Week 6 Influence Lines-Part 2 - موقع المهندس حمادة شعبان · 2017-11-18 · Influence Lines-Part 2 Week 6 ... Influence Lines For Floor Girders ىييلإ لاوأ

Dec 2015

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طالما أن الموت ليس هدفا في حد

ذاته، فلماذا نعيش حياة األموات،

وكأنه الهدف األوحد في حياتنا؟

Influence

Lines-Part 2

Week 6

استراكشرنوت

. تتكون النوت من عشرة أسابيع

على شرح كل نوتيحتوي

من ألمثلة وتمارينوحلول

سابقة.هوموركات وامتحانات

(hs.net-engتابع )

النوت يتم تنقيحها وتحديثها

وإضافة الجديد إليها بشكل دوري،

داوم على التواصل معنا.

Dec 2015




Influence Lines For Floor Girders

ىيل أوال إليييا تنتقييي( فإنهييaل )يي الشكيييا فييي( كمييSlabsي )يييي* فييي حييال وأييود أحمييال عل

(floor beams( ومن ثم تنتقل إلي )girdersعل )ى ( هيئةConcentrated Loads )

.(c)كما في الشكل

حال طلب حساب في *(Shear) ( طة ند أي نق في pع )(panel) مة (Shear)فإن قي

لي لي (Shear)عند هذه النقطة تكون مساوية ل هذا ا لي (panel)في (Shear)حيث أن ا

الواحد ال يختلف من نقطة ألخرى. (Panel)في الي

فإنييه يمكيين اسييتخدام طريقيية (P)عنييد أي نقطيية (Moment)* فييي حييال طلييب حسيياب

(Tabulate Values( أو طريقة )Equations لحسياب تيأثيير )(Load Unit) ىعلي

(Floor Panels أوال ومن ثم حساب )(M) بدراسة المقطع عند النقطة المطلوب حساب

(M) لها كما في الشكل(d).

أنت اآلن مجموع

خيارات عمرك السابقة.

Dec 2015




Example 6-13

Draw the influence line for the shear in

panel CD of the floor girder in Fig. (a)

Solution:

Tabulate values. The unit load is placed at each floor beam

location and the shear in panel CD is calculated. A table of

the results is shown in Fig. (b) The details for the calculation

when x = 0 and x = 6m are given in Figs. (c)

& (d), respectively. Notice how in each case

the reactions of the floor beams on the girder

are calculated first, followed by a

determination of the girder support reaction

at F (Gy is not needed), and finally, a

segment of the girder is considered and the

internal panel shear VCD is calculated. In the

same way, we can determine the values for

VCD when x = 3m, 9m, and 12m.

(b)

(d)

(c)

بعض الناس طغاة، فقط ألن

البعض اآلخر وافق أن يركع لهم.

(a)

Dec 2015




Example 6-14

Draw the influence line for the moment

at point F of the floor girder in Fig. (a)

Solution:

Tabulate values. The unit load is placed at x = 0 and each panel

point thereafter. The corresponding values for MF are the table Fig.

(b). Details of the calculations for x = 2m are shown in Fig. (c). It’s

first necessary to determine the reactions of the floor beams on the

girder, followed by a determination of the girder support reaction

Gy (Hy is not needed), and finally segment GF of the girder is

considered and the internal moment MF is calculated. In the same

way determine the other values of MF listed in Fig. (b)

(a)

(b)

(c)

(d)

بعض الناس ال يتغيرون،

أحيانا فقط يغيرون أزيائهم.

Dec 2015




Solution:

a)

Vertical shear in panel BC:

Unit load @: 0 ≤ x ≤ 5 m

+ ∑ 𝐹𝑦 = 0

𝑉𝐵𝐶 + 𝐷𝑦 = 0

𝑉𝐵𝐶 = − 𝐷𝑦

Unit load @: 10 ≤ x ≤ 20 m

+ ∑ 𝐹𝑦 = 0

− 𝑉𝐵𝐶 + 𝐵𝑦 = 0

𝑉𝐵𝐶 = 𝐵𝑦

b)

Moment at B:


+ ∑ 𝑀𝐵 = 0

𝑀𝐵 + 1 ∗ (5 − 𝑥) = 0

𝑀𝐵 = 𝑥 − 5


𝑀𝐵 = 0

دع بعض أخطاء اآلخرين

تمر دون أن تالحظها.

