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problem 2 determine the tensile capacity of

2. Design of Welded Connections - American Welding

2.2.2 Joint Welding Sequence. Drawings of those joints or groups of joints in which it is especially impor-tant that the welding sequence and technique be carefully controlled to minimize shrinkage stresses and distortion shall be so noted. 2.2.3 Weld Size and Length. Contract design drawings shall specify the effective weld length and, for partial


Version II 5-2 The tension members can have a variety of cross sections. The single angle and double angle sections [Fig 2(a)] are used in light roof trusses as in industrial buildings. The tension members in bridge trusses are made of channels or I sections, acting individually or built-up [Figs. 2(c) and 2(d)]. Assignment 5 solutions(c) Determine the yield strength at a strain offset of 0.002. (d) Determine the tensile strength of this alloy. (e) What is the approximate ductility, in percent elongation? (f) Compute the modulus of resilience. Solution This problem calls for us to make a stress-strain plot for aluminum, given its tensile


tensile strength of the bolt in kgf/mm2 and the second number, y, indicates one-tenth of the ratio of the yield stress to ultimate stress, eed as a percentage. Thus, for example, grade 4.6 bolt will have a minimum ultimate strength 40 kgf/mm2 (392 Mpa) and minimum yield strength of 0.6 times 40, which is 24 kgf/mm2 (235 Mpa). Basic Concepts , Rectangular and T BeamsFIGURE 1.3. Transformed section for flexure just prior to cracking. 1.4.2 Cracked, Linear Stage . When the moment is increased beyond M cr, the tensile stresses in concrete at the tension zone increased until they were greater than the modulus of rupture f ctr, and cracks will develop.The neutral axis shifts upward, and cracks extend close to the level of the shifted neutral axis.

Chap. 6 Structural Analysis -

2. Draw the free-body diagram of a joint with one or two unknowns. Assume that all unknown member forces . act in tension (pulling the pin) unless you can determine by inspection that the forces are compression loads. 3. Apply the scalar equations of equilibrium, F. X = 0 and F. Y = 0, to determine the unknown(s). If the answer is positive Chapter 2 Stress and Strain- Axial Loading2-24 Example A brass bolt with a diameter of 0.375" is fitted inside a 7/8" diameter steel tube with a wall thickness of 1/8". After the nut has been snugged, it is tightened 1/4 turn. The bolt is single threaded and has a pitch of 0.1". Determine the normal stress in the bolt and the tube. E (brass)= 15,000 ksi and E (steel)= 29,000 ksi.

Chapter 5 Stresses In Beams -

Substituting the eion for 1/from Eq.(5.2) into Eq. (5.1), we get the flexure formula:(5.3) Note that a positive bending moment M causes negative (compressive) stress above the neutral axis and positive ( tensile) stress below the neutral axis I My = [ ] I M max c max = Chapter 7 SIMPLE CONNECTIONSThe load capacity of connection = fastener capacity * no. of fastener or The load capacity of connection = weld capacity per inch * length of weld. Fu = specified tensile strength of the connected edge of the adjacent hole or edge of the material (in.). t = thickness of connected material Example 7.1 Calculate and check the design


Problem 3.1 Department of Mechanical Engineering 11 An eye bolt is to be used for lifting a load of 10 kN. The eye bolt is screwed into the frame of the motor. The eye bolt has coarse threads. It is made of plain carbon steel 30C8 (S yt = 400N/mm2) and factor of safety is 6. Determine the size of the bolt Problem 3.2 Two plates are fastened by Design of Structures:LESSON 9. Design of Tension MemberArea of 2 rivet holes 2 x (16 + 1.5) x 8 = 280 mm2. Net area of the member 1712 280 = 1,432 mm2. Therefore safe tension for the member = 150 x 1432 = 2,14,800 N = 214.8 kN. Example 9.4:The tension member of a roof truss carries a maximum axial tension of 250 kN. Design the section.

