24.02.2017 Views

Structural Concrete - Hassoun

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

4.6 Additional Examples 175<br />

Example 4.11<br />

The simply supported beam shown in Fig. 4.12 carries a uniform dead load of 2.8 K/ft (including<br />

self-weight) in addition to a service load of 1.6 K/ft. Also, the beam supports a concentrated dead load<br />

of 16 K and a concentrated live load of 7 K at C, 10 ft from support A.<br />

a. Determine the maximum factored moment and its location on the beam.<br />

b. Design a rectangular section to carry the loads safely using a steel percentage of about 1.5%,<br />

b = 20 in., f c ′ = 4ksi,andf y = 60 ksi.<br />

Solution<br />

a. Calculate the uniform factored load: w u = 1.2(2.8) + 1.6(1.6) = 5.91 K/ft. Calculate the concentrated<br />

factored load: P u = 1.2(16) + 1.6(7) = 30.4 K. Calculate the reaction at A by taking moments<br />

about B:<br />

R A = 5.91(30) 30∕2<br />

30 + 30.4(20) = 108.92K<br />

30<br />

R B = 5.91(30)+30.4 − 108.92 = 98.78K<br />

Maximum moment in the beam occurs at zero shear. Starting from B,<br />

V = 0 = 98.78 − 5.91x and x = 16.71 ft from B at D<br />

( ) 16.71<br />

M u (atD) =98.72(16.71)−5.91(16.71) = 825.5K⋅ ft (design moment)<br />

2<br />

( ) 20<br />

M u (atC) =98.78(20)−5.91(20) = 793.6K⋅ ft<br />

2<br />

7<br />

16<br />

2.8 1.6<br />

Shearing<br />

force<br />

diagram<br />

8<br />

26.5"<br />

Bending<br />

moment<br />

diagram<br />

Section at D<br />

Figure 4.12 Example 4.11.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!