Problem 7-8: Use moment area theorems to calculate the slope at A and deflection at C for the beam shown in the following figure 7-8(a). The beam cross-section is rectangular with a width of 5 cm and depth of 10 cm. The modulus of elasticity of beam material is 200 GPa.

 

Solution:

According to first moment-area theorem, the change in slope of two points on the elastic curve is equal to the area of M/EI diagram between these two points.

Second Moment area theorem says the tangential deviation of a point B with respect to tangent at A is equal to the moment of area of M/EI diagram between A and B taken about point B.

In this problem, the loading is not symmetric as the concentrated load is not applied at the center of the beam. Therefore in this case the point of maximum deflection (point of zero slope) will not be under the load but it will be somewhere between A and C.

First draw the Bending moment diagram for the given beam as shown in figure 7-8(b) then draw the elastic curve (deflection curve) of the beam as in figure 7-8(c). In Figure (c) draw a tangent from A then draw projections of points B and C on the tangent from A and let these projections meet the tangent at B’ and C’ respectively.

From Figure (c) we find that;

Slope at A = tan BAB’ = BB’/AB = t_B/A/AB

and deflection at C = CC’’ = CC’ – C’C’’

t_B/A = tangential deviation of point B w.r.t. tangent at A.

t_C/A = tangential deviation of point C w.r.t. tangent at A.

According to moment-area theorem II, the tangential deviation of a point B w.r.t. tangent at A is equal to the moment of area of M/EI diagram between A and B taken about the point B.

Similarly the tangential deviation of a point C w.r.t. tangent at A is equal to the moment of area of M/EI diagram between A and C taken about the point C.

Moment of inertia of the section = BD³/12 = 5 x 10³/12 = 416.67 cm⁴ = 416.67 x 10^-8 m⁴

Modulus of Elasticity of material, E = 200 GPa = 200 x 10⁹ N/m² = 200 x 10⁶ kN/m²

t_B/A = Moment of area between A and B about point B / EI = ((0.5 x 1 x 22.5) x (2/3) x (1) + (0.5 x 3 x 22.5) x (1 + (2/3) x 3)) / EI = 108.75 kNm³/EI

Therefore Slope at A = t_B/A/AB = (108.75 kNm³/ EI)/4 = 108.75 kNm³ / (4 x (200 x 10⁶ kN/m²) x (416.67 x 10^-8 m⁴)) = 0.0325

C’C” = t_C/A = (0.5 x 3 x 22.5 x (1/3 x 3))/EI = 33.75 kNm³/EI

From the similarlity of triangles ABB’ and ACC’ we get

CC’/3 = BB’/4 ;

Therefore CC’ = (3/4) x BB’ = (3/4) x t_B/A = (3/4) x 108.75 kNm³/EI = 81.56 kNm³/EI

Hence Deflection at C = CC’’ = CC’ – C’C’’ = (81.56 - 33.75) kNm³/EI = 47.81 kNm³/EI = 47.81 kNm³/((200 x 10⁶ kN/m²) x (416.67 x 10^-8 m⁴)) = 0.057 m = 5.7 cm

Other solved example on Moment Area Theorems

You can also solve this problem by using our online calculator for Slope and Deflection of simply supported beam