Reinforced Concrete Design

centroid of the tensile steel d c == shear force capacity in concrete V s = shear force capacity in steel shear stirrups V u = shear at a distance of d away from repeat from 2. until a found from step 3 matches a used in step 2. Design Chart Method:1. calculate bd2 M R n n SOLUTION (17.3) Known:A simply supported steel shaft 17-20 SOLUTION (17.10) Known:The dimensions of a steel shaft are given. Find:Determine the critical speed of rotation for the steel shaft. Schematic and Given Data:25 mm dia. 50 kg 600 mm 600 mm Assumptions:1. Bearing friction is negligible. 2. The bearings supporting the

SOLUTION (6.19) Known:A machine frame made of steel

Find:Determine the tensile strength a ductile material must have in order to provide a safety factor of 2 with respect to initial yielding at the locations investigated in the above listed problems. Determine the answer using both the maximum-shear-stress theory and the maximum-distortion-energy theory. Assumption:The materials are homogeneous. STRIPPING STRENGTH OF TAPPED HOLES - HobsonExample 2. Calculate the length of engagement for the above conditions if only 140,000 psi is to be applied. (This is the same as using a bolt with a maximum tensile strength of 140,000psi.) From E506 obtain value of 1.06D Minimum length of engagement = (0.500) (1.06) = 0.530. Example 3. Suppose in Example 1 that minimum

Strength of Materials Basics and Equations Mechanics of

The constant, E, is the modulus of elasticity, Young's modulus or the tensile modulus and is the material's stiffness. Young's modulus is in terms of 10 6 psi or 10 3 kg/mm 2. If a material obeys Hooke's Law it is elastic. The modulus is insensitive to a material's temper. Normal force is directly dependent upon the elastic modulus. TENSION CAPACITY OF BOLTED SINGLE ANGLE SHEAR capacity of the connection is found to be nearly 90 kips. The results show that the tension capacity of the single angle connection improves very slightly (~6%) when a lateral restraint is present. Two primary failure modes are identified:1) shear fracture of the angle and 2) tensile failure of the bolts.


Determine the design strength of the splice between two W-shapes. Solution:W14 X43 A992 (Fy = 50 ksi) Ag = 12.6 sq. in t f = 0.53 in 1. For the limit state of yielding, Design tensile strength, t P n = t F y A g = (0.9) (50)(12.6) = 567 kips 2. Net Area Area to be deducted from each flange = 2 TRUSS GUSSET PLATES11 2 1 50.0% 10 6 1 16.7% 9 94 12 12.8% 8 75 11 14.7% 7 70 26 37.1% 6 28 3 10.7% 5 44 8 18.2% 4 87 47 54.0% 3 41 7 17.1% 2 50 8 16.0% 1 108 21 19.4% Population Analysis Analysis Region Truss Requiring % Requiring

Tensile Strength Formula Ultimate Tensile Strength Formula

Ultimate tensile strength is shortened to tensile strength or ultimate strength. It is defined as the capacity of a material to break under tension. The tensile strength is measured by the maximum stress that a material can withstand while being stretched or pulled before breaking. Tensile Stress Area of Bolt Calculatorgetcalc's Tensile Stress Area of Bolt Calculator is an online mechanical engineering tool to calculate the tensile (critical) stress area or the mimimum area of threaded section of the bolt, in both US customary & metric (SI) units.

5. Flexural Analysis and Design of Beams 5.1. Reading

5.9. Example. Calculate Nominal Moment Capacity of a Beam f c=4,000 psi Determine the nominal moment Mn at which the beam given below will fail. Given f y = 60,000 psi Solution à = A s bd = 2.35 10×23 = 0.0102 M n = à f y bd2 1 0.59 à f y f c M n = (0.0102) ×(60 ksi) ×(10 in)×(23 in)2 × 1 0.59 ×(0.0102) ×60 4 = 2,950,000

Other steel

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