Homework 6 (Due Monday, October 22)

— David Wagner 2007/10/17 11:37

Solve the following problems in your textbook:

Problem 6.14

24F-1.7E SP (southern Pine, assume no wane)

a.	2400	210	1.7	1.3
	F_bx⁺	F_v	E_x	E_y
CD	1.15	1.15	—	—
CM	1	1	1	1
Ct	1	1	1	1
Cl	1	—	—	—
CV	0.98	—	—	—
Cfu	1	—	—	—
Cc	1	—	—	—
	F'_b	F'_v	E'_x	E'_y
b.	2705	241.5	1.7	1.3

D+S → CD=1.15

$C_V = root{20}{{21*12*5.125}/{L*d*b}} = root{20}{{1291.5}/{20*5*19.25}}=$ 0.98

r=5.557; A=96.25; S=308.8 ; I=2972;

$w_{S}=(300)/12=$ 25 lb/in; $w_{D+S}=(200+300)/12=$ 41.7 lb/in.

$M = {w L^2}/8 = {41.7*(20*12)^2}/8=$ 300,000 in-lb; $f_b = M/S = {300000/{308.8}}=$ 971.5 ≤ 2705 psi✔.

V = {w L}/2 ={500*20}/2= 5000 lb; $f_v = {{3V}/{2A}} = {{3*5000}/{2*96.25}}=$ 77.9 ≤ 241.5 psi✔.

$Delta_{Sok}=L/360={20*12}/360=$ 0.67 in; $Delta_{D+Sok}=L/240={20*12}/240=$ 1 in.

$Delta_S ={5 w_S L^4}/{384EI}={5*25*(20*12)^4}/{384*1.7e6*2972}=$ 0.21 ≤ 0.67 in✔; $Delta_{D+S} ={500/300}Delta_S=$ 0.36 ≤ 1 in✔.

c.	971.5 psi	77.9 psi	0.21 in	0.36 in
	f_b	f_v	Δ_S	Δ_(D+S)

d.	Adequate by ASD.

	2400	210	1.7	1.3
KF/φ	2.16	2.16	—	—
1/φ	1/0.85	1/0.75	—	—
	F_bn	F_vn	E_x	E_y
e.	6099	605	1.7	1.3
CM	1	1	1	1
Ct	1	1	1	1
Cl	1	—	—	—
CV	0.98	—	—	—
Cfu	1	—	—	—
Cc	1	—	—	—
φx	0.85	0.75	—	—
λ	0.8	0.8	—	—
	F'_bn	F'_vn	E'_x	E'_y
f.	4064	363	1.7	1.3

M=S f_b=308.8*4064= 1,255,000 in-lb.

$V={2f_v A}/{3} = {2*363*96.25}/{3}=$ 23,292 lb.

g.	1,255,000 in-lb	23,292 lb
	M'_n	V'_n

w_u=(1.2*200 + 1.6*300)/12= 60 lb/in

$M_u={w_u L^2}/8 = {60*(20*12)^2}/8=$ 432,000 ≤ 1,255,000 in-lb✔.

$V_u={w_u L}/2={60*20*12}/2=$ 7200 ≤ 23,292 lb✔.

$Delta_{D+S} ={5 w_u L^4}/{384EI}={5*60*(20*12)^4}/{384*1.7e6*2972}=$ 0.51 ≤ 1 in✔; $Delta_{S} ={{1.6*300}/{1.2*200 + 1.6*300}}Delta_{D+S}={{480}/{720}}*0.51=$ 0.34 ≤ 0.67 in✔.

h.	432,000 in-lb	23,292 lb	0.34 in	0.51 in
	M_u	V_u	Δ_S	Δ_(D+S)

Camber: $1.5*{{1.2*200}/{1.2*200 + 1.6*300}}Delta_{D+S}=1.5*{{240}/{720}}*0.51=$ 0.255 in.

i.	Adequate by LRFD. Camber: 0.255 in.

Problem 6.18

a. Size Category	Dimension Lumber

4×12, Select Structural, Southern Pine

b.	1900	175	1.8e6
	F_b	F_v	E
CD	1.25	1.25	—
CM	1	1	1
Ct	1	1	1
CL	1	—	—
CF	1.1	—	—
Cfu	1	—	—
Ci	1	1	1
Cr	1	—	—
	F'_b	F'_v	E'
c.	2612	219	1.8e6

P=2000 lb, L₁=96 in, L₂=48 in, A=39.38, S=73.83, I=415.3.

$Delta_{1,max} = L_1/360=96/360=$ 0.27 in.
$Delta_{2,max} = L_2/180=48/180=$ 0.27 in.
$V_1 = {P L_2}/L_1 ={2000*4}/8 =$ 1000 lb;
>2000 → V=2000.
96,000 in-lb.
1300 ≤ 2612✔.
$f_V={3V}/{2A} = {3*2000}/{2*39.38}=$ 76.2 ≤ 219✔.
$Delta_1 approx {0.0642PL_2 {L_1}^2}/{EI}={0.0642*2000*48*96^2}/{1.8e6*415.3}=$ 0.076 ≤ 0.27 in✔.
$Delta_2 = {{P{L_2}^2}/{3EI}}(L_1+L_2) ={{2000*48^2}/{3*1.8e6*415.3}}(96+48) =$ 0.30 > 0.27 in✘.

d.	1300 psi	76.2 psi	0.076 in	0.30 in
	f_b	f_v	Δ₁	Δ₂

Unless free-end deflection Δ₂ is close enough (0.3 ≤ 0.3 ≈ 0.27),

e.not quite adequate by ASD.

