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In mechanics of solids, understanding how internal forces act inside a material is essential to predict its behavior under load. When a force acts perpendicular (normal) to a surface, it produces normal stress. However, if the force acts parallel, or tangential, to the surface, it causes a different kind of stress known as shear stress.
Imagine sliding a deck of cards: forces parallel to the deck cause one card to slide over the next. This relative sliding tendency inside materials underlies the concept of shear stress. Mechanical components like beams, shafts, and fasteners often experience such forces, making shear stress crucial to analyze for safe and efficient design.
Why focus on shear stress? Because materials can fail or deform not only by pulling or pushing (normal stress) but also by being "cut" or "slid" internally under tangential forces. Grasping shear stress helps predict these behaviors and informs choices about shapes, materials, and sizes to prevent failures.
To define shear stress, let's start with a simple scenario. Consider a force \( F \) acting tangentially on a surface of area \( A \). The internal resistance offered by the material over that area is called shear stress, denoted by \( \tau \) (the Greek letter tau).
Mathematically, the average shear stress \( \tau \) is expressed as:
Here, \( \tau \) has units of pressure, such as Pascals (Pa) or Newtons per square meter (N/m²). This formula gives the average shear stress assuming the force is uniformly distributed over the area.
Important: In reality, shear stress is rarely uniform; it varies over the cross-section depending on geometry and loading. Still, this formula is foundational and widely used in basic design and analysis.
Beams commonly carry loads that introduce transverse shear forces. These forces produce internal shear stresses that vary across the beam's cross-section. Understanding this variation is vital for safe beam design.
Consider a rectangular beam section subjected to a shear force \( V \). The shear stress at any vertical position \( y \) from the neutral axis depends on how the internal forces are carried. The maximum shear stress does not occur at the outermost fibers but rather at the neutral axis (mid-height) where shear stress concentration is highest.
For rectangular sections, the maximum shear stress \( \tau_{\text{max}} \) due to a shear force \( V \) is given by:
This indicates the shear stress is not simply average shear force divided by area, but intensified due to stress distribution.
Different cross-sectional shapes, such as circular or I-beams, have different shear stress distributions. For example, circular beams under transverse shear exhibit a parabolic distribution with maximum at the neutral axis. Shear flow is the rate of transverse shear force transfer per unit length along the cross-section, useful for analyzing thin-walled structures.
Shafts used in machinery frequently transmit power by rotation, leading to torsion - twisting forces along the shaft axis. Torsional loads create shear stresses distributed over the cross-section of the shaft.
Imagine turning a screwdriver: the handle applies a torque \( T \) that twists the shaft. The internal fibers resist this twist via shear stress which varies from zero at the shaft center to a maximum at the surface.
The shear stress at a distance \( \rho \) from the center is given by the formula:
The polar moment of inertia \( J \) quantifies the shaft's resistance to torsion. For a solid circular shaft of diameter \( d \):
The maximum shear stress occurs at the surface where \( \rho = \frac{d}{2} \).
Step 1: Convert dimensions to meters for consistency:
Width \( b = 100 \text{ mm} = 0.1 \text{ m} \), height \( h = 200 \text{ mm} = 0.2 \text{ m} \)
Step 2: Calculate cross-sectional area \( A = b \times h = 0.1 \times 0.2 = 0.02 \text{ m}^2 \)
Step 3: Use maximum shear stress formula for rectangular beam:
\[ \tau_{\text{max}} = \frac{3V}{2A} \]
Where \( V = 10 \times 10^3 \text{ N} \), \( A = 0.02 \text{ m}^2 \)
Step 4: Calculate shear stress
\[ \tau_{\text{max}} = \frac{3 \times 10,000}{2 \times 0.02} = \frac{30,000}{0.04} = 750,000 \text{ Pa} = 0.75 \text{ MPa} \]
Answer: Maximum shear stress is 0.75 MPa at the neutral axis.
