A steel rod of 15- mm diameter is held snugly (but without any initial stresses) between rigid walls by the arrangement shown in the figure part (a). (For the steel rod, use α = 12 x 10-6/˚C and E = 200 GPa.) Calculate the temperature drop ΔT (degrees Celsius) at which the average shear stress in the 12 – mm diameter bolt becomes 45 MPa. Also, what

A solid brass shaft is loaded as shown below by two torques TB= 1100 Nm and TC= 400 Nm. The lengths of the two segments are LAB= 750 mm and LBC= 600 mm. For brass, G = 41 GPa. Neglect the effect of any stress concentrations. Determine the smallest diameters dAB and dBC for which an allowable shear stress of 60 MPa is not exceeded. Express each diameter as an integer in units of mm.

The assembly shown in figure below consists of a light rigid bar AB, pinned at O, that is attached to the steel and aluminum rods. In the position shown, bar AB is horizontal and there is a gap, A = 5 mm, between the lower end of the steel rod and its pin support at C. Compute the stress in the aluminum rod when the lower end of the steel rod is touch to its support. A 0.75 m Steel A = 250 mm E = 200 GPa *i* O 1.5 m Aluminum L = 2m A = 300 mm' E = 70 GPa B D

A steel pipe is subjected to the 16-kN load shown below. Its outside diameter is 140 mm and its thickness is 7 mm. Consider that the yield strength of steel is 240 MPa. Analyze point H ONLY. Due to the pressure in the pipe, there is an 11.25-MPa (T) normal stress along x and a 22.50-MPa (T) normal stress along z of point H. A. Determine the internal forces and moments at the pertinent section through an FBD. B. Determine the stresses experienced by point H. Illustrate using a stress element. C. Determine the FS by MSST at point H D. Determine the von Mises equivalent stress at point H E. Determine the FS by MDET at point H 700 mm 16 KN 1,300 mm