unknown. The reaction is a force that acts perpendicular to the surface at the point of contact. (3) Smooth surface F One unknown. The reaction is a force that acts perpendicular to the surface at the point of contact. 3
Type of onnection Idealized Symbol Reaction Number of Unknowns (4) Smooth pin or hinge F y F x Two unknowns. The reactions are two force components. (5) F x F y Two unknowns. The reactions are a force and moment. Internal pin (6) Fixed support F x M F y Three unknowns. The reactions are the moment and the two force components. 4
Equation of Equilibrium Σ F 0 Σ M O 0 Σ F x 0 Σ F y 0 Σ F z 0 Σ M x 0 Σ M y 0 Σ M z 0 5
Internal Resultant Loadings F section F 3 F 1 F 4 y F N M O x M O N V F 3 F 1 V F 4 6
Example 1 Determine the resultant internal loadings acting on the cross section at of the beam shown. 70 N/m 3 m 6 m 7
Support Reactions 70 N/m (1/)(9)(70) 115 N M 3645 N m 3 m 6 m 540 N Internal Loading y 115 N 180 N/m 540 N V 135 N 70 N/m 90 N/m M 3645 N m 3 m y 115 N -540() -1080 N m M 540 N 0 180 N/m 0 540 N 1080 N m N m 4 m 8
Shear and bending moment diagram (1/)(9)(70) 115 N 70 N/m 3645 N m 115 N V (N) 3 m 6 m 115 540 + M (N m) - 3645 1080 b Side View 540 N 1080 N m flanges t w web 540 N Front View h 9
Example Determine the resultant internal loadings acting on the cross section at of the machine shaft shown. The shaft is supported by bearings at and, which exert only vertical forces on the shaft. 00 mm 5 N 800 N/m D 100 mm 100 mm 50 mm 50 mm 10
Support Reactions (800 N/m)(0.15 m) 10 N/m 5 N x D 0.75 m 0.15 0.1 m y y + Σ M 0: -(10)(0.75) + y (0.4) - (5)(0.5) 0, y 363.75 N + Σ F x 0: x 0 x 0, + Σ F y 0: y - 10 +363.75-5 0 y -18.75 N, Internal Loading 800(0.05) 40 N + ΣF x 0: N 0 0. m 18.75 N 0.05 m M V N + ΣF y 0: + ΣM 0: -18.75-40 - V 0, V -58.75 N 18.75(0.5) + 40(0.05) + M 0 M -5.69 N m 11
Internal Loading 800(0.05) 40 N 0. m M -5.69 N m N 0 V -58.75 N 18.75 N 0.05 m 800(0.05) 40 N 0. m 5.69 N m 58.75 N 0 18.75 N 0.05 m 1
Example 3 The hoist consists of the beam and attached pulleys, the cable, and the motor. Determine the resultant internal loadings acting on the cross section at if the motor is lifting the kn load W with constant velocity. Neglect the weight of the pulleys and beam. 0.15 m 1 m m 0.5 m 0.15 m D W 13
0.15 m 1 m m 0.5 m 0.15 m D W W kn 0.15 m 1 m T kn N M V + + Σ F x 0: Σ F y 0: + N 0 N - kn - - V 0, V - kn W kn + Σ M 0: M - (0.15) + (1.15) 0 M - kn m 14
Example 4 Determine the resultant internal loadings acting on the cross section at G of the wooden beam shown. ssume the joints at,,, D, and E are pin-connected. 6.5 kn G D E 1 m 4 kn/m 0.6 m 0.6 m m 15
Support Reactions F Two - force 6.5 kn G D E E y E x 1 m 4 kn/m (1/)()(4) 4 kn 0.6 m 0.6 m /3 m (/3)() 4/3 m + Σ M E 0: 4(4/3) + 6.5(3.) - F (1) 0, F 6.1 kn + Σ F x 0: 6.13 + E x 0 E x -6.1 kn, + Σ F y 0: -6.5-4 + E y 0 E y 10.5 kn 16
