Question 5.4: The beam ABC shown in Fig 5-15a has simple supports at A and...

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  mm^2

Area 2:                                                          y_2=h/2=40  mm

A_2=ht=(80  mm)(12  mm)=960  mm^2

Area 3:                                                          y_3=y_2          A_3=A_2

c_1=\frac{\sum{y_iA_i} }{\sum{A_i}}=\frac{y_1A_1  +  2y_2A_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_1d^2_1                               (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 S_1=\frac{I}{c_1}                  S_2=\frac{I}{c_2}). With the cross-sectional properties determined, we can now proceed to calculate the maximum stresses from Eqs. (5-14a and b).

σ_1=-\frac{M_{C_1}}{I}=-\frac{M}{S_1}                    σ_2=\frac{M_{C_2}}{I}=\frac{M}{S_2}

Maximum stresses. At the cross section of maximum positive bending moment, the largest tensile stress occurs at the bottom of the beam  (σ_2)  and the largest compressive stress occurs at the top  (σ_1).  Thus, from Eqs. (5-14b) and (5-14a), respectively, we get

σ_t=σ_2=\frac{M_{pos}}{S_2}=\frac{2.025  kN·m}{40,100  mm^3}=50.5  MPa

σ_c=σ_1=-\frac{M_{pos}}{S_1}=-\frac{2.025  kN·m}{133,600  mm^3}=-15.2  MPa

Similarly, the largest stresses at the section of maximum negative moment are

σ_t=σ_1=-\frac{M_{neg}}{S_1}=-\frac{-3.6  kN·m}{133,600  mm^3}=26.9  MPa

σ_c=σ_2=\frac{M_{neg}}{S_2}=\frac{-3.6  kN·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,

(σ_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:

(σ_c)_{max}=-89.8  MPa

Thus, we have determined the maximum bending stresses due to the uniform load acting on the beam.