A steel bar 600mm long and having 30mm diameter, is turned down to 25mm diameter for one fourth of its length. It is heated at 30 C above room temperature, clamped at both ends and then allowed to cool to room temperature. If the distance between the clamps is unchanged, the maximum stress in the bar (a = 12.5 x 10- 6 per C and E = 200 GN/m2) is * || 25 MN/m2 40 MN/m2 75 MN/m2 50 MN/m2

Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part Several forces are acting on a bolt as shown in the diagram below. It is desired to determine the resultant of all of the forces acting on the bolt. The axis of the bolt is at a 45° angle with respect to the x-axis. The line of action of force Pis parallel to the x-axis. The magnitudes of the forces and the angles labeled are as follows: P= 35 N S= 60 N T=31 N a=15° B= 65° Q = 53 N 8= 25° T S Using the x-and y-components of the resultant of the forces acting on the bolt, determine the magnitude of the resultant force. The magnitude of the resultant force is N.

A long metal plate (E = 210 GPa) with some cutouts in it is mounted between two rigid plates at 6 am. By noon, the temperature has risen by 20°C. The structure is made from aluminium with a coefficient of thermal expansion of a = 23 x 10-6/°C. Please use the stress concentration chart shown below for any relevant calculations. a10 h L b O Use the following values in your working out: K 3.2 3.0 2.8 2.6 2.4 2.2 2.0 L = 13 m 0 0.1 P 0.2 W W Javg 0.3 2r P (w2r)t 0.4 P ▬▬▬▬▬▬▬ 0.5 h = 5.3 m a = 0.8 m b = 1.6 m a) What is the average normal stress throughout the plate at noon? b) What is the maximum stress at any point in the structure? c) The material the plate is made from has a tensile yield stress of 416 MPa and a compressive yield stress of 200 MPa. Using these values, do you expect the plate to yield at/before noon?

An aluminum alloy [E = 70 GPa; v = 0.33; a = 23.0x10-6/°C] pipe is subjected to a tensile load P. The pipe has an outside diameter of D = 280 mm, a cross-sectional area of A = 7550 mm², and a length of L = 9.5 m. The initial longitudinal normal strain in the pipe is zero. After load P is applied and the temperature of the pipe has been increased by AT = 40°C, the longitudinal normal strain in the pipe is found to be 2260 με. Calculate the magnitude of load P. Answer: P = M. L kN D P