Internal loadings acting on the cross section at G F 6.13 kn 34.0 kn 6.5 kn G 0.6 m Member G F V F 39.81 o 50.19 o 34.0 kn M N F D -1.78 kn + ΣM G 0: M G -34.0 sin39.81 o (0.6) + 6.5(0.6) 0 M G 9.17 kn m + + Joint ΣF x 0: ΣF y 0: -F sin50.19 o + 6.13 0 F 34.0 kn, (T) -F cos50.19 o -F D 0 F D -1.8 kn, () + ΣF x 0: 34.0 cos39.81 o + N G 0 N G -6.1 kn + ΣF y 0: -6.5 + 34.0 sin39.81 o - V G 0 V G 15.3 kn 17
Example 5 Determine the resultant internal loadings acting on the cross section at of the pipe shown. The pipe has a mass of kg/m and is subjected to both a vertical force of 50 N and a couple moment of 70 N m at its end. It is fixed to the wall at. z x 50 N 70 N m 0.75 m 0.5 m D y 1.5 m 18
Free-ody Diagram z (M ) z (M ) y (M ) x x 50 N 70 N m (F ) x (F ) y 0.65 m (F ) z W D 9.81 N 0.5 m W D D 0.65 m y W D ( kg/m)(0.5 m)(9.81 N/kg) 9.81 N W D ( kg/m)(1.5 m)(9.81 N/kg) 4.5 N Equilibrium of Equilibrium Σ F x 0: (F ) x 0 Σ F y 0: (F ) y 0 Σ F z 0: (F ) z - 9.81-4.55-50 0, (F ) z 84.3 N Σ (M ) x 0: (M ) x + 70-50(0.5) - 4.55(0.5) - 9.81(0.5) 0, (M ) x -30.3 N m Σ (M ) y 0: (M ) y + 4.55(0.65) + 50(1.5) 0, (M ) y -77.8 N m Σ (M ) x 0: (M ) z 0 19
Free-ody Diagram (F ) x 0 (F ) y 0 (F ) z 84.3 N (M ) x -30.3 N m (M ) y -77.8 N m (M ) z 0 z 77.8 N m 84.3 N 9.81 N 0.5 m 30.3 N m 4.5 N D x 50 N 0.65 m y 70 N m 0.65 m 0
Stress z +z Face +y Face y x +x Face 1
z σ z y compatibility, τ zx τ zy σ' x -x Face σ x σ' x σ y σ' y τ' xy τ yz σ z σ z -y Face x σ' y τ' yz τ' yx τ xz σ x τ' zy -z Face τ' xz τ xy τ' zx σ' z σ' y x z τ yx τ' yx x z σ y y τ' xy y z Top View τ xy τ' xy τ yx τ' yx τ zx τ xz σ' x y z τ yx x z y σ y x z x σ x y z τ xy y z
σ' y x z τ' O yx x z τ' xy y z O σ' x y z τ yx x z y σ y x z x σ x y z τ xy y z + ΣF y 0: - σ' y x z + σ y x z 0 σ' y σ y + ΣF x 0: σ x y z - σ' x y z 0 σ x σ' x + ΣM O 0: (τ xy y z)( x) - (τ yx x z)( y) 0 τ xy τ yx 3
z σ τ x y z τ zy x y zx x y z σ z x y z τ xz y z τ xy y z σ x y z τ yz x z σ y x z τ yx x z y x σ' y x z σ x y z σ' x y z σ y x z y x y x σ' z x y τ' yz x z z τ zy x y τ yz x z y τ' yx x z z τ' xy y z τ yx x z y τ xz y z z τ zx x y τ' xz y z y x τ' zy x y x τ xy y z x τ' zx x y 4
verage Normal Stress in an xially Loaded ar ssumptions The material must be - Homogeneous material - Isotropic material Internal force ross-sectional area External force 5
σ z F σ σ x y σ σ x y + F Rz ΣF ; z df σd σ σ 6
Example 6 The bar shown has a constant width of 35 mm and a thickness of 10 mm. Determine the maximum average normal stress in the bar when it is subjected to the loading shown. 1 kn 9 kn 4 kn D kn 35 mm 9 kn 4 kn 7
1 kn 9 kn 4 kn D 9 kn 4 kn 35 mm (kn) kn 1 30 x 10 mm 30 kn 35 mm σ σ σ 30 kn (0.035 m)(0.01 max m) 85.7 Ma 8
1 kn 9 kn 4 kn D 9 kn 4 kn 35 mm σ (Ma) kn 34.3 85.7 6.9 x 10 mm 30 kn 35 mm σ σ σ 30 kn (0.035 m)(0.01 max m) 85.7 Ma 9
Example 7 The 80 kg lamp is supported b two rods and as shown. If has a diameter of 10 mm and has a diameter of 8 mm, determine which rod is subjected to the greater average normal stress. 60 o 5 4 3 30
