### 101. Prove that the sum of the normal strains in

2 sin 60°d(10-6) e y A-36 steel shaft in the manner shown. When the shaft is rotating with an angular velocity of , the reading on the strain gauge is Determine The state of strain at the point has components of and Use the strain-transformation equations to determine the 97. B AB. - Auburn UniversitySolve the problem using the stress transformation equations. Show the result on the (45 - (-60))>2 30 =-1.75 s avg = s x + s y 2 = 45 + (-60) 2 =-7.50 MPa t max in-plane = Aa s x - s y 2 b 2 + t xy 2 = Aa 45 -( 60) 2 b 2 A point on a thin plate is subjected to the two

### Analysis of Statically Determinate Trusses

Common Types of Trusses Bridge Trusses In particular, the Pratt, Howe,and Warren trusses are normally used for spans up to 61 m in length. The most common form is the Warren truss with verticals. For larger spans, a truss with a polygonal upper cord, such as the Parker truss, is CHAPTER 1 Analysis of Stress - Pearson Higher Edcated at point . Q. on the cut surface is acted on by force Let the origin of coor-dinates be placed at point . Q, with . x. normal and . y, z. tangent to In general, does not lie along . x, y, or . z. Decomposing into components parallel to . x, y, and. z (Fig.1.1c),we define the . normal stress. and the . shearing stresses. and (1.2) Chap. 6 Structural Analysisby taking an imaginary cut (shown here as a-a) through the truss. Since truss members are subjected to only tensile or compressive forces along their length, the internal forces . at the cut member will also be either tensile or compressive with the same magnitude. This result is based on the equilibrium principle . and Newtons third law.

### Chapter 1 Tension, Compression, and Shear

A = C (d2 2 - d 1 2) = C (1302 - 902) = 6,912 mm2 4 4 P 240,000 N " = C = CCCCC = 34.7 MPa (comp.) A 6,912 mm2 the compressive strain is 0.55 mm = C = CCCC = 550 x 10-6 = 550 m/m L 1,000 mm Example 1-2 a circular steel rod of length L and diameter d hangs and holds a weight W at its lower end Chapter 6:Analysis of Structures6.2 -6.3 Trusses Trusses are used commonly in Steel buildings and bridges. All straight members connected together with pin joints connected only at the ends of the members and all external forces (loads & reactions) must be applied only at the joints . Definition:A truss is a structure that consists of Every member of a truss is a 2 force member. HW 17 SOLUTIONS - University of Utah2 kip C 2 kip 6(100) 80 in. 5 kip S kip 150 in. 2(150) 5MD 5 kip B -8(80) AE in. Ans The posiüve sign indicates that end A moves away from end D. 41. The ship is pushed through the water using an A-36 steel propeller shaft that is 8 m long, measured from the

### HW 21 SOLUTIONS - University of Utah

60(2) = psi = 300 psi 2(0.2) Ans Ans 86. The open-ended polyvinyl chloride pipe has an inner diameter of 4 in. and thickness of 0.2 in. If it carries flowing water at 60 psi pressure, determine the state of stress in the walls of the pipe. pr = 60<2) = 600 psi 0.2 A ns Ans is ME 230 Kinematics and Dynamics - University of point O, the bodys center of gravity G moves in a circular path of radius r G. Thus, the acceleration of point G can be represented by a tangential component (a G) t = r G and a normal component (a G) n = r G 2. Equations of motion for pure rotation (17.4) W. Wang 9 ME 303 MANUFACTURING ENGINEERINGQ2.2 A steel tensile specimen with an initial diameter of 20mm and gage length of 120mm is subjected to a load of 125 kN and a gage length of 135mm is observed. Assuming uniform deformation at this point, calculate the followings:a) calculate the true stress, strain and the instantaneous diameter

### Question 1:A steel machine part is statically loaded and

2 60.21 320 n = => n = 2.66 n = => n = 4.72 Question 2:A steel LPG tank is shown in the figure. The wall thickness of the tank is 15 mm and has a yield strength of 340 MPa. The full weight of the tank is 6500 kg and the internal pressure is 3 MPa. Calculate the factor of safety of the tank according to the distortion energy theory Solved:P.4.1 (40 Points). A Material Has A Fatigue Limit If the safe range of stress is determined by the Goodman prediction calculate its value. P.4.2 (60 points). A steel component is subjected to a reversed cyclic loading of 100 cycles/day over a period of time in which 160 N/mm2 is applied for 200 cycles, 140 N/mm2 is applied for 200 cycles and +100 N/mm2 is applied for 600 cycles.