A beam having a T section with the dimensions shown in Fig. 3–15 is subjected to a bending moment of 1600 N · m that causes tension at the top surface. Locate the neutral axis and find the maximum tensile and compressive bending stresses.

Question

Accepted Answer

The area of the composite section is [latex]A = 1956 mm^{2}[/latex]. Now divide the T section into two rectangles, numbered 1 and 2, and sum the moments of these areas about the top edge. We then have

[latex]1956c_{1} = 12(75)(6) + 12(88)(56)[/latex]

and hence [latex]c_{1}[/latex] = 32.99 mm. Therefore [latex]c_{2}[/latex]= 100 − 32.99 = 67.01 mm.
Next we calculate the second moment of area of each rectangle about its own centroidal axis. Using Table A-18, we find for the top rectangle

Table A–18
Geometric Properties (Continued)

Part 1 Properties of Sections

A = area
G = location of centroid
[latex]I_{x} =\int {y^{2} dA }[/latex]= second moment of area about x axis
[latex]I_{y} =\int {x^{2} dA }[/latex] second moment of area about y axis
[latex]I_{xy} =\int {xy d A}[/latex] = mixed moment of area about x and y axes
[latex]J_{G} =\int {r^{2} d A} =\int{(x^{2}+ y^{2}) d A} = I_{x} + I_{y}[/latex]= second polar moment of area about axis through G
[latex]k^{2}_{x} = I_{x} /A [/latex]= squared radius of gyration about x axis

Rectangle

[latex]A = bh I_{x} =\frac {bh^{3}}{12} I_{y} =\frac {b^{3}h}{12} I_{xy} = 0[/latex]

Circle

[latex]A =\frac {πD^{2}}{4} I_{x} = I_{y} =\frac {πD^{4}}{64} I_{xy}= 0 J_{G} =\frac {πD^{4}}{32}[/latex]

circle

[latex]A =\frac {π}{4}(D^{2} − d^{2}) I_{x} = I_{y} =\frac {π}{64} (D^{4} − d^{4}) I_{xy} = 0 J_{G} =\frac {π}{32} (D^{4} − d^{4})[/latex]

Right triangles

[latex]A =\frac {bh}{2} I_{x} =\frac {bh^{3}}{36} I_{y} =\frac {b^{3}h}{36} I_{xy} = \frac {−b^{2}h^{2}}{72}[/latex]

Right triangles

[latex]A =\frac {bh}{2} I_{x} =\frac {bh^{3}}{36} I_{y} =\frac {b^{3}h}{36} I_{xy} = \frac {−b^{2}h^{2}}{72}[/latex]

Quarter-circles

[latex]A =\frac {πr^{2}}{4} I_{x} = I_{y} = r^{4}(\frac {π}{16} −\frac {4}{9π}) I_{xy} = r^{4} (\frac {1}{8} −\frac {4}{9π})[/latex]

Quarter-circles

[latex]A =\frac {πr^{2}}{4} I_{x} = I_{y} = r^{4}(\frac {π}{16} −\frac {4}{9π}) I_{xy} = r^{4} (\frac {4}{9π} −\frac {1}{8})[/latex]

Part 2 Properties of Solids ( Density, Weight per Unit Volume)

Rods

[latex]m =\frac {πd^{2}lρ}{4g} I_{y} = I_{z} =\frac {ml^{2}}{12}[/latex]

Round disks

[latex]m =\frac {πd^{2}tρ}{4g} I_{x} =\frac {md^{2}}{8} I_{y} = I_{z} =\frac {md^{2}}{16}[/latex]

Rectangular prisms

[latex]m =\frac {abcρ}{g} I_{x} =\frac {m}{12}(a^{2} + b^{2}) I_{y} =\frac {m}{12}(a^{2} + c^{2}) I_{z} =\frac {m}{12}(b^{2} + c^{2})[/latex]

Cylinders

[latex]m =\frac {πd^{2}lρ}{4g} I_{x} =\frac {md^{2}}{8} I_{y} = I_{z} =\frac {m}{48}(3d^{2} + 4l^{2})[/latex]

Hollow cylinders

[latex]m =\frac {π(d^{2}_{o} − d^{2}_{i})lρ}{4g} I_{x} =\frac {m}{8}(d^{2}_{o} +d^{2}_{i}) I_{y} = I_{z} =\frac {m}{48}(3d^{2}_{o} + 3d^{2}_{i} + 4l^{2})[/latex]

[latex]I_{1} =\frac {1}{12}bh^{3} =\frac {1}{12}(75)12^{3} = 1.080 × 10^{4} mm^{4}[/latex]

For the bottom rectangle, we have

[latex]I_{2} =\frac {1}{12}(12)88^{3} = 6.815 × 10^{5} mm^{4}[/latex]

We now employ the parallel-axis theorem to obtain the second momen of area of the composite figure about its own centroidal axis. This theorem states

[latex]I_{z} = I_{cg} + Ad^{2}[/latex]

where [latex]I_{cg}[/latex] is the second moment of area about its own centroidal axis and [latex]I_{z}[/latex] is the second moment of area about any parallel axis a distance d removed. For the top rectangle, the distance is

[latex]d_{1} = 32.99 − 6 = 26.99 mm[/latex]

and for the bottom rectangle,

[latex]d_{2}= 67.01 − 44 = 23.01 mm[/latex]

Using the parallel-axis theorem for both rectangles, we now find that

[latex]I = [1.080 × 10^{4} + 12(75)26.99^{2}] + [6.815 × 10^{5} + 12(88)23.01^{2}][/latex]
[latex]= 1.907 × 106 mm^{4}[/latex]

Finally, the maximum tensile stress, which occurs at the top surface, is found to be

[latex]σ =\frac {Mc_{1}}{I}=\frac {1600(32.99)10^{−3}}{1.907(10^{−6})} = 27.68(10^{6}) Pa = 27.68 MPa[/latex]

Similarly, the maximum compressive stress at the lower surface is found to be

[latex]σ =-\frac {Mc_{2}}{I}=−\frac {1600(67.01)10^{−3}}{1.907(10^{−6})}[/latex]

[latex] = −56.22( 10^{6} ) Pa = −56.22 MPa[/latex]

Question 3.5: A beam having a T section with the dimensions shown in Fig. ...

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