transformed beam. Thus,/= 22,000/25 = 880 lb/in
2
(6067.6 kPa) < 1200 lb/in
2
(8274
kPa).
In the actual beam, the maximum timber stress, which occurs at the back of the chan-
nel, is even less than 880 lb/in^2 (6067.6 kPa). Therefore, the strength of the member is
controlled by the allowable stress in the steel.
- Compare the capacity of the original and reinforced beams
Let subscripts 1 and 2 denote the original and reinforced beams, respectively. Compute
the capacity of these members, and compare the results. Thus M 1 -flic = 1200(1152)/6 =
230,000 lb-in (25,985.4 N-m); M 2 = 880(7626)/8.40 = 799,000 lb-in (90,271.02 N-m);
M 2 IM 1 = 799,000/230,000 = 3.47. Thus, the reinforced beam is nearly 3V 2 times as strong
as the original beam, before reinforcing.
BEAM SHEAR FLOWAND
SHEARING STRESS
A timber beam is formed by securely bolting a 3 x 6 in (76.2 x 152.4 mm) member to a
6 x 8 in (152.4 x 203.2 mm) member (exact size), as shown in Fig. 30. If the beam carries
a uniform load of 600 Ib/lin ft (8.756 kN/m) on a simple span of 13 ft (3.9 m), determine
the longitudinal shear flow and the shearing stress at the juncture of the two elements at a
section 3 ft (0.91 m) from the support.
Calculation Procedure:- Compute the vertical shear at the given section
Shear flow is the shearing force acting on a unit distance. In this instance, the shearing
force on an area having the same width as the beam and a length of 1 in (25.4 mm) meas-
ured along the beam span is required.
Using dimensions and data from Fig. 30, we find R = /2(60O)(O) - 3900 Ib (17,347.2
N); V= 3900 - 3(600) = 2100 Ib (9340.8 N). - Compute the moment of inertia of the cross section
I= (
1
Xi 2 )(W) = (
1
X^)(I I)
3
= 666 in
4
(2.772 dm
4
)FIGURE 30