Step 1: Convert diameter to meters:
\( d = 50 \text{ mm} = 0.05 \text{ m} \)
Step 2: Calculate the polar moment of inertia:
\[ J = \frac{\pi d^{4}}{32} = \frac{3.1416 \times (0.05)^4}{32} = \frac{3.1416 \times 6.25 \times 10^{-7}}{32} = 6.136 \times 10^{-8} \text{ m}^4 \]
Step 3: Calculate radius \( \rho = \frac{d}{2} = 0.025 \text{ m} \)
Step 4: Apply shear stress formula at surface:
\[ \tau_{\text{max}} = \frac{T \rho}{J} = \frac{200 \times 0.025}{6.136 \times 10^{-8}} = \frac{5}{6.136 \times 10^{-8}} = 8.15 \times 10^{7} \text{ Pa} = 81.5 \text{ MPa} \]
Answer: The maximum shear stress in the shaft is 81.5 MPa.
Step 1: Calculate cross-sectional area:
\( A = b \times h = 0.15 \times 0.3 = 0.045 \text{ m}^2 \)
Step 2: Maximum shear stress:
\[ \tau_{\text{max}} = \frac{3V}{2A} = \frac{3 \times 20,000}{2 \times 0.045} = \frac{60,000}{0.09} = 666,667 \text{ Pa} = 0.667 \text{ MPa} \]
Step 3: Calculate moment of inertia \( I \) for rectangle:
\[ I = \frac{b h^{3}}{12} = \frac{0.15 \times (0.3)^3}{12} = \frac{0.15 \times 0.027}{12} = 3.375 \times 10^{-4} \text{ m}^4 \]
Step 4: Calculate bending stress at outer fiber:
\[ \sigma = \frac{M c}{I}, \quad c = \frac{h}{2} = 0.15 \text{ m} \]
\[ \sigma = \frac{12,000 \times 0.15}{3.375 \times 10^{-4}} = \frac{1,800}{3.375 \times 10^{-4}} = 5.33 \times 10^{6} \text{ Pa} = 5.33 \text{ MPa} \]
Step 5: Combined stresses: bending is normal stress (\( \sigma \)), shear stress is perpendicular. For conservative design, use Von Mises or maximum shear stress criteria, but here approximate maximum combined stress:
\[ \sigma_{\text{max, combined}} = \sqrt{\sigma^{2} + 3\tau^{2}} = \sqrt{(5.33)^{2} + 3 \times (0.667)^{2}} \approx \sqrt{28.4 + 1.34} = \sqrt{29.74} = 5.45 \text{ MPa} \]
Answer:
Maximum shear stress = 0.667 MPa, bending stress = 5.33 MPa, combined stress = 5.45 MPa.
Step 1: Calculate area of each material:
\( A = b \times h = 0.1 \times 0.2 = 0.02 \text{ m}^2 \) for both.
Step 2: Calculate shear flow using:
\[ q = \tau \times t \]
Shear flow is constant at interface, and shear flow \( q = \frac{VQ}{I} \), where \( Q \) is the first moment of area.
Step 3: Calculate \( Q \) (first moment of area) for one material about neutral axis.
Step 4: Calculate moment of inertia \( I \) for combined section.
Due to complexity, steps would involve section analysis students will learn later; key insight is shear stress is not uniform and depends on individual materials' properties and geometry.
Answer: Shear stress at interface must be calculated from shear flow and thickness; exact numeric solution is problem-specific but follows this method.
Step 1: Convert rpm to revolutions per second (rps):
\( N = \frac{1200}{60} = 20 \text{ rps} \)
Step 2: Calculate maximum torque using shear stress formula:
Radius \( r = \frac{d}{2} = 0.02 \text{ m} \)
Polar moment of inertia:
\[ J = \frac{\pi d^{4}}{32} = \frac{3.1416 \times (0.04)^4}{32} = 8.042 \times 10^{-7} \text{ m}^4 \]
Using \(\tau_{\max} = \frac{T r}{J}\), rearranged to find torque:
\[ T = \frac{\tau_{\max} J}{r} = \frac{40 \times 10^6 \times 8.042 \times 10^{-7}}{0.02} = 1608.4 \text{ N·m} \]
Step 3: Calculate power transmitted:
\[ P = 2 \pi N T = 2 \times 3.1416 \times 20 \times 1608.4 = 202,056 \text { W } = 202 \text{ kW} \]
Answer: Maximum power transmitted by the shaft is approximately 202 kW.
When to use: Solving shear stress distribution problems in beams.
When to use: Power transmission calculations involving rotating shafts.
When to use: Analyzing shear in beams with regular cross-sections like rectangles and circles.
When to use: Shaft torsion and shear stress calculations.
When to use: Final steps of any numerical problem.
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