Question 5.4: The beam ABC shown in Fig 5-15a has simple supports at A and...
The beam ABC shown in Fig 5-15a has simple supports at A and B and an overhang from B to C. The length of the span is 3.0 m and the length of the overhang is 1.5 m. A uniform load of intensity q = 3.2 kN/m acts throughout the entire length of the beam (4.5 m).
The beam has a cross section of channel shape with width b = 300 mm and height h = 80 mm (Fig. 5-16a). The web thickness is t = 12 mm, and the average thickness of the sloping flanges is the same. For the purpose of calculating the properties of the cross section, assume that the cross section consists of three rectangles, as shown in Fig. 5-16b.
Determine the maximum tensile and compressive stresses in the beam due to the uniform load.
Reactions, shear forces, and bending moments .We begin the analysis of this beam by calculating the reactions at supports A and B, using the techniques described in Chapter 4. The results are
R_A = 3.6 kN R_B = 10.8 kN
From these values, we construct the shear-force diagram (Fig. 5-15b). Note that the shear force changes sign and is equal to zero at two locations: (1) at a distance of 1.125 m from the left-hand support, and (2) at the right-hand reaction.
Next, we draw the bending-moment diagram, shown in Fig. 5-15c. Both the maximum positive and maximum negative bending moments occur at the cross sections where the shear force changes sign. These maximum moments are
M_{pos} = 2.025 kN·m M_{neg} = -3.6 kN·m
respectively.
Neutral axis of the cross section (Fig. 5-16b). The origin O of the yz coordinates is placed at the centroid of the cross-sectional area, and therefore the z axis becomes the neutral axis of the cross section. The centroid is located by using the techniques described in Chapter 10, Section 10.3 (available online), as follows.
First, we divide the area into three rectangles (A_1, A_2, and A_3). Second, we establish a reference axis Z-Z across the upper edge of the cross section, and we let y_1 and y_2 be the distances from the Z-Z axis to the centroids of areas A_1 and A_2, respectively. Then the calculations for locating the centroid of the entire channel section (distances c_1 and c_2) are as follows:
Area 1: y_1 = t/2 = 6 mm
A_1 = (b – 2t)(t) = (276 mm)(12 mm) = 3312 mm²
Area 2: y_2 = h/2 = 40 mm
A_2 = ht = (80 mm)(12 mm) = 960 mm²
Area 3: y_3 = y_2 A_3 = A_2
c_1=\frac{\sum{y_i A_i} }{\sum{A_i} }=\frac{y_1 A_1+2y_2 A_2}{A_1+2A_2}
=\frac{(6 mm)(3312 mm^2)+2(40 mm)(960 mm^2)}{3312 mm^2+2(960 mm^2)} =18.48 mm
c_2 = h – c_1 = 80 mm – 18.48 mm = 61.52 mm
Thus, the position of the neutral axis (the z axis) is determined.
Moment of inertia. In order to calculate the stresses from the flexure formula, we must determine the moment of inertia of the cross-sectional area with respect to the neutral axis. These calculations require the use of the parallel-axis theorem (see Chapter 10, Section 10.5 available online).
Beginning with area A_1, we obtain its moment of inertia (I_z)_1 about the z axis from the equation
(I_z)_1= (I_c)_1 + A_1 d_1^2 (c)
In this equation, (I_c)_1 is the moment of inertia of area A_1 about its own centroidal axis:
(I_c)_1=\frac{1}{12} (b-2t)(t)^3=\frac{1}{12} (276 mm)(12 mm)^3=39,744 mm^4
and d_1 is the distance from the centroidal axis of area A_1 to the z axis:
d_1 = c_1 – t/2 = 18.48 mm – 6 mm = 12.48 mm
Therefore, the moment of inertia of area A_1 about the z axis (from Eq. c) is
(I_z)_1=39,744 mm^4 + (3312 mm^2)(12.48 mm^2) = 555,600 mm^4
Proceeding in the same manner for areas A_2 and A_3, we get
(I_z)_2= (I_z)_3= 956,600 mm^4
Thus, the centroidal moment of inertia I_z of the entire cross-sectional area is
I_z = (I_z)_1+(I_z)_2+(I_z)_3 = 2.469 \times 10^6 mm^4
Section moduli. The section moduli for the top and bottom of the beam, respectively, are
S_1=\frac{I_z}{c_1} =133,600 mm^3 S_2=\frac{I_z}{c_2} =40,100 mm^3
(see Eqs. 5-15a and b). With the cross-sectional properties determined, we can now proceed to calculate the maximum stresses from Eqs. (5-14a and b).
\sigma _1=-\frac{Mc_1}{I}=-\frac{M}{S_1} \sigma _2=\frac{Mc_2}{I}=\frac{M}{S_2} (5-14a,b)
S_1=\frac{I}{c_1} S_2=\frac{I}{c_2} (5-15a, b)
Maximum stresses. At the cross section of maximum positive bending moment, the largest tensile stress occurs at the bottom of the beam (\sigma_2) and the largest compressive stress occurs at the top (\sigma_1). Thus, from Eqs. (5-14b) and (5-14a), respectively, we get
\sigma _t=\sigma _2=\frac{M_{pos}}{S_2} =\frac{2.025 kN\cdot m}{40,100 mm^3} =50.5 MPa
\sigma _c=\sigma _1=-\frac{M_{pos}}{S_1} =-\frac{2.025 kN\cdot m}{133,600 mm^3} =-15.2 MPa
Similarly, the largest stresses at the section of maximum negative moment are
\sigma _t=\sigma _1=-\frac{M_{neg}}{S_1} =-\frac{-3.6 kN\cdot m}{133,600 mm^3} =26.9 MPa
\sigma _c=\sigma _2=\frac{M_{neg}}{S_2} =\frac{-3.6 kN\cdot m}{40,100 mm^3} =-89.8 MPa
A comparison of these four stresses shows that the largest tensile stress in the beam is 50.5 MPa and occurs at the bottom of the beam at the cross section of maximum positive bending moment; thus,
(\sigma_t)_{max} = 50.5 MPa
The largest compressive stress is -89.8 MPa and occurs at the bottom of the beam at the section of maximum negative moment:
(\sigma_c)_{max} = -89.8 MPa
Thus, we have determined the maximum bending stresses due to the uniform load acting on the beam.