Dec 2015




Solution:

Vertical shear in panel DE:

a)


+ ∑ 𝐹𝑦 = 0

𝑉𝐷𝐸 + 𝐹𝑦 = 0

𝑉𝐷𝐸 = −𝐹𝑦

Unit load @: 12 ≤ x ≤ 15 m

+ ∑ 𝐹𝑦 = 0

−𝑉𝐷𝐸 + 𝐴𝑦 = 0

𝑉𝐷𝐸 = 𝐴𝑦

b)

Moment at C:


+ ∑ 𝑀𝐶 = 0

−𝑀𝐶 + (𝐹𝑦 ∗ 9) = 0

𝑀𝐶 = 9 𝐹𝑦

c)


+ ∑ 𝑀𝐶 = 0

𝑀𝐶 – (𝐴𝑦 ∗ 6) = 0, 𝑀𝐶 = 6

كأن الحاسد إنما خلق ليغتاظ.

Dec 2015




Solution:

a)

Vertical shear in panel CD:


+ ∑ 𝐹𝑦 = 0

𝑉𝐶𝐷 + 𝐹𝑦 = 0

𝑉𝐶𝐷 = −𝐹𝑦


+ ∑ 𝐹𝑦 = 0

−𝑉𝐶𝐷 + 𝐴𝑦 = 0

𝑉𝐶𝐷 = 𝐴𝑦

b)

Moment at E:


+ ∑ 𝑀𝐸 = 0

𝑀𝐸 + (𝐹𝑦 ∗ 3) = 0

𝑀𝐸 = 3 𝐹𝑦


+ ∑ 𝑀𝐸 = 0

𝑀𝐸 – (𝐴𝑦 ∗ 12) = 0, 𝑀𝐸 = 12 𝐴𝑦

0.2

2.4

0.8 1.00

0.4

-0.4

yF

yA

CDV

EM

6-34. Draw the influence lines for (a) the shear in

panal CD of the girder, and (b) the moment at E.

يكتمون األسرار عدا كل الناس

أولئك الذين نفضي لهم بأسرارنا.

Dec 2015




1.00

-4.5

DM

CDV

6-35. A uniform live load of 6 kN/m and a concentrated live force of

9 kN are to slabs. Determine (a) the maximum live shear in panel

CD, and (b) the maximum live moment at B.

Solution:

Vertical shear in panel CD:

Unit load @: 0 ≤ x ≤ 1.5 m

+ ∑ 𝐹𝑦 = 0

𝑉𝐶𝐷 − 1 = 0

𝑉𝐶𝐷 = 1


+ ∑ 𝐹𝑦 = 0

−𝑉𝐶𝐷 = 0

𝑉𝐶𝐷 = 0

Moment at B:

Unit load @: 0 ≤ x ≤ 4.5 m

+ ∑ 𝑀𝐵 = 0

− 𝑀𝐵 − 1 ∗ (4.5 − 𝑥) = 0

𝑀𝐵 = 𝑥 − 4.5

a)

(𝑉𝐶𝐷)𝑚𝑎𝑥 = (6)(0.5)(1.5)(1) + (1)(1.5)(6) + 9 (1) = 22.5 𝑘𝑁

b)

(𝑀𝐵)𝑚𝑎𝑥 = (6) (0.5) (4.5) (−4.5) + (9) (−4.5) = − 101.25 𝑘𝑁. 𝑚

ال توقف طفلك عند حدود تعلمك،

في زمن مختلف.فهو مولود

Dec 2015




-2.00

-1.00

BCV

MC

6-36. A uniform live load of 27 kN/m and a single concentrated live

force of 60 kN are placed on the top beams. If the beams also support a

uniform dead load of 5.25 kN/m, determine (a) the maximum shear in

panel BC of the girder and (b) the maximum moment in the girder at C.

Solution:



+ ∑ 𝐹𝑦 = 0

𝑉𝐵𝐶 + 1 = 0

𝑉𝐵𝐶 = − 1


+ ∑ 𝐹𝑦 = 0

𝑉𝐵𝐶 = 0

𝑉𝐵𝐶 = 0

Moment at C:


+ ∑ 𝑀𝐶 = 0

−𝑀𝐶 − 1 ∗ (2 − 𝑥) = 0

𝑀𝐶 = 2 − 𝑥

a)

(𝑉𝐵𝐶)𝑚𝑎𝑥

= (5.25 + 27)(−1)(0.5)(1) + (5.25 + 27)(−1)(1) + (60)(1) = −108.375 𝑘𝑁

b)

(𝑀𝐶)𝑚𝑎𝑥

= (60)(−2) + (5.25 + 27)(−2)(2)(0.5) = −184.5 𝑘𝑁. 𝑚

يستطيع أحد ركوب ظهرك ال

إال إذا كنت منحنيا .