Internal Loading y F 60 o 5 4 3 F x verage Normal Stress d 10 mm 5 3 4 60 o d 8 mm ΣF + + ΣF y x 80(9.81) 784.8 N 4 0 ; F ( ) cos 60 o F 0 5 3 o 0 ; F ( ) + F sin 60 784.8 0 5 F 395. N, F 63.4 N σ σ F 395. N π (0.004 m) F 63.4 π (0.005 N m) 7.86 8.05 Ma Ma The average normal stress distribution acting over a cross section of rod. 8.05 Ma 8.05 Ma 8.05 Ma 63.4 N 31
Example 8 Member shown is subjected to a vertical force of 3 kn. Determine the position x of this force so that the average compressive stress at is equal to the average tensile stress in the tie rod. The rod has a cross-sectional area of 400 mm and the contact area at is 650 mm. x 3 kn 00 mm 3
F x 3 kn 00 mm Internal Loading + ΣFy 0; F + F 3 0 (1) F + Σ M 0: -3(x) + F (00) 0 ------() verage Normal Stress σ F F 400 650 F 1.65F (3) Substituting (3) into (1), solving for F, the solving for F, we obtain F 1.143 kn F 1.857 kn The position of the applied load is determined from (); x 14 mm 33
onnections Simple Shear D τ avg V V V V τ avg 34
Single Shear Top View Side View Top View Side View V V 35
Double Shear / / / / Top View Top View Side View Side View V / V / V / V / 36
Normal Stress: ompression and Tension Load Tension Load a b a b d t b b bt t σ @ a a bt (b-d)t d σ @ b b ( b d) t Section a-a b t Section b-b 37
earing Stress b t d b b / / t d earing Stress bearing dt σ bearing dt 38
σ normal ( b d) t t θ dθ d b σ bearing 90 + ΣF x 0: d σ b( ) t cosθdθ 0 90 dt d 90 ( ) tσ b sinθdθ 90 σ bearing σ bearing td sin 90 td o * 39
xial Force Diagram for ompression Load on late b d t Tension - - ompression Normal Stress: σ, ( bd) t earing Stress σ bearing dt compression 40
xial Force Diagram for Tension Load on late a b d t Tension ompression Normal Stress: max + σ, ( b d) t earing Stress σ bearing dt Shearing Stress τ at tension t a τ / τ / 41
Shearing Stress on pin Single Shear τ 4
Double Shear / / / / / / / / / / / / / / τ τ / / 43
ure Shear z Section plane τ zy z τ yz y x y τ yzτzy x ure shear τ zy τ yz 44
Example 9 The bar shown a square cross section for which the depth and thickness are 40 mm. If an axial force of 800 N is applied along the centroidal axis of the bar s cross-sectional area, determine the average normal stress and average shear stress acting on the material along (a) section plane a-a and (b) section plane b-b. 800 N a a b b 0 mm 30 o 0 mm 45
40 mm a b 0 mm 800 N a b 30 o 0 mm (a) section plane a-a a 800 N N 800 N a σ N (0.04 800 N m)(0.04 m) 500 ka τ avg 0 500 ka 46
40 mm a b 0 mm 800 N (b) section plane b-b a b 30 o 0 mm 800 N x 30 o V 800 cos30 o 69.8 N b b 30 o 30 o 800 N N 800sin30 o 400 N 40 mm o 40 sin 30 ; x x 80 mm σ State of Stress @ b-b b τ avg σ b 16.51 ka 15 ka τ avg N 400 N 15 ka (0.04 m)(0.08 m) V 69.8 N 16.51 ka (0.04 m)(0.08 m) 47
Example 10 The wooden strut shown is suspended form a 10 mm diameter steel rod, which is fastened to the wall. If the strut supports a vertical load of 5 kn, compute the average shear stress in the rod at the wall and along the two shaded planes of the strut, one of which is indicated as abcd. b a c d 0 mm 40 mm 5 kn 48
