Additional Example 5 - sf sdfd sfg dsfg dsaf g dfg dsfg dsfg sdg sad fasd f PDF

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Question 1 The portal frame in Figure 1 is pinned at A, C and E, but welded at B and D. Sketch the bending moment diagrams for the sides and top of the frame and insert the principal values.

Figure 1 Solution:

10 kN D 10 kN

B

C

RAy A

REy RAx

E

REx

R Ey ( 4 + 4 + 4 + 4 + 4) + 10 10 −10  4 −10 8 −10 12 −10 16 = 0 , R Ey = 15 kN RAy + REy − 10 −10 − 10 −10 = 0 , RAy = 40 − REy = 40 −15 = 25 kN

1

RCy

10 kN D 10 kN

C

RCx

REy E

REx  10 + REy  10 − 10  2 −10  6 = 0 , R Ex =

REx

80 − 15  10 = −7 kN 10

R Ax + R Ex − 10 = 0 , R Ax = 10 + 7 = 17 kN

M V

RAy RAx x

M + R Ax  10 − R Ay  x = 0

M = R Ay  x − RAx 10 = 25 x −170

2

10 kN M

V

RAy RAx x

M + 10( x − 4) + R Ax  10 − RAy  x = 0 M = R Ay  x − RAx 10 − 10( x − 4) = 15 x −130 10 kN

M

V

RAy

RAx x

M + 10( x − 4) + 10( x − 8) + R Ax 10 − RAy  x = 0 M = R Ay  x − R Ax 10 − 10( x − 4) −10( x − 8) = 5 x − 50

3

M

10 kN

V

REy REx x1

− M + R Ex 10 + REy  x1 = 0 M = REy  x1 + REx 10 = 15 x1 − 70

M

10 kN

V

10 kN

REy REx x1

− M + REx 10 + REy  x1 −10( x1 − 4) = 0 M = REy  x1 + REx 10 = 5 x1 − 30 Let M = 0, x1 = 6 m

4

10 kN M V

10 kN

REy REx x1

− M + REx 10 + REy  x1 −10( x1 − 4 ) −10 (x1 − 8 ) = 0 M = REy  x1 + REx 10 = −5 x1 + 50

-170 kNm

-70 kNm

-70 kNm -170 kNm

-10 kNm -10 kNm 10 kNm

5

6m

-70 kNm

Question 2 A 12 ft long beam made of a timber with an allowable normal stress of 2.4 ksi is designed to carry two 4.8 klb loads, as shown in Figure 2. The cross-section of the beam is rectangular, 4 in wide and 4.5 in deep. Since such a beam would not satisfy the allowable normal stress requirement, it will be reinforced by gluing planks of the same timber, 4 in wide and 1.25 in thick, to the top and bottom of the beam in a symmetric manner. (1) Determine the required number of pairs of planks in consideration of the allowable normal stress. (2) Determine the length of the planks in each pair that will yield the most economical design.

4.8 klb

4.8 klb

Figure 2 Solution: (1) 4.8 klb 4.8 klb

4.8 klb

4.8 klb

M

6

V

4.8 klb

4.8 klb M

V

4.8 klb Since the beam is symmetric along its length, only left half is discussed. For the section between A and B

M − R A  x = 0 , M = R A  x = 4.8x For the section between B and C

M − R A  x + 4.8( x − 4) = 0 M = RA  x − 4.8( x − 4) = 4.8 x − 4.8( x − 4) = 19.2 klbft (Maximum) M

19.2 klb·ft

0

x x = 4 ft

x = 8 ft

7

x = 12 ft

I=

 max

4h 3 12 h 19.2 1000  12    My 2= = = 2400 , h = 12 in 3 I 4h 12

Since the beam is 4.5 in deep and each plank is 1.25 in thick, to have h = 12 in, three pairs of planks are required, which provide a height 3  2 1.25 = 7.5 in, then 4.5 + 7.5 = 12 in. (2) The bending moment varies within the range between A and B, M = 4.8 x , then

 max

h  4.8  1000x   2 My  2  = 2400 , x = h in, which defines the maximum distance x = = 3 I 4h 3 12

from which a given depth h of the cross section is acceptable. For h = 4.5 in, x1 =

4.5 2 = 6.75 in = 0.5625 ft 3

72 = 16.33 in = 1.36 ft 3 9.5 2 Using two pair of planks, h = 4.5 + 4  1.25 = 9.5 in, x3 = = 30.08 in = 2.5 ft 3

Using a pair of planks, h = 4.5 + 2  1.25 = 7 in, x2 =

The length of the outer planks lo = 12 − 2  2.5 = 7 ft The length of the middle planks lm = 12 − 2 1.36 = 9.28 ft The length of the inner planks li = 12 − 2  0.5625 = 10.875 ft

8

Question 3 Beam AB is made of three planks glued together and is subjected, in its plane of symmetry, to the loading shown in Figure 3. The width of each glued joint is 20 mm. Determine the shear stress in each joint at section n-n of the beam.

1.5 kN

1.5 kN

n

A

B n 0.4 m

0.2 m

0.4 m

Beam cross-section 100 20 Joint 1 80

20 Joint 2

20 60

Units: mm Figure 3

9

Solution: 1.5 kN

1.5 kN

RA

RB

R A = RB = 1.5 kN

V

1.5 kN

B

A

V n −n = 1.5 kN

1.5 kN

100 20 S C 80

20

20 60

10

x

S=

I=

=

S 1 A1 + S 2 A2 + S 3 A3 100  20  10 + 80  20  60 + 60  20 110 = 51.67 mm = A1 + A2 + A3 100  20 + 80  20 + 60  20 100  20 3 20  80 3 2 2 + (51.67 − 10)  100  20 + + (60 − 51.67 )  80  20 12 12 3 60  20 2 + (110 − 51.67 )  60 20 = 8.63 106 mm4 + 12 V  Q' I  b( y )

For Joint 1,  1 =

1.5  1000  100  20  (51.67 −10) = 0.725 MPa 8.63 10 6  20

For Joint 2,  2 =

1.5  1000  60  20  (110 − 51.67) = 0.608 MPa 8.63 10 6  20

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