Question

ep, dimensional analysis #3

1. the horizontal shear stress in a material under load is given by the equation,

v = 3/2(v/(bd^2))
where v is the maximum unit horizontal shearing stress in psi (m/l^2), v is the total vertical shear in lbs (m), h is the width of the cross - section (l), and d is the depth of the cross - section(l). determine if this equation is dimensionally consistent.

1. the design of a wood beam for strength in bending is given by the equation,

s = m/f_b
where s = required section modulus (mm^3), f_b = allowable extreme fiber (bending) stress (kg/m^2), m = maximum bending moment (kg - m). determine if this equation is dimensionally consistent.

1. determine the dimensions and units for q if q = f l/(v^2) where f is a force, l is a length, and v is a velocity.

Explanation:

Step1: Recall dimensional - analysis rules

Dimensional - analysis involves checking the dimensions of each term in an equation. The dimensions of force $F$ are $[MLT^{- 2}]$, length $L$ is $[L]$, mass $M$ is $[M]$, time $T$ is $[T]$.

Step2: Analyze the first equation $v=\frac{3}{2}\frac{V}{(bd)}$

• The dimension of $v$ (shear stress) is $\frac{F}{A}=\frac{[MLT^{-2}]}{[L^{2}]}=[ML^{-1}T^{-2}]$.
• The dimension of $V$ (vertical shear force) is $[MLT^{-2}]$, $b$ (width) is $[L]$, and $d$ (depth) is $[L]$. Then $\frac{V}{bd}$ has dimension $\frac{[MLT^{-2}]}{[L][L]}=[ML^{-1}T^{-2}]$. And $\frac{3}{2}\frac{V}{bd}$ also has dimension $[ML^{-1}T^{-2}]$. So the first equation is dimensionally consistent.

Step3: Analyze the second equation $S = \frac{M}{F_{b}}$

• The dimension of $S$ (section modulus) is $[L^{3}]$.
• The dimension of $M$ (bending - moment) is $[ML^{2}T^{-2}]$, and the dimension of $F_{b}$ (allowable stress) is $\frac{F}{A}=\frac{[MLT^{-2}]}{[L^{2}]}=[ML^{-1}T^{-2}]$. Then $\frac{M}{F_{b}}=\frac{[ML^{2}T^{-2}]}{[ML^{-1}T^{-2}]}=[L^{3}]$. So the second equation is dimensionally consistent.

Step4: Analyze the third equation $Q=\frac{FL}{V^{2}}$

• The dimension of $F$ is $[MLT^{-2}]$, $L$ is $[L]$, and $V$ is $\frac{d}{t}=[LT^{-1}]$.
• Then $Q=\frac{FL}{V^{2}}$, substituting the dimensions: $\frac{[MLT^{-2}][L]}{[LT^{-1}]^{2}}=\frac{[ML^{2}T^{-2}]}{[L^{2}T^{-2}]}=[M]$. The unit of force $F$ is $N = kg\cdot m/s^{2}$, length $L$ is $m$, and velocity $V$ is $m/s$. Then $Q=\frac{F\times L}{V^{2}}=\frac{(kg\cdot m/s^{2})\times m}{(m/s)^{2}}=kg$.

Answer:

1. The first equation is dimensionally consistent.
2. The second equation is dimensionally consistent.
3. The dimension of $Q$ is $[M]$ and the unit is $kg$.