HW 14 - Homework set 14 Solutions PDF

Preview

Summary

Homework set 14 Solutions...

Description

PROBLEM 6.4

s s s

A square box beam is made of two 20 80-mm planks and two 20  120-mm planks nailed together as shown. Knowing that the spacing between the nails is s 30 mm and that the vertical shear in the beam is V  1200 N, determine (a) the shearing force in each nail, (b) the maximum shearing stress in the beam.

20 mm 80 mm 20 mm

120 mm

SOLUTION 1 1 b2 h23  b1 h13 12 12 1 1  (120)(120) 3  (80)(80) 3  13.8667  106 mm4 12 12

I

 13.8667  10 6m 4

(a)

A1  (120)(20)  2400 mm 2 y1  50 mm 3 3 6 3 Q1  A1 y1  120 10 mm  120 10 m

VQ (1200)(120  10 6 )   10.3846  103 N/m I 13.8667 106 qs  2F nail q

Fnail 

(b)

qs (10.3846 103 )(30 103 )  2 2

Fnail  155.8 N 

Q  Q1  (2)(20)(40)(20)  120  103  32  103  152  103 mm3  152  106 m3

 max 

VQ (1200)(152  106 )  It (13.8667  10 6 )(2  20  10 3 )

 329  10 3Pa

 max  329 kPa 

0.3 m

PROBLEM 6.10 n 10 kN

40 mm a

100 mm

12 mm 150 mm 12 mm

n 1.5 m

200 mm

For the beam and loading shown, consider section n-n and determine (a) the largest shearing stress in that section, (b) the shearing stress at point a.

SOLUTION At section n-n, V  10 kN. I  I1  4 I 2 

1 3 1  b1 h1  4  b2 h32  A2 d22  12 12 



 1   1 (100)(150)3  4   (50)(12)3  (50)(12)(69)2  12 12   

 28.125  10 6  4 0.0072  10 6  2.8566  106   39.58  106 mm4  39.58  106 m4

(a)

Q  A1y1  2 A2y 2  (100)(75)(37.5)  (2)(50)(12)(69)  364.05  10 3 mm 3  364.05  10 6 m 3 t  100 mm = 0.100 m

max 

(b)

(10  103 )(364.05  10 6 ) VQ   920  103 Pa 6 It (39.58  10  )(0.100)

 max  920 kPa 

Q  A1y 1  2 A2 y 2  (100)(40)(55)  (2)(50)(12)(69) 3 3 6 3  302.8  10 mm  302.8  10 m

t  100 mm  0.100 m

a 

VQ (10  103 )(302.8  10 6 )   765  103 Pa It (39.58  106 )(0.100)

 a  765 kPa 

1.5 in.

PROBLEM 6.15 For a timber beam having the cross section shown, determine the largest allowable vertical shear if the shearing stress is not to exceed 150 psi.

2 in.

4 in. w = 2.5 in. 2 in.

1.5 in.

SOLUTION I 

1 (1.5  83  2(0.5)(4)3 ) 12

I  69.333 in4

 all  150 psi

Q  (1.5 in.)(2 in.)(3 in.)  9 in 3;

At point a:

m 

VQ ; It

150 psi 

t  1.5 in.

V (9 in3 ) ; (69.333 in4 )(1.5 in.)

V  1733 lb 

At neutral axis:

Q  (1.5 in.)(2 in.)(3 in.)  (2.5 in.)(2 in.)(1in.)  14 in3 , t  2.6 in.

m 

VQ ; It

150 psi 

We choose smaller shear.

V(14 in 3 ) ; (69.333 in4 )(2.5 in.)

V  1857 lb  V  1733 lb ◄...