Question 2.1: Figure 2.29a shows a diagram of the bones and biceps muscle ...
Figure 2.29a shows a diagram of the bones and biceps muscle of a person’s arm supporting a mass; Figure 2.29b shows a biomechanical model of the arm, in which the biceps muscle AB is represented by a bar with pin supports. The suspended mass is m = 2 kg, and the weight of the forearm is 9 N. If the crosssectional area of the tendon connecting the biceps to the forearm at A is 28 mm², what is the average normal stress in the tendon?
Given: Dimensions of and loading on truss system.
Find: Average normal stress in tendon AB.
Assume: Equilibrium; planar system; neglect weight of muscle and tendon AB.
We first need to find the internal axial force in AB and then calculate the normal stress by dividing this force by the cross-sectional area. We must construct an FBD of the system (Figure 2.30)
Equilibrium requires that the sum of moments taken about point C be zero, where a counterclockwise moment is taken to be positive:
ΣM_C = 0 = 19.62 \textrm{N} (0.35 m) + 9 \textrm{N} (0.15 \textrm{m}) – P_{AB} \sin θ (0.05 \textrm{m}).Solving for P_{AB},
P_{AB}= \frac{8.22 \textrm{N}\cdot \textrm{m}}{(0.05 \textrm{m})\sin \theta } =166.76 N.
The average normal stress σ_{_{AB}} is then
\sigma _{_{AB}}=\frac{P_{AB}}{A_{AB}} = \frac{166.76 \textrm{N}}{28\times 10^{-6} \textrm{m}^2}= 5.95 MPa.
