You are asked to design the manipulator arm for a fire truck for the New York City Fire Department. The arm will be 140 inches long. During manipulation operations, the arm may be subjected to an end load (perpendicular to its length) of \( 240 \quad \) Lbs., and an axial pull (parallel to its length) as large as \( 640 \quad \mathrm{Lb} \). You must also design for bending of the arm while it is in operation.
During manipulation, rotating objects may need to be stopped. Thus the arm may be subjected to a torque of \( 11,400 \quad \) inch-Lbs. In order to ensure precise placement or removal of objects manipulated, the shaft must not elongate more than 0.0004 inches nor twist more than \( \quad .8^{\circ} \) degrees, nor deflect more than 2.0 inches during its use.
NYFD requires that the arm must be designed with a safety factor of 1.66 for normal stress and 2.5 for shear stress. NYFD wants you to specify the radius ( \pm 0.05 in.) of the circular shaped solid manipulator arm, so that weight will be minimized.
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Answer
Step 4: Determine the minimum radius based on normal stress, torque stress, and deflection criteria, and minimize the weight of the manipulator arm.
Step 1 ：Step 1: Calculate normal stress: \(\sigma = \cfrac{P_{max}}{A} \), where \(P_{max} = 640 \mathrm{Lb} \cdot 1.66 \) and A is the cross-sectional area.
Step 2 ：Step 2: Calculate torque stress: \(\tau = \cfrac{T_{max}}{J} \), where \(T_{max} = 11,400 \mathrm{in\cdot Lb} \cdot 2.5 \) and J is the polar moment of inertia.
Step 3 ：Step 3: Determine maximum deflection: \(\Delta = \cfrac{PL^3}{48EI} \), where P is the axial load, L is the length, and E is the modulus of elasticity. For deflection <= 2 inches, solve for I (the moment of inertia).
Step 4 ：Step 4: Determine the minimum radius based on normal stress, torque stress, and deflection criteria, and minimize the weight of the manipulator arm.
