Determine the maximum shear stress in the beam section shown in Fig. P.11.19 stating clearly the point at which it occurs. Calculate also the rate of twist of the section if the shear modulus G = 25000 N/mm^2.

Question

Determine the maximum shear stress in the beam section shown in Fig. P.11.19 stating clearly the point at which it occurs. Calculate also the rate of twist of the section if the shear modulus G = 25000 N/mm².

Accepted Answer

The loading is equivalent to a torque of 1×25 = 25 kNmm and a shear load of 1 kN applied as shown in Figs S.11.19(a) and (b).
Torque: The maximum shear stress due to the applied torque is given by Eq. (11.31) in which

[latex] au_{\max }=\perp G t \frac{ d heta}{ d x}=\perp \frac{T t}{J}[/latex] (11.31)

[latex] J=\frac{100 imes 3^{3}}{3}+\frac{80 imes 2^{3}}{3}=1113.3 mm ^{4} [/latex]

Then

[latex] au_{\max (324)}=\pm \frac{25 imes 10^{3} imes 3}{1113.3}=\pm 67.4 N / mm ^{2} [/latex]

and

[latex] au_{\max (12)}=\pm \frac{25 imes 10^{3} imes 2}{1113.3}=\pm 44.9 N / mm ^{2} [/latex]

The rate of twist of the section is given by Eq. (11.26) and is

[latex] T=G J \frac{ d heta}{ d x} [/latex] (11.26)

[latex] \frac{ d heta}{ d x}=\frac{25 imes 10^{3}}{25000 imes 1113.3}=9.0 imes 10^{-4} rad / mm [/latex]

Shear: The shear stress distribution is determined using the method described in Ch. 10. Since the section has a vertical axis of symmetry [latex] I_{z y}=0 ext { and since } S_{z}=0 [/latex] also, Eq. (10.22) reduces to

[latex] q_{s}=\frac{\left(S_{y} I_{z y}-S_{z} I_{z} ight)}{I_{z} I_{y}-I_{z y}^{2}} \int_{0}^{s} t z d s+\frac{\left(S_{z} I_{z y}-S_{y} I_{y} ight)}{I_{z} I_{y}-I_{z y}^{2}} \int_{0}^{s} t y d s [/latex] (10.22)

[latex] q_{s}=-\frac{S_{y}}{I_{z}} \int_{0}^{s} t y d s [/latex] (i)

The position of the centroid of area and the second moment of area [latex] I_{z} [/latex] are found in the usual way and are

[latex] \bar{y}=13.9 mm , \quad I_{z}=252290 mm ^{4} [/latex]

Then, from Eq. (i)

[latex] q_{12}=-\frac{\left(-1 imes 10^{3} ight)}{252290} \int_{0}^{S_{1}} 2\left(-66.1+s_{1} ight) d s_{1} [/latex]

which gives

[latex] q_{12}=7.93 imes 10^{-3}\left(-66.1 s_{1}+0.5 s_{1}^{2} ight) [/latex] (ii)

The maximum value of [latex] q_{12} ext { occurs when } d q_{12} / d s_{1}=0 [/latex], that is when, from Eq. (ii)

[latex] s_{1}=66.1 mm [/latex]

Substituting this value in Eq. (ii) gives

[latex] q_{12(\max )}=-17.4 N / mm [/latex]

so that

[latex] au_{12(\max )}=-17.4 / 2=-8.7 N / mm ^{2} [/latex]

The value of the shear flow in the web 12 at 2 is, from Eq. (ii)

[latex] q_{2}=7.93 imes 10^{-3}\left(-66.1 imes 80+0.5 imes 80^{2} ight)=-16.6 N / mm [/latex]

Therefore the maximum shear flow in the flange is -8.3 N/mm at 2 and the corresponding shear stress is -8.3/3=-2.8 N/mm² and occurs in the directions 32 and 42. The maximum shear stress in the section due to torsion and shear is then

[latex] au_{\max }=67.4+2.8=70.2 N / mm ^{2} [/latex]

and occurs on the underside of 24 at 2 and on the upper surface of 32 at 2.

Question 11.19: Determine the maximum shear stress in the beam section shown...

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