b a c d 0 mm 40 mm verage shear stress in the rod at the wall 5 kn V 5 kn d 10 mm τ avg V 5 kn π (0.005 m) 5 kn 63.7 Ma 5 kn 63.7 Ma verage shear stress along the two shaded plane V.5 kn b a 5 kn V.5 kn c d τ avg V.5 kn (0.04 m)(0.0 m) 3.1 Ma 5 kn 3.1 Ma 49
Example 11 The inclined member show is subjected to a compressive force of 3 kn. Determine the average compressive stress along the areas of contact defined by and, and the average shear stress along the horizontal plane defined by ED. 5 3 3 kn 4 5 mm 40 mm 80 mm 50 mm 50
The average compressive stress along the areas of contact defined by and : 3 kn 5 4 3 5 mm 5 3 3 kn 4 40 mm 80 mm V 1.8 kn 50 mm F 3(3/5) 1.8 kn F 3(4/5).4 kn 5 3 3 kn 4 σ F 1.8 kn (0.05 m)(0.040 m) 1800 ka 1.8 Ma 1.8 Ma σ F.4 kn (0.050 m)(0.040 m) 100 ka 1. Ma 1. Ma 51
The average shear stress along the horizontal plane defined by ED : 3 kn 5 4 3 5 3 3 kn 4 5 mm 40 mm 80 mm 50 mm F 3(3/5) 1.8 kn V 1.8 kn F 3(4/5).4 kn τ avg V 1.8 kn (0.04 m)(0.08 0.56 Ma m) 56.5 ka 1.8 kn 0.56 Ma 5
llowable Stress F. S fail allow F. S σ σ fail allow F. S τ τ fail allow 6. Design of Simple onnections σ allow V τ allow ross-sectional rea of a Tension Member a τ allow a σ allow 53
ross-sectional rea of a onnector Subjected to Shear V τ allow σ allow Required rea to Resist earing (σ b ) allow (σ ) b allow 54
Required rea to Resist Shear by xial Load Uniform shear stress (τ allow ) d l τ allow πd 55
Example 1a The control arm is subjected to the loading shown. (a) Determine the shear stress for the steel at all pin (b) Determine normal stress in rod, plate D and E. The thickness of plate D and E is 10 mm. φ 10 mm φ 5 mm D E φ 5 mm 00 mm D E 50 mm 50 mm 13 kn 5 kn pin φ 0 mm 13 kn 75 mm 50 mm 3 4 5 5 kn 56
F 00 mm 00 mm 13 kn 75 mm 50 mm 3 4 5 5 kn x y 13 kn 75 mm 50 mm 36.87 o 5 kn y Reaction + ΣM 0: F ( 0.) + 13(0.075) 5sin 36.87(0.15) 0 F 4.5 kn, + ΣF x 0: -4.5 - x + 5cos36.87 o 0 x 15.5 kn, x ΣF y 0: + y + 13-5sin36.87 o 0 (15.5) + () 15.63 y kn, kn 57
φ 35 mm (a) Shear stress 4.5 kn 00 mm 00 mm φ 5 mm φ 5 mm D E 5 3 pin φ 0 mm 13 kn 4 5 kn 75 mm 50 mm 15.63 kn 13 kn 75 mm 50 mm 5 kn in (Double shear) in D and E (Single shear) 15.63 kn 7.815 kn τ D VD 13 kn 6.48 Ma ( π / 4)(0.05 ) D τ 7.815 kn V 7.815 kn 4.88 Ma ( π / 4)(0.0) τ E VE 5 kn 50.93 Ma ( π / 4)(0.05 ) E 58
φ rod 10 mm (b) Normal stress 4.5 kn φ 5 mm D E φ 5 mm 00 mm D E 50 mm 50 mm 13 kn 5 kn 15.63 kn 13 kn 75 mm 50 mm 5 kn σ ale σ 4.5 kn 56.7 Ma ( π / 4)(0.010) late D 13 kn 13 kn σ D 13 kn t 0.01 m 0.05 m σ D 13 kn (0.05)( 0.01) 6 Ma 59
φ rod 35 mm 4.5 kn φ 5 mm D E φ 5 mm 00 mm D E 50 mm 50 mm 13 kn 5 kn 15.63 kn 13 kn 75 mm 50 mm 5 kn late E 5 kn 50 mm σ E 50 mm 5 kn 5 kn 0.05 m t 0.01 m 0.05 m σ E 5 kn (0.05 0.05)(0.01) 100 Ma 60
