Question 12.8: A 5-m-long, simply supported steel beam AD is to carry the d...
A 5-m-long, simply supported steel beam AD is to carry the distributed and concentrated loads shown. Knowing that the allowable normal stress for the grade of steel is 160 MPa, select the wide-flange shape to be used.
STRATEGY: Draw the bending-moment diagram to find the absolute maximum bending moment. Then, using this moment, you can determine the required section modulus that satisfies the given allowable stress.
MODELING and ANALYSIS:
Reactions. Consider the entire beam to be a free body (Fig. 1).
+\circlearrowleft M_A=0: \quad D(5 \text{ m})-(60 \text{ kN})(1.5 \text{ m})-(50 \text{ kN})(4 \text{ m})=0 \\ \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad D=58.0 \text{ kN} \quad \quad \pmb{\text{D}}=58.0 \text{ kN}\uparrow \\ \underrightarrow{+} \sum F_x=0: \quad \quad A_x=0 \\ +\uparrow \sum{F_y}=0: \quad A_y+58.0 \text{ kN}-60 \text{ kN}-50 \text{ kN}=0 \\ \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad A_y=52.0 \text{ kN} \quad \quad \pmb{\text{A}}=52.0 \text{ kN}\uparrowShear Diagram. The shear just to the right of A is V_A=A_y=+52.0 \text{ kN}.
Since the change in shear between A and B is equal to minus the area under the load curve between these two points,
V_B=52.0 \text{ kN}-60 \text{ kN}=-8 \text{ kN}
The shear remains constant between B and C, where it drops to –58 kN, and keeps this value between C and D. Locate the section E of the beam where V = 0 by
V_E-V_A=-wx \\ 0-52.0 \text{ kN}=-(20 \text{ kN/m})x
So, x = 2.60 m.
Determination of \pmb{|M|_{\text{max}}}. The bending moment is maximum at E, where V = 0. Since M is zero at the support A, its maximum value at E is equal to the area under the shear curve between A and E. Therefore, |M|_{\text{max}} = M_E = 67.6 \text{ kN.m}.
Minimum Allowable Section Modulus. Substituting the values of \sigma_{\text{all}}\text{ and } |M|_{\text{max}}=M_E=67.6 \text{ kN.m} into Eq. (12.9) gives
S_{\text{min}}=\frac{|M|_{\text{max}}}{\sigma_{\text{all}}} =\frac{67.6 \text{ kN.m}}{160 \text{ MPa}} =422.5 \times 10^{-6}\text{ m}^3=422.5 \times 10^3 \text{ mm}^3
Selection of Wide-Flange Shape. From Appendix B, compile a list of shapes that have a section modulus larger than S_{\text{min}} and are also the lightest shape in a given depth group (Fig.2).
The lightest shape available is W360 × 32.9
REFLECT and THINK: When a specific allowable normal stress is the sole design criterion for beams, the lightest acceptable shapes tend to be deeper sections. In practice, there will be other criteria to consider that may alter the final shape selection.
