# Ha Long Bay

Topics: Beam, Second moment of area, Bending Pages: 92 (6058 words) Published: October 11, 2012
304

CHAPTER 5

Stresses in Beams (Basic Topics)

Design of Beams
P

Problem 5.6-1 The cross section of a narrow-gage railway
bridge is shown in part (a) of the figure. The bridge is constructed with longitudinal steel girders that support the wood cross ties. The girders are restrained against lateral buckling by diagonal bracing, as indicated by the dashed lines.

The spacing of the girders is s1 50 in. and the spacing
of the rails is s2 30 in. The load transmitted by each rail to a single tie is P 1500 lb. The cross section of a tie, shown in part (b) of the figure, has width b 5.0 in. and depth d.
Determine the minimum value of d based upon an allowable
bending stress of 1125 psi in the wood tie. (Disregard the weight of the tie itself.)
Solution 5.6-1

P

Railway cross tie

P

s2

Steel rail
Wood
tie

d
b
Steel
girder

(b)

s1
(a)
P

s2

Steel rail
Wood
tie

d
b

s1

s1
d

50 in. b 5.0 in. s2 30 in.
depth of tie
P 1500 lb
1125 psi
allow
P ( s1 s2 )
Mmax
15,000 lb-in.
2
bd 2 1
5d 2
S
(50 in.)( d 2 )
d inches
6
6
6

Mmax

sallow S

Solving, d 2

15,000
16.0 in.

( 1125) ¢

dmin

4.0 in.

3P ( s1 s2 )
bsallow

Note: Symbolic solution: d 2

Problem 5.6-2 A fiberglass bracket ABCD of solid circular cross section has the shape and dimensions shown in the figure. A vertical load P 36 N acts at the free end D.
Determine the minimum permissible diameter dmin of the bracket if the allowable bending stress in the material is 30 MPa and b 35 mm. (Disregard the weight of the bracket itself.)

5b
A

B

2b
D
P

Solution 5.6-2 Fiberglass bracket
DATA P 36 N
30 MPa
allow
CROSS SECTION
d = diameter

I

b

35 mm

d4
64

MAXIMUM BENDING MOMENT
MAXIMUM BENDING STRESS
Mmax c
d
smax
c
sallow
I
2

MINIMUM DIAMETER
( 96)(36 N)(35 mm)
96Pb
d3
sallow
( 30 MPa)
1,283.4 mm3

Mmax
3Pbd
2I

P(3b)

96 Pb
d3

dmin

5d 2

6

10.9 mm

C
2b

SECTION 5.6

Design of Beams

P

Problem 5.6-3 A cantilever beam of length L 6 ft supports a uniform load of intensity q 200 lb/ft and a concentrated load P 2500 lb (see figure).
Calculate the required section modulus S if allow 15,000 psi. Then select a suitable wide-flange beam (W shape) from Table E-1, Appendix E, and recalculate S taking into account the weight of the beam. Select a new beam size if necessary.

q

2500 lb

200 lb/ft

L = 6 ft

Solution 5.6-3

Cantilever beam

P

2500 lb q 200 lb/ft
15,000 psi
allow

L

6 ft

S

REQUIRED SECTION MODULUS
qL2
Mmax PL
15,000 lb-ft 3,600 lb-ft
2
18,600 lb-ft 223,200 lb-in.
Mmax
sallow

S

223,200 lb-in.
15,000 psi

TRIAL SECTION W 8
18.2 in.3
q0 L
2

M0
Mmax

3

14.88 in.

15.2 in.3

15.2 in.3

Beam is satisfactory.

21

P = 4000 lb
7.5 ft
q = 400 lb/ft

L = 15 ft

TRIAL SECTION W 8
S

24.3

REQUIRED SECTION MODULUS
qL2
8

26,250 lb-ft
S

227,700 lb-in.
15,000 psi

18.2 in.3

W8

227,700 lb-in.

Simple beam

4000 lb q 400 lb/ft
L 15 ft
16,000 psi
use an 8-inch W shape
allow

Mmax
sallow

4536 lb-in.

4,536

Mmax
sallow

Required S

P

Mmax

378 lb-ft

223,200

Problem 5.6-4 A simple beam of length L 15 ft carries a uniform load of intensity q 400 lb/ft and a concentrated load P 4000 lb (see figure).
Assuming allow 16,000 psi, calculate the required section
modulus S. Then select an 8-inch wide-flange beam (W shape) from Table E-1, Appendix E, and recalculate S taking into account the weight of the beam. Select a new 8-inch beam if necessary.

PL
4

21 lb/ft

2

Use

Solution 5.6-4

q0

21

28
28 lb/ft

q0

2

M0
15,000 lb-ft

11,250 lb-ft

q0 L
8

Mmax

315,000 lb-in.

315,000 lb-in.
16,000 psi

in.3

3

19.69 in.

787.5 lb-ft

315,000

Required S