Example 1b The control arm is subjected to the loading shown. Determine the required diameter of the steel pin at if the allowable shear stress for the steel is τ allow 55 Ma. Note in the figure that the pin is subjected to double shear. 00 mm 13 kn 75 mm 50 mm 3 4 5 kn 61
F 00 mm 00 mm 13 kn 75 mm 50 mm 3 4 5 kn x y 13 kn 75 mm 50 mm kn y Internal Shear Force + ΣM 0: F ( 0.) 13(0.075) sin 36.87(0.15) 0 F 13.15 kn + ΣF x 0: -13.15 - x + cos36.87 o 0 x 4.47 kn x ΣF y 0: + y - 13 - sin36.87 o 0 (4.47) + (6.) y 6. kn 6.58 kn 6
13.15 kn 00 mm 00 mm 13 kn 75 mm 50 mm 3 4 5 kn 6.58 kn 13 kn 75 mm 50 mm kn Required rea 6.58 kn τ V allow 13.9 10 6 55 10 3 41.6 10 6 m 4 mm 13.9 kn 13.9 kn π d ( ) 4 mm d 17.55 mm Use d 0 mm 63
Example 13a The two members are pinned together at as shown below. Top views of the pin connections at and are also given. If the pins have an allowable shear stress of τ allow 86 Ma, the allowable tensile stress of rod is (σ t ) allow 11 Ma and the allowable bearing stress is (σ b ) allow 150 Ma, determine to the smallest diameter of pins and,the diameter of rod and the thickness of necessary to support the load. Top view 13 kn 5 4 3 1 m 0.6 m 64
Internal Force y 13 kn F x 36.87 o 1 m 0.6 m + Σ M 0: 13 (1) + F sin 36.87(1.6) 0 F 13.54 kn + Σ F y 0: y 13 + 13.54 sin 36.87 0 y 4.88 kn + Σ F x 0: x +13.54 cos 36.87 0 x 10.83 kn 65
Diameter of the ins 11.88 kn y 4.88 kn 13 kn F 13.54 kn 36.87 o x 10.83 kn Top view 1 m 0.6 m in 11.88 kn in F 13.54 kn V 5.94 kn V 5.94 kn V 13.54 kn V τ π 4 allow ( d ) 5.94 kn 86 10 kn/ m 69.07 mm 69.07 mm 3 d 9.38 mm, Use d 10 mm π ( 4 V τ d allow ) 13.54 kn 86 10 kn / m 157.4 mm 157.4 mm 3 d 14.16 mm, Use d 15 mm 66
Diameter of Rod 11.88 kn y 4.88 kn 13 kn 36.87 o F 13.54 kn x 10.83 kn 1 m 0.6 m 13.54 kn 13.54 kn ( σ t ) allow 11 13.54 kn 10 kn/ m 10.9 mm 3 π 4 ( d ) 10.9 mm d 1.4 mm, Use d 15 mm 67
Thickness 11.88 kn y 4.88 kn 13 kn F 13.54 kn 36.87 o x 10.83 kn 1 m 0.6 m φ 10 mm t 0 kn φ 15 mm ( σ ) 150 10 t b allow 6 11.88 10 (0.010 ) t 0.0079 m 3 11.88 kn 13.54 kn t ( σ b ) allow 150 10 6 13.54 10 (0.015) t 3 (σ b ) (σ b ) t 0.0060 m y comparison use t 8 mm 68
Example 13b The two members are pinned together at as shown below. Top views of the pin connections at and are also given. If the pins have an allowable shear stress of τ allow 86 Ma, the allowable tensile stress of rod is (σ t ) allow 11 Ma and the allowable bearing stress is (σ b ) allow 150 Ma, determine to the maximum load that the beam can supported. Top view φ 10 mm t 8 mm φ 15 mm 5 3 4 1 m 0.6 m φ 15 mm t 8 mm 69
Internal Force y F x 36.87 o 1 m 0.6 m + Σ M 0: ( 1) + F sin 36.87(1.6) 0 + ΣF 0 ; + 1.04 sin 36.87 0 y y F 1.04 (T) y 0.375 + ΣF x 0 ; + 1.04 cos 36.87 0 x x 0.834 kn 70
0.914 0.834 Top view in (double shear) 0.914 y 0.375 F 1.04 36.87 o 1 m 0.6 m φ 15 mm φ 10 mm in (single shear) F 1.04 V 0.457 τ allow V 0.457 V 3 0.457 86 10 ka ( π / 4)(0.01) 14.78 kn V 1.04 τ allow V 3 1.04 86 10 ka ( π / 4)(0.015 ) 14.58 kn 71
