Question

QUESTION 9 20 16a 6а T k A beam with the hollow square cross-section is shown in the Figure above. Young Modulus of this beam is 208.59 GPa. Pis a point of observation. If a is 8.48 mm, and applied shear Force is 3.845 kN, the shear stress at point P is ... MPa.

Answer 1

Given data:

a = 8.48 mm

shear force, V = 3.845 kN

Solution:

The equation of shear stress is

...... Eq. (1)

T = * Q I *t

where,

V = shear force

Q = first moment of area

I = moment of inertia of the entire cross-sectional area

t = width of the beam at depth of specific section.

х ба Neutral axis a ба

Calculate the Q considering area 1, 2 and 3.

$Q =6a*a*2.5a + a*a*a*2 = 17a^{3}=17*8.48^{3}=103666.6 \text{ mm}^{3}$

The centroid is symmetrical to the x and y-axis.

Using the theorem of parallel axis

$\\I =\frac{6a*a^{3}}{12}+6a*a*(2.5a)^{2}+\frac{6a*a^{3}}{12}+6a*a*(2.5a)^{2}+\frac{a*(4a)^{3}}{12}+\frac{a*(4a)^{3}}{12} \\ I =a^{4}+75a^{4}+10.67a^{4}\\ I =86.67a^{4}\\ I =86.67*(8.48)^{4} I =448179.7 \text{ mm}^{4}$

t =a+a = 2a = 2*8.48 = 16.96 mm

substitute the values of V, Q, I and t in Eq. (1)

$\\\tau =\frac{3.845*103666.6}{448179.7*16.96}\\ \\ \tau =0.05243 \text{ kN/mm}^{2}\\ \tau =0.05243\text{ kN/mm}^{2}\left ( \frac{1000 \ MPa}{1 \ kN/mm^{2}} \right )\\ \\ \tau =52.43 \text{ MPa}$

Therefore, shear stress at point P is 52.43 MPa.