Question

Step 1: Identify a truss (in this case quick search on Google) Step 2: look up average freight car length (55') Step 3: look up average loaded freight car weight (130 tons = 260,000 lbs) Step 4: Estimate truss dimensions based on assumed freight car langth 120 55' 30' 20 Step 5: Figure out worst loading condition (train on entire span at the same time) Step 6: calculate distributed load of train (260,0001b7 55ft = 4,727 lb/ft = 4.7 kip/ft Step 7: Apply load tracing to find truss loading Focus

Answer 1

Given data:

Length of the bridge, L = 120 ft

height of the bridge, H = 30 ft.

Horizontal distance between the joints = 20 ft.

Solution;

Draw schematic diagram of one side truss of the bridge.

E H 30 ft I M L K J 20 ft 20 ft 20 ft 20 ft 20 ft 20 ft

Average freight car length is 55 ft.

average loaded freight car weight, W = 260000 lb

Considering worst loading condition ( train on entire span at the same time).

The uniformly distributed load of train si

260000 w 4.7 lb/ft 55

There are two trusses .

The uniformly distributed load on one truss will be

4.7 0 = 2.35 lb/ft 2

Apply the load tracing to find truss loading.

considering the uniformly distributed load , the load on joints will be as below.

23.5 16 23.5 lb H 30 ft 20 ft 20 ft 20 ft 20 ft 20 ft 20 ft 47 16 47 1b 47 16 47 lb 47 1b

The free body diagram the truss is as below.

E 23.5 1b 23.5 16 H 30 ft I N M K Lupus X 20 ft 20 ft 20 ft 20 ft 20 ft 20 ft 47 16 47 16 47 16 47 lb 47 lb Ay y

Write the equilibrium equation of forces along the x-axis.

ΣF = 0 Α, = 0

It is symmetrical truss, therefore, The reaction force at both supports will be same.

Ay = ly 23.5+ 47 x 5 + 23.5 2 141 lb

The truss analysis.

In a three member joint: If two of those members are parallel and there are no other external loads (or reactions) at the joint Tthen the member that is not parallel is a zero force member.

so,

Member BN, EL, and HJ are zero force members.

Consider joint A.

23.5 lb 10 F AN A x Ay

30 0 = tan-1 20 A = 56.31°

Fy = 0 Ay – 23.5+ FAB X sin 0 141 – 23.5+ FAB X sin 56.1=0 FAB = –141.56 lb (compression)

Fx = 0 Ar + FAB X cos 0 + FAN = 0 0 - 141.56 x cos 56.1° + FAN = 0 FAN = 78.95 lb (tension)

Consider joint B.

B F BN

FBN = 0 FAB = FBC = -141.56 lb (compression)

Consider joint N.

F. CN BN 0 F AN N F MN 47 1b

Fy=0 FCN - 47 + FBN x0=0 FCN – 47 +0 x sin 56.1° = 0 FCN = 47 lb (tension)

0 Fr=0 - FAN - FBN X 0 + FMN – 78.95 – 0 x 0 + FMN = 0 FMN = 78.95 lb (tension)

Consider joint C.

F CD o BC A F F CN CM

Fy=0 - FCN – FBC X sin 8 - FCM X sin 0 = 0 – 47 + 141.56 x sin 56.1º – FCM X sin 56.1° = 0 FCM 84.93 lb (tension)

Fr= 0 FCD – FBC X cos 0 + FCM X cos 0 = 0 Fcp + 141.56 x cos 56.1° +84.93 x cos 56.1° = 0 Fcp = -126.32 lb (compreesion)

consider joint M

F CM F DM 6 F, M MIN F INI 47 lb

Fy=0 FDM - 47 + FCM X sin 0 = 0 FDM – 47 + 84.93 x sin 56.1° = 0 FDM= –23.49 lb (compresion)

Fr=0 FLM - FMN – FCM X cos 0 = 0 FLM – 78.95 – 84.93 x cos 56.1° = 0 FLM = 126.32 lb (tension)

Consider joint D.

D FDE e DL F DM

Fy=0 – FDM - FDL X sin 0 = 0 23.49 - FDL X sin 56.1° = 0 28.3 lb (tension) FDL

Fr=0 FCD + FDE + FDL X cos 0 = 0 126.32 + FDE +28.3 x cos 56.1° = 0 FDE = -142.1 lb (compression)

It is symmetrical truss.

The list of forces in members are as follows.

$\\F_{AB}=F_{HI}= 141.56 \text{lb (compression)}\\ F_{BN}=F_{HJ}=F_{EL}= 0\\ F_{BC}=F_{GH}= 141.56 \text{lb (compression)} \\ F_{AN}=F_{IJ}= 78.95 \text{lb (tension)}\\ F_{CN}=F_{GJ}= 47 \text{lb (tension)}\\ F_{CM}=F_{GK}= 84.93 \text{lb (tension)} \\ F_{CD}=F_{FG}= 126.32 \text{lb (compression)}\\F_{MN}=F_{JK}= 78.95 \text{lb (tension)}\\ F_{DM}=F_{FK}= 23.49 \text{lb (compression)} \\ F_{DE}=F_{EF}= 142.1 \text{lb (compression)}\\ F_{DL}=F_{FL}= 28.3 \text{lb (tension)} \\ F_{ML}=F_{KL}= 126.32 \text{lb (tension)}\\$