## B

120N/cm

60N/cm

Figure P.5.3a-c. Distributed force systems acting on a beam.

Figure P.5.4. A man performing seated machine rows.

Figure P.5.5. A rod under axial load. The cross-sectional area of the circular rod varies linearly from one end to the other.

a distance of 0.8 m from point B. The length of the rod is 2.2 m. Its radius at B is 0.3 m and at D is 0.2 m. The rod is made of homogeneous material so that its Young's modulus E does not vary with position. Determine the horizontal support forces at B and D. Hint: Let e1 be the unit vector in the direction of P. Equation of motion in the e1 direction dictates that

where B and D are the horizontal support forces at B and D, respectively. They are positive when directed in - e1 direction. The section of the rod that lies between B and force P is in tension and the remaining section is in compression. As the distance between B and D does not change, elongation S1 of BP must be equal to the amount of compression 82 of PD. The length change 8 of a rod under a uniaxial force is given by

where the integration is over the length of the rod, F is the axial force acting on the rod, dx is a small length element along the axis of the rod, A = A (x) is the cross-sectional area, and E is Young's modulus. The parameter 8 is positive when F is tension and negative when F is compression. Show that in this case A (x) = w (0.3 - 0.045 x)2. Furthermore, the constant length condition reduces to the following equation:

B [0/08(0.3 - 0.045x)-2 dx] = (7,000 - B) [0.8/22(0.3 - 0.045x)-2 dx] Answer: B = 5,075 N, D = 1,925 N.

Problem 5.6. Discuss the potential benefit of heel cushion cups in alleviating the heel pain that afflicts many runners.