Dec 2015




1.00

0.5

-0.5

1.5

1.5

1.00

0.5

-0.5

0.5

-0.5

0.5

-2.5 -2.5

YD

YB

BCV

CM

6-37. Draw the influence lines for (a) the shere in panal BC

of the girder, and (b) the moment at C. Assume the support at

B is a pin and D is roller.

Solution:

a)



+ ∑ 𝐹𝑦 = 0

𝑉𝐵𝐶 + 𝐵𝑦 = 0

𝑉𝐵𝐶 = −𝐵𝑦

Unit load @: 10 ≤ x ≤ 20 m

+ ∑ 𝐹𝑦 = 0

−𝑉𝐵𝐶 − 𝐷𝑦 = 0

𝑉𝐵𝐶 = −𝐷𝑦

b)

Moment at C:


+ ∑ 𝑀𝐶 = 0

𝑀𝐶 + (𝐵𝑦 ∗ 5) = 0

𝑀𝐶 = 5 𝐵𝑦

Unit load @:15 ≤ x ≤ 20 m

+ ∑ 𝑀𝐶 = 0

𝑀𝑐 – (𝐴𝑦 ∗ 5) = 0, 𝑀𝐶 = 5 𝐴𝑦

ال يمكن ألحد أن يشعرك

بالوضاعة من دون موافقتك؟

Dec 2015




-1.98

1.98

FM

EFV

DV 0.67

-0.33

0.33

1.67

1 1.33

0.33 0.67

6-38. A uniform live load of 5 kN/m and a single concentrated live force

of 18 kN are to be placed on the floor slabs. Determine (a) the maximum

positive live vertical reaction at support D, (b) the maximum positive live

shear in panel EF of the girder, and (c) the maximum positive live

moment at F. Assume the support at D is a roller and G is a pin.

Solution:

Unit load @: 0 ≤ x ≤12 m

+ ∑ 𝐹𝑦 = 0

𝑉𝐸𝐹 + 𝐷𝑦 − 1 = 0

𝑉𝐸𝐹 = 1 − 𝐷

Unit load @:15 ≤ x ≤ 18 m

+ ∑ 𝐹𝑦 = 0

−𝑉𝐸𝐹 − 𝐷𝑦 = 0

𝑉𝐸𝐹 = 𝐷𝑦

Moment at F:


+ ∑ 𝑀 = 0

𝑀𝐹 − 𝐷𝑦 (6) + 1(15 − 𝑥) = 0

𝑀𝐹 = 𝑥 + 𝐷𝑦 (6) − 15

𝑎) (𝐷𝑦)𝑚𝑎𝑥(𝑇)

= (5)(1.67)(0.5) (18) + (18) (1.67) = 105.21 𝑘𝑁

𝑏) (𝑉𝐸𝐹)𝑚𝑎𝑥(𝑇)

= (5)(9)(0.5)(0.67) + (5)(0.33)(0.5)(4.5) + (18)(0.67) = 30.85 𝑘𝑁

𝑐) (𝑀𝐹)𝑚𝑎𝑥(𝑇)

= (5) (1.98) (0.5) (9) + (18) (1.98) = 80.24 𝑘𝑁. 𝑚

الطيور على أشكالها تقع.

Dec 2015




1 0.67

0.33

-0.33

1 1.33

0.67 0.33

0.67 0.33 0.33

2.68

-1.32

BCV

DM

BV

EV

6-39. Draw the influence lines for (a) the moment at D in the

girder, and (b) the shear in panal BC.

Solution:

a)

Moment at D:


+ ∑ 𝑀𝐷 = 0

−𝑀𝐷 + 𝐸𝑦 (4) = 0

𝑀𝐷 = 𝐸𝑦 (4)

b)



+ ∑ 𝐹𝑦 = 0

𝑉𝐵𝐶 + 𝐸𝑦 = 0

𝑉𝐵𝐶 = − 𝐷𝑦


+ ∑ 𝐹𝑦 = 0

−𝑉𝐸𝐹 − 𝐵𝑦 = 0

𝑉𝐸𝐹 = 𝐵𝑦

لو لم تكن أزءا من الحل، فعلى

األقل ال تكن أزءا من المشكلة.

Dec 2015




Influence Lines For Trusses

قوى Section Method( أو )Joint Methodيمكن استخدام أي من طريقتي ) * ( إليجاد ال

المطلوبة، وكذلك يمكن الدمج بينهما. Membersفي الي

( لتحديد Equations( أو )Tabulate Valuesيمكن تطبيق إحدى الطريقتين السابقتين ) *

. Influence Linesرسم المطلوبة ومن ثم Members قيمة القوي في الي

بدء Support Reactionsللي ) Influence Linesأحيانا يكون من األفضل رسم * بل ال ( ق

غالبييا مييا تكييون Members( ألن معييادالت القييوي فييي الييي Influence Linesفييي إيجيياد )

بداللتها.

