To prevent aseptic loosening in total knee replacements due to chronic and
advanced osteoarthritis (OA) in the distal femur of Mr. David's, the following
students Obi, Jessica, Sam, and Icool in our

Question

To prevent aseptic loosening in total knee replacements due to chronic and
advanced osteoarthritis (OA) in the distal femur of Mr. David's, the following
students Obi, Jessica, Sam, and Icool in our BME 2502 course recommended
additional layer of polyethylene/polycarbonate spacer. If the state of stress in the
element of his femur is represented in Figure 1 below. To enable the design of the
robust femoral-knee implant replacement, the following must be determined:
| 40 MPa
LL _ = 35MPa
Ir Lo a a | 60 MPa
= —_— —
SL
/ 5 >.
i, 3 Conponenc
Figure 1

a. Calculate the equivalent state of stress on an element at the same point
oriented 30° counterclockwise with respect to the element shown. Sketch
the results on the rotated element.

b. Calculate the state of stress that represents the principal stresses, the angle
of orientation, and sketch the results on the updated rotated element.

c. Calculate the state of stress that represents the maximum in-plane shear
stress (Tmaxin-plane) and associated average normal stress, the angle of
orientation, and sketch the results on the updated rotated element.

d. Quickly sketch the Mohr’s circle (MC) for the original stat

VIPSolutions · Accepted Answer

Let's break down how to approach each part of this problem step by step, using the values from Figure 1.

### **Given:**
- Stresses on the element:
  - $\sigma_x = 60$ MPa (horizontal, right)
  - $\sigma_y = 35$ MPa (vertical, up)
  - $\tau_{xy} = 40$ MPa (shear, assume positive in the direction shown)

---

### **a. State of Stress on an Element Rotated 30° Counterclockwise**

The stress transformation equations for a rotated element are:

$$
\sigma_x' = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos 2\theta + \tau_{xy} \sin 2\theta
$$
$$
\sigma_y' = \frac{\sigma_x + \sigma_y}{2} - \frac{\sigma_x - \sigma_y}{2} \cos 2\theta - \tau_{xy} \sin 2\theta
$$
$$
\tau_{x'y'} = -\frac{\sigma_x - \sigma_y}{2} \sin 2\theta + \tau_{xy} \cos 2\theta
$$

Where $\theta = 30^\circ$.

- $\cos 2\theta = \cos 60^\circ = 0.5$
- $\sin 2\theta = \sin 60^\circ = 0.866$

Substitute values:
- $\sigma_x = 60$ MPa
- $\sigma_y = 35$ MPa
- $\tau_{xy} = 40$ MPa

**Calculations:**

$$
\frac{\sigma_x + \sigma_y}{2} = \frac{60 + 35}{2} = 47.5 \text{ MPa}
$$
$$
\frac{\sigma_x - \sigma_y}{2} = \frac{60 - 35}{2} = 12.5 \text{ MPa}
$$

Now:
$$
\sigma_x' = 47.5 + (12.5)(0.5) + (40)(0.866)
$$
$$
= 47.5 + 6.25 + 34.64 = 88.39 \text{ MPa}
$$

$$
\sigma_y' = 47.5 - (12.5)(0.5) - (40)(0.866)
$$
$$
= 47.5 - 6.25 - 34.64 = 6.61 \text{ MPa}
$$

$$
\tau_{x'y'} = - (12.5)(0.866) + 40(0.5)
$$
$$
= -10.825 + 20 = 9.18 \text{ MPa}
$$

**Sketch**: Draw a rotated element (30° CCW from the original), label the new faces with $\sigma_x' = 88.39$ MPa, $\sigma_y' = 6.61$ MPa, and $\tau_{x'y'} = 9.18$ MPa.

---

### **b. Principal Stresses and Orientation**

Principal stresses ($\sigma_1, \sigma_2$) are given by:

$$
\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2}
$$

$$
\sqrt{(12.5)^2 + (40)^2} = \sqrt{156.25 + 1600} = \sqrt{1756.25} = 41.92
$$

So,
$$
\sigma_1 = 47.5 + 41.92 = 89.42 \text{ MPa}
$$
$$
\sigma_2 = 47.5 - 41.92 = 5.58 \text{ MPa}
$$

Angle $\theta_p$ to principal planes (from x-axis):

$$
\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} = \frac{2 \times 40}{60 - 35} = \frac{80}{25} = 3.2
$$
$$
2\theta_p = \tan^{-1}(3.2) = 72.65^\circ
$$
$$
\theta_p = 36.33^\circ
$$

**Sketch**: Draw the element rotated by $\theta_p = 36.33^\circ$, label $\sigma_1 = 89.42$ MPa, $\sigma_2 = 5.58$ MPa, and no shear.

---

### **c. Maximum In-Plane Shear Stress**

Maximum in-plane shear stress ($\tau_{max}$):

$$
\tau_{max} = \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } = 41.92 \text{ MPa}
$$

Associated average normal stress ($\sigma_{avg}$):

$$
\sigma_{avg} = \frac{\sigma_x + \sigma_y}{2} = 47.5 \text{ MPa}
$$

Orientation angle to max shear:

$$
\theta_s = \theta_p + 45^\circ = 36.33^\circ + 45^\circ = 81.33^\circ
$$

**Sketch**: Draw the element rotated by $81.33^\circ$, label shear $\tau_{max} = 41.92$ MPa, and normal stress $\sigma_{avg} = 47.5$ MPa.

---

### **d. Mohr’s Circle**

- Center: ($\sigma_{avg}$, 0) = (47.5 MPa, 0)
- Radius: $\tau_{max} = 41.92$ MPa

**Key Points:**
- Principal stresses: $89.42$ MPa, $5.58$ MPa (on $\sigma$ axis)
- Max shear: $(47.5, 41.92)$, $(47.5, -41.92)$

**Quick Sketch:**
1. Draw the $\sigma$-axis (horizontal), $\tau$-axis (vertical).
2. Mark the center at (47.5, 0).
3. Plot principal stresses at (89.42, 0) and (5.58, 0).
4. Mark max shear at (47.5, ±41.92).

---

## **Summary Table**

| Parameter                        | Value           |
|-----------------------------------|-----------------|
| Rotated Stress ($\sigma_x'$)    | 88.39 MPa       |
| Rotated Stress ($\sigma_y'$)    | 6.61 MPa        |
| Rotated Shear ($\tau_{x'y'}$)   | 9.18 MPa        |
| Principal Stress ($\sigma_1$)   | 89.42 MPa       |
| Principal Stress ($\sigma_2$)   | 5.58 MPa        |
| Principal Angle ($\theta_p$)    | 36.33°          |
| Max Shear ($\tau_{max}$)        | 41.92 MPa       |
| Avg Normal ($\sigma_{avg}$)     | 47.5 MPa        |
| Shear Angle ($\theta_s$)        | 81.33°          |

---

If you need the explicit sketches, let me know and I can describe or create them!

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