0.914 y 0.375 36.87 o F 1.04 0.834 1 m 0.6 m φ rod 15 mm Rod 1.04 1.04 σ allow 3 1.04 11 10 ka ( π / 4)(0.015 ) 19 kn 7
0.914 y 0.375 F 1.04 36.87 o 0.834 Top view φ 10 mm t 10 mm ( σ b ) allow 0.914 6 0.914 150 10 (0.010 )(0.010 ) 16.41 kn φ 10 mm t 10 mm 1 m 0.6 m 0 kn F 1.04 (σ b ) (σ b ) φ 15 mm t 10 mm ( σ ) 150 10 b allow 6 1.60 1.04 (0.015 )( 0.01) kn t 10 mm φ 15 mm y comparison all 14.58 kn 73
Example 14 The suspender rod is supported at its end by a fixed-connected circular disk as shown. If the rod passes through a 40-mm diameter hole, determine the minimum required diameter of the rod and the minimum thickness of the disk needed to support the 0 kn load. The allowable normal stress for the rod is σ allow 60 Ma, and the allowable shear stress for the disk is τ allow 35 Ma. 40 mm t 40 mm τ allow d 0 kn 0 kn 74
40 mm t 40 mm τ allow d 0 kn 0 kn Diameter of Rod Thickness of Disk σ allow 60 0 10 3 kn kn/ m τ V allow 35 0 10 3 kn kn/ m π 0 d 4 60 10 3 kn kn/ m π (0.0 m) t 35 0 10 3 kn kn/ m d 0.006 m 0.6 mm t 4.55x10-3 m 4.55 mm 75
Example 15 n axial load on the shaft shown is resisted by the collar at, which is attached to the shaft and located on the right side of the bearing at. Determine the largest value of for the two axial forces at E and F so that the stress in the collar does not exceed an allowable bearing stress at of (σ b ) allow 75 Ma and allowable shearing stress of the adhesive at of τ allow 100 Ma, and the average normal stress in the shaft does not exceed an allowable tensile stress of (σ t ) allow 55 Ma. F E 60 mm 0 mm 80 mm 76
F E 60 mm 0 mm 3 80 mm Load 3 osition xial Stress earing Stress 55 10 3 (σ t ) allow kn/ m 3 π (0.03 1 51.8 kn m) 3 3 (σ b ) allow 3 3 75 10 kn/ m [ π (0.04 m) π (0.03 m) ] 55 kn 60 mm 80 mm 77
F E 60 mm 0 mm 3 80 mm Load 100 Shearing Stress 60 mm 3 τ 10 allow 3 kn/ m shear 3 osition 80 mm 0 mm 3 [π (0.04 m)(. 00 )] xial Stress 1 51.8 kn earing Stress 55 kn Shearing Stress 3 55 kn 3 55 kn The largest load that can be applied to the shaft is 51.8 kn. 78
Example 16 The rigid bar shown supported by a steel rod having a diameter of 0 mm and an aluminum block having a cross-sectional area of 1800 mm. The18-mmdiameter pins at and are subjected to single shear. If the failure stress for the steel and aluminum is (σ st ) fail 680 Mpa and (σ al ) fail 70 Ma, respectively, and the failure shear stress for each pin is τ fail 900 Ma, determine the largest load that can be applied to the bar. pply a factor of safety of F.S.0. Steel 0.75 m luminum m 79
Steel 0.75 m F luminum 0.75 m 1.5 m m F + ΣM 0: (1.5 m) - F ( m) 0 ---------> F 0.65 + ΣM 0: (0.75 m) - F ( m) 0 ----------> F 0.375 Rod ( σ st ) allow ( σ st ) F. S fail F 3 680 10 ka 0.65 π(0.01 1 171 kn m) lock 70 ( σ al ) allow ( σ al ) F. S fail 3 10 ka 0.375 1800 10 168 kn 6 F m 80
Steel 0.75 m F 0.65 luminum 0.75 m 1.5 m m F 0.375 in or τ allow τ fail F. S F pin 3 900 10 ka 0.65 π(0.009 m) 3 183 kn Summary Rod 1 171 kn lock 168 kn in or 3 183 kn Largest load 168 kn 81