Page 59 - Handbook of Civil Engineering Calculations, Second Edition

1.42 STRUCTURAL STEEL ENGINEERING AND DESIGN

2. Determine the reaction at each support
Take moments with respect to the other support. Thus M D 25R A 6(21) 20(20)
45(7.5) 7(2.5) 4.2(5) 0; M A 6(4) 20(5) 45(17.5) 7(27.5) 4.2(30)
25R D 0. Solving gives R A 33 kips (146.8 kN); R D 49.2 kips (218.84 kN).
3. Verify the computed results and determine the shears
Ascertain that the algebraic sum of the vertical forces is zero. If this is so, the computed
results are correct.
Starting at A, determine the shear at every significant section, or directly to the left or right
of that section if a concentrated load is present. Thus V A at right 33 kips (146.8 kN); V B at
left 33 8 25 kips (111.2 kN); V B at right 25 6 19 kips (84.5 kN); V C 19
12 7 kips (31.1 kN); V D at left 7 45 38 kips ( 169.0 kN); V D at right
38 49.2 11.2 kips (49.8 kN); V E at left 11.2 7 4.2 kips (18.7 kN); V E at right
4.2 4.2 0.
4. Plot the shear diagram
Plot the points representing the forces in the previous step in the shear diagram. Since the
loading between the significant sections is uniform, connect these points with straight
lines. In general, the slope of the shear diagram is given by dV/dx w, where w unit
load at the given section and x distance from left end to the given section.
5. Determine the bending moment at every significant section
Starting at A, determine the bending moment at every significant section. Thus M A 0;
M B 33(4) 8(2) 116 ft·kips (157 kN·m); M C 33(10) 8(8) 6(6) 12(3)
194 ft·kips (263 kN·m). Similarly, M D 38.5 ft·kips ( 52.2 kN·m); M E 0.
6. Plot the bending-moment diagram
Plot the points representing the values in step 5 in the bending-moment diagram (Fig. 25).
Complete the diagram by applying the slope equation dM/dx V. where V denotes the
shear at the given section. Since this shear varies linearly between significant sections, the
bending-moment diagram comprises a series of parabolic arcs.
7. Alternatively, apply a moment theorem
Use this theorem: If there are no externally applied moments in an interval 1-2 of the
2
span, the difference between the bending moments is M 2 M 1 1 Vdx the area
under the shear diagram across the interval.
Calculate the areas under the shear diagram to obtain the following results: M A 0;
M B M A /2(4)(33 25) 116 ft·kips (157.3 kN·m); M C 116 /2(6)(19 7)
1
1
1
194 ft·kips (263 kN·m); M D 194 /2(15)(7 38) 38.5 ft·kips ( 52.2 kN·m); M E
38.5 /2(5)(11.2 4.2) 0.
1
8. Locate the section at which the bending moment is maximum
As a corollary of the equation in step 6, the maximum moment occurs where the shear is
zero or passes through zero under a concentrated load. Therefore, CF 7/3 2.33 ft
(0.710 m).
9. Compute the maximum moment
1
Using the computed value for CF, we find M F 194 /2(2.33)(7) 202.2 ft·kips
(274.18 kN·m).

BEAM BENDING STRESSES
A beam having the trapezoidal cross section shown in Fig. 26a carries the loads indicated
in Fig. 26b. What is the maximum bending stress at the top and at the bottom of this
beam?