Assume L from occupancy

	1900	175	1.8e6
KF/φ	2.16	2.16	—
1/φ	1/0.85	1/0.75	—
	F_bn	F_vn	E
f.	4828	504	1.8e6
CM	1	1	1
Ct	1	1	1
CL	1	—	—
CF	1.1	—	—
Cfu	1	—	—
Ci	1	1	1
Cr	1	—	—
φx	0.85	0.75	—
λ	0.8	0.8	—
	F'_bn	F'_vn	E'
g.	3612	302	1.8e6

M=S f_b=73.83*3612= 266,674 in-lb.

$V={2f_v A}/{3} = {2*302*39.38}/{3}=$ 7,929 lb.

h.	266,674 in-lb	7,929 lb
	M'_n	V'_n

P_u=1.2D+1.6L=1.2*400+1.6*1600=480+2560= 3040 lb

3040 ≤ 7,929 lb.✔
145,920 ≤ 266,674 in-lb.✔
$Delta_1={3040/2000}*0.076=$ 0.12 ≤ 0.27 in.✔
$Delta_2={3040/2000}*0.30=$ 0.46 > 0.27 in.✘

i.	145,920 in-lb	3040 lb	0.12 in.	0.27 in.
	M_u	V_u	Δ₁	Δ₂

Although fine for shear, bending, and deflection between supports,

j.not LRFD adequate for free-end deflection.

Problem 6.27

128 lb/ft = 10.7 lb/in.
168 in.
899 lb.
$M={wL^2}/8={10.7*168^2}/8=$ 37,750 in-lb.

V: 899 lb	M: 37,750 in-lb

$A ge {3V}/{2 f_v} ={3*899}/{2*207}=$ 6.5
$S ge M/{f_b}=37750/1323=$ 28.5

Try a 2×12, A=16.88, S=31.64

a. Size Category

Dimension Lumber

	1000	180
b.	F_b	F_v
CD	1.15	1.15
CM	1	1
Ct	1	1
CL	1	—
CF	1	—
Cfu	1	—
Ci	1	1
Cr	1.15	—
c.	F'_b	F'_v
	1323	207

1193 ≤ 1323✔.
$f_V={3V}/{2A} = {3*899}/{2*16.88}=$ 80 ≤ 207✔*.

*Shear strength is much greater than shear force, so the small end notches are OK.

d.	1193	80
	f_b	f_v

	1000	180
KF/φ	2.16	2.16
1/φ	1/0.85	1/0.75
	F_bn	F_vn
e.	2541	518
CM	1	1
Ct	1	1
CL	1	—
CF	1.1	—
Cfu	1	—
Ci	1	1
Cr	1.15	—
φx	0.85	0.75
λ	0.8	0.8
	F'_bn	F'_vn
f.	2186	311

M=S f_b=31.64*2168= 68,596 in-lb.

$V={2f_v A}/{3} = {2*311*16.88}/{3}=$ 3500 lb.

g.	68,596 in-lb	3,500 lb
	M'_n	V'_n

$w_u=2*(1.6*50+{{1.2*12}/sqrt{144+16}}*15)=2*(80+17)=2*97.1=$ 194 lb/ft = 16.2 lb/in.

1361 ≤ 3,500 lb.✔*
57154 ≤ 68,596 in-lb.✔

*The amount of shear the member can support is much greater than the ultimate shear load expected, so the small end-notches are OK.

h.	57,154 in-lb	1361 lb
	M_u	V_u

Problem 6.31

A_c=1.5*1.5= 2.25 in²

$f_{c ortho}=(140+560)/2.25=$ 311 psi

a. f_c⊥	311 psi

	335	1050
CM	1	1
Ct	1	1
CF	—	1
Ci	1	1
Cb	1*	—
	F'_c⊥	F*_c
b.	335	1050

*Assume the bearing as at the end of the member. (Distance not given.)

theta=atan(6/12)= 26.6°

$F prime_{c theta} = {{{F^circ}_c F_{c ortho}prime}/{{F^circ}_c sin^2 theta + F_{c ortho}prime cos^2 theta}}={335*1050}/{1050*0.20 + 335*0.80}=351750/478=$ 736

c.	F'_cθ	736 psi

P_u=1.2*140+1.6*560= 1064

d.	P_u	1064 lb

{335*2.25}= 754

e.	P'_n⊥	754 lb

736*2.25= 1656

f.	P'_nθ	1656 lb

Homework 6 (Due Monday, October 22)

Problem 6.14

Problem 6.18

Problem 6.27

Problem 6.31

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