1. FASTENERS FASTENERS - BOLTED CONNECTIONS A set of n bolts is to be used to provide a clamping force of F between two components. The load is shared equally among the bolts. Specify suitable bolts, including the grade of the material, if each is to be stressed to K % of its proof strength. The variable K is called the demand factor. The load on each screw is to be F P = n Specify a bolt made from a SAE grade steel, having a proof strength [ σ] MPa. Then the allowable stress is σa = K [ σ] The required tensile stress area for the bolt is then At = P σa From a table find the required tensile stress area for the thread. The required tightening torque will be T = K D P where D - nominal outside diameter of threads P - clamping load K- constant dependent on the lubrication present 1.1 Constant Dependent On The Lubrication Present For average commercial conditions, use k1 = 0.15 if any lubrication at all is present. Even cutting fluids or other residual deposits on the threads will produce conditions consistent with k1 = 0.15. If the threads are well cleaned and dried, k1 = 0.20 is better. Of course, these values are approximate, and variations among seemingly identical assemblies should be expected. Testing and statistical analysis of the results are recommended. -1-
MDESIGN SOFTWARE - Example Fasteners Input Data: Clamping force F = 10000 Ν Number of bolts n = 4 Demand factor k = 50 % Constant dependent on the lubrication present k1 = 0.15 Results SAE steel grade = 2 Proof strength [σ] = 379.2 MPa Allowable stress σa = 189.606 MPa Required tensile stress area At = 13.18 mm² Thread type 12-24 UNC Basic major diameter D = 5.48 mm Tensile stress area At(table) = 15.6 mm² Required tightening torque T = 2057.7 N.mm -2-
2. BOLTED CONNECTIONS The basic approach to the analysis and design of eccentrically loaded joints is to determine the forces that act on each bolt because of all the applied loads. Then, by a process of superposition, the loads are combined vectorially to determine which bolt carries the greatest load. The following equations are used: Determination of the direct shear force on the bolt pattern and on each individual bolt, assuming that all bolts share the shear load equally: F S = P / N Computation of the force on each bolt required to resist the bending moment from the relation Fi = Mri 2 r where r i = radial distance from the centroid of the bolt pattern to the ith bolt F i = force on the ith bolt due to the moment. The force acts perpendicular to the radius. A vector summation of forces acting on each bolt can be performed either analytically or graphically, or each force can be resolved into horizontal and vertical components. The components can be summed and the resultant can be computed. Let s use the latter approach for this problem. The x- and y-components of Fi are Fix = Fi sin θ Fiy = Fi cos θ The total force in the x- (y-) direction is then F +, or F iy + Fsy ix F sx Then the resultant force on bolt I is ( F + F ) + ( F F ) 2 R i = ix sx 2 iy + sy The required area for the bolt is R A i s = τa The required diameter of the bolt would be 4A D = s π This module determines the minimum required diameter of the bolts in a connection subjected to direct shear and shear due to a moment applied to the member. See the graphic aid Geometry of bolted joint. -3-
This module facilitates the design of a bolted connection comprised of any array of bolts for which the distance from the centroid of the array to any individual bolt is the same. Examples are a rectangular array of four bolts or a circular array of any number of bolts. The analysis considers only a single force to be applied to the connection. If more than one force is applied, the resultant of all applied forces must be determined by the user for input to the program. Both the magnitude and the orientation of the line of action of the force must be known. The program assumes that the bolts are subjected to single shear. If they are in double shear, the applied force should be divided by 2.0. 2.1 Allowable stresses for bolts ASTM grade Allowable shear stress Allowable tensile stress A307 69 MPa 138 MPa A325 and A449 121 MPa 303 MPa A490 52 MPa 372 MPa 2.2 Distance to the Centroid, a The user must determine the perpendicular distance, a, from the centroid of the bolt array to the line of action of the applied force. If the line of action of the applied force acts through the centroid, a = 0 and the bolts carry only direct shear. 2.3 Angle of inclination, α The user must determine the angle of inclination, α, of the line of action of the applied force relative to a perpendicular to the centerline through the array. -4-
MDESIGN SOFTWARE - Example bolted connection Input data: Bolted Connections Bolt material type = A307 Allowable shear stress for bolt σa = 68 MPa Shear load P = 15564 N Number of bolts N = 4 Distance to the centroid a = 304.8 mm Radial distance for bolt(s) r = 127 mm x-distance from bolt to centroid Δx = 101.6 mm y-distance from bolt to centroid Δy = 76.2 mm Angle of inclination α = 36 Results Load per bolt in x-direction Fsx = 2286.4 Ν Load per bolt in y-direction Fsy = 3147 Ν Moment to be resisted M = 4.74 10 6 Ν mm Force required to resist the bending moment Fi = 9335.8 N Total force in x-direction Ftx = 7887.9 N Total force in y-direction Fty = 10615.7 N Required area for the bolt As = 194.4 mm 2 Required diameter Dr = 15.7 mm Resultant force on bolt Rs = 13225.5 N Nearest standart bolt diameter D = 15.8 mm -5-