Chapter 6, Problem 97RP

The composite beam shown is made by welding C200 × 17.1 rolled-steel channels to the flanges of a W250 × 80 wide-flange rolled-steel shape. Knowing that the beam is subjected to a vertical shear of 200 kN, determine (a) the horizontal shearing force per meter at each weld, (b) the shearing stress at point a of the flange of the wide-flange shape.

Fig. p6.97

To determine

The horizontal shearing force per meter at each weld.

Answer to Problem 97RP

The horizontal shearing force per meter at each weld is $\underset{\_}{146.1\text{kN}/\text{m}}$.

Explanation of Solution

Given information:

The composite beam is made by welding $\text{C}200\times 17.1$ rolled steel channels to the flanges of a $\text{W}250\times 80$ wide-flanged rolled steel shape.

The beam is subjected to a vertical shear of $200\text{kN}$.

Calculation:

Provide the section properties of $\text{C}200\times 17.1$ as shown below.

The Area of the section is $A=2,170{\text{mm}}^2$.

The width of the flange is $b_f=57.4\text{mm}$.

The thickness of flange is $t_f=9.91\text{mm}$.

The moment of inertia of the section is $I_y=0.545\times {10}^6{\text{mm}}^4$.

The centroid of the section is $\overline{x}=14.5\text{mm}$.

Provide the section properties of $\text{W}250\times 80$ as shown below.

The overall depth of the section is $d=257\text{mm}$

Thickness of flange is $t_f=15.6\text{mm}$.

Moment of inertia of the section is $I_x=126\times {10}^6{\text{mm}}^4$.

Sketch the channel section above the neutral axis as shown in Figure 1.

Refer to Figure 1.

Calculate the location of the centroid $\left({\overline{y}}_c\right)$ above the neutral axis for the  omposite section as shown below.

$\begin{array}{c}{\overline{y}}_c=\frac{257}{2}+57.4-14.5\\ =171.4\text{mm}\end{array}$

Calculate the moment of inertia (I) for the composite beam as shown below.

$I=I_x+2\times \left(I_y+A{\overline{y}}_c^2\right)$

Substitute $0.545\times {10}^6{\text{mm}}^4$ for $I_y$, $126\times {10}^6{\text{mm}}^4$ for $I_x$, $2,170{\text{mm}}^2$ for A, and $171.4\text{mm}$ for ${\overline{y}}_c$.

$\begin{array}{c}I=126\times {10}^6+2\times \left(0.545\times {10}^6+2,170\times {171.4}^2\right)\\ =126\times {10}^6+128.59\times {10}^6\\ =254.59\times {10}^6{\text{mm}}^4\end{array}$

Calculate the first moment of area as shown below.

$Q={\displaystyle \sum A\overline{y}}$ (1)

Calculate the first moment for the two welds (Q) as shown below.

$\begin{array}{c}Q=2,170\times 171.4\\ =371,938{\text{mm}}^3\end{array}$

Calculate the horizontal shear per unit length (q) as shown below.

$q=\frac{VQ}{I}$

Substitute $200\text{kN}$ for V, $371,938{\text{mm}}^3$ for Q, and $254.59\times {10}^6{\text{mm}}^4$ for I.

$\begin{array}{c}q=\frac{200\text{kN}\times \frac{{\text{10}}^3\text{N}}{1\text{kN}}\times 371,938{\text{mm}}^3\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^3}{254.59\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4}\\ =\frac{74.3876}{254.59\times {10}^{-6}}\\ =292.2\times {10}^3\text{N}/\text{m}\end{array}$

Calculate the shearing force per meter of weld for one weld as shown below.

$\begin{array}{c}q=\frac{292.2\times {10}^3}{2}\\ =146.1\times {10}^3\text{N}/\text{m}\times \frac{1\text{kN}}{1,000\text{N}}\\ =146.1\text{kN}/\text{m}\end{array}$

Therefore, the horizontal shearing force per meter at each weld is $\underset{\_}{146.1\text{kN}/\text{m}}$.

To determine

The shearing stress at point a of the flange.

Answer to Problem 97RP

The shearing stress at point a of the flange is $\underset{\_}{19.98\text{MPa}}$.

Explanation of Solution

Given information:

The beam is subjected to a vertical shear of $200\text{kN}$.

Calculation:

Refer to part (a).

The moment of inertia is $I=254.59\times {10}^6{\text{mm}}^4$.

Sketch the channel section through point a as shown in Figure 2.

Refer to Figure 2.

The thickness of the section is $t=2t_f$

Substitute $15.6\text{mm}$ for $t_f$.

$\begin{array}{c}t=2\times 15.6\\ =31.2\text{mm}\end{array}$

Calculate the location of the centroid at point a $\left({\overline{y}}_a\right)$ for the section as shown below.

$\begin{array}{c}{\overline{y}}_a=\frac{257}{2}-\frac{15.6}{2}\\ =120.7\text{mm}\end{array}$

Calculate the first moment of area $\left(Q_a\right)$ as shown below.

$\begin{array}{c}{Q}_a=2,170\times 171.4+2\times 112\times 15.6\times \left(\frac{257}{2}-\frac{15.6}{2}\right)\\ =371,938+421,774.08\\ =793,712.08{\text{mm}}^3\end{array}$

Calculate the shear stress $\left(\tau \right)$ as shown below.

$\tau =\frac{VQ}{It}$

Substitute $200\text{kN}$ for V, $793,712.08{\text{mm}}^3$ for Q, $31.2\text{mm}$ for t, and $254.59\times {10}^6{\text{mm}}^4$ for I.

$\begin{array}{c}\tau =\frac{200\text{kN}\times \frac{{\text{10}}^3\text{N}}{1\text{kN}}\times 793,712.08{\text{mm}}^3}{254.59\times {10}^6{\text{mm}}^4\times 31.2\text{mm}}\\ =\frac{158.74\times {10}^9}{7.943\times {10}^9}\\ =19.98\text{N}/{\text{mm}}^2\times \frac{1\text{MPa}}{1\text{N}/{\text{mm}}^2}\\ =19.98\text{MPa}\end{array}$

Therefore, the shearing stress at point a of the flange is $\underset{\_}{19.98\text{MPa}}$.