Solution to Problem 575 | Horizontal Shearing Stress

$f_v = \dfrac{VQ}{Ib}$
 

 

Where
$V = 100(1000) = 100\,000 \, \text{N}$

$I = \dfrac{120(200^3)}{12} - \dfrac{100(160^3)}{12} = 45\,866\,666.67 \, \text{mm}^4$

$b = 20 \, \text{mm}$

Maximum horizontal shear stress occurs at the neutral axis
$Q_{NA} = 120(20)(90) + 20(80)(40) = 280\,000 \, \text{mm}^3$
 

Thus,
$(\,f_v\,)_{max} = \dfrac{100\,000(280\,000)}{45\,866\,666.67(20)}$

$(\,f_v\,)_{max} = 30.52 \, \text{ MPa}$           answer
 

Minimum horizontal shear stress in the web occurs at the junction of flange and web
$Q = 120(20)(90) = 216\,000 \, \text{mm}^3$
 

$(\,f_v\,)_{min} = \dfrac{100\,000(216\,000)}{45\,866\,666.67(20)}$

$(\,f_v\,)_{min} = 23.55 \, \text{ MPa}$           answer
 

The horizontal shearing stresses vary parabolically from the top to the bottom of the web. Recall that the average height of parabolic segment is 2/3 of its altitude measured from its base. Thus,
$(\,f_v\,)_{ave} = (\,f_v\,)_{min} + \frac{2}{3}[\,(\,f_v\,)_{max} - (\,f_v\,)_{min}\,]$

$(\,f_v\,)_{ave} = 23.55 + \frac{2}{3}[\,30.52 - 23.55\,]$

$(\,f_v\,)_{ave} = 28.20 \, \text{ MPa}$
 

Shear force carried by web alone
Force = Stress × Area
$V_{web} = (\,f_v\,)_{ave} \, A_{web}$

$V_{web} = 28.20 (160 \times 20)$

$V_{web} = 90 229.33 \, \text{N}$

$V_{web} = 90.23 \, \text{ kN}$
 

Percentage of shear force carried by web alone
$\%V_{web} = \dfrac{V_{web}}{V} \times 100\%$

$\%V_{web} = \dfrac{90.23}{100} \times 100\%$

$\%V_{web} = 90.23\%$           answer