Please correct my understanding of this question. -In the equation of the moment at point A, there is a value, 36(11.5). I assume that is a force from the distributed load. -If you are finding the moment at point A, wouldn't you only need one value from the distributed load which is 27(10)? So my question is how do you get the value of 36 and 11.5, separately. And what is it for?

T·r T (from torsion) = Tr4 J ="- (Solid rod) F Onormal = A o (from bending) M•y - (Solid rod) The solid rod shown in the figure below has diameter of 1.25 in. If it is subjected to the force of 60 lb, determine the state of stress at point A using the principle of superposition. Neglect the weight of the rod, and the stress due to the I %3D I 4 transverse shear force. 60 lb k-10 in.- 15 in. Front View -20 in. В

Problem 8.61 Part A The two blocks used in a measuring device have negligible weight. The spring is compressed 4 in. when in the position shown. The end of the screw is smooth and the coefficient of static friction at all other points of contact is μ= 0.29. (Figure 1) Determine the smallest axial force P which the adjustment screw must exert on B in order to start the movement of B downward. Express your answer to three significant figures and include the appropriate units. Figure k = 20 lb/in. A 60° 45° B 1 of 1 P= Value Submit Request Answer Provide Feedback ? Units

Question 5 T/2 В 1200 Nm A homogeneous steel shaft (G=83GPA) is loaded as shown in the above Figure. Assuming T = 300 N-m, determine the torsional force in N-m in Member BC. Note: Include sign (+/-), assuming the direction of Torsion (T) at D is positive Given: AB: Length = 1.75m, Diameter = 75mm BC: Length = 1.50m, Diameter = 60mm AB: Length = 1.00m, Diameter = 40mm