ئة Influence Lines* الي ها علي هي لي )أي أن التي سنتعامل معها ستكون من الدرأة األو

(.Curvesخطوط وليست

( إليجاد Maximum influence at a pointخطييوات )* يتيم اتبيياع نفييس

(Maximum influence in a member.)

ليست الحقيقة دوما التي

يموت من أألها الناس.

(c)

Dec 2015




Example 6-15

Draw the influence line for the force

in the member GB of the bridge truss

shown in Fig. (a)

Solution:

Tabulate Values. Here each successive

joint at the bottom cord is loaded with a unit

load and the force in member GB is

calculated using the method of sections,

Fig, (b).

For example, placing the unit load at x = 6m

(joint B), the support reaction at E is

calculated first, Fig.(a), then passing a

section through HG, GB, BC and isolating

the right segment, the force in GB is

determined, Fig. (c).

In the same manner, we can determine the

other values listed in the table.

حياتهم محاولين معظم الناس يقضون

إظهار محاسن ليست فيهم.

Dec 2015




Example 6-16

Draw the influence line for the

force in the member CG of the

bridge truss shown in Fig. (a)

Solution:

Tabulate Values.

A table of unit-load position at the joints of the bottom cord

versus the force in member CG is shown in Fig. (b).

These values are easily obtained by isolating joint C, Fig. (c).

Here it’s seen that CG is a zero-force member unless the unit

load is applied at joint C, in which FCG = 1 (T).

من السيء أدا أن يستوي وأودك

وعدمه عند بعض المقربين منك.

Dec 2015




Example 6-17

Determine the largest force that

can be developed in member BC

of the bridge truss shown in Fig.

(a) due to a moving force of 100 kN and a moving distributed

load of 7.5 kN / m. The loading is applied at the top cord.

Solution:

Tabulate Values. A table of unit-load position x at the joints along

the top cord versus the force in member BC is shown in Fig. (b).

The method of sections can be used for the calculations. For

example, when the unit load is at joint I (x = 8m). Fig (a), the

reaction is Ey determined first (Ey = 0.25).

Then the truss is sectioned through BC, HB,

and HI, and the right segment is isolated Fig.

(c), obtain FBC by summing moments about

point H to eliminate FHI and FHB. In similar

manner we can determine the other values in

Fig. (b).

Concentrated Live Force. The largest force in

member BC occurs when the moving force of 100

kN is placed at x = 16m. Thus,

𝐹𝐵𝐶 = (1.33) ∗ (100) = 133 𝑘𝑁

Distributed Live Load. The uniform live load must

be placed over the entire deck of the truss to create

the largest tensile force in BC.

𝐹𝐵𝐶 = [1

2∗ (32) ∗ (1.33)] ∗ 7.5 = 160 𝑘𝑁

Total Maximum Force

(𝐹𝐵𝐶)𝑚𝑎𝑥 = 133 𝑘𝑁 + 160 𝑘𝑁 = 293 𝑘𝑁

من النادر أن نفكر فيما نملك،

األغلب أننا مشغولون بما ال نملك.

Dec 2015




1.00 0.67 0.33

-0.77

-0.38

yA

GFF

Solution:


+ ∑ 𝑀𝐵 = 0

𝐴𝑦 (6) + 𝐹𝐺𝐹 (5.2) = 0

𝐹𝐺𝐹 = − 𝐴𝑦 (1.15)

نحن مجانين إذا لم نستطع أن

نفكر، متعصبون إذا لم نرد أن

نفكر. نفكر، عبيد إذا لم نجرؤ أن

6-45. Draw the influence lines for the force in member GF of the

Warren truss. Indicate numerical values for the peaks. All members

have the same length.

Dec 2015




FCF

yA

yD

0.33 0.67

100

1.00 0.67 0.33

-0.38

0.38

6-46. Draw the influence lines for the force in member FC

of the Warren truss. Indicate numerical values for the

peaks. All members have the same length.

Solution:


+ ∑ 𝐹𝑦 = 0

𝐷𝑦 – 𝐹𝐹𝐶 𝑠𝑖𝑛 60° = 0

𝐹𝐹𝐶 = 1.15 𝐷𝑦

Unit load @: 12 ≤ x ≤ 18 m

+ ∑ 𝐹𝑦 = 0

𝐴𝑦 + 𝐹𝐹𝐶 𝑠𝑖𝑛 60° = 0

𝐹𝐹𝐶 = −1.15 𝐴𝑦

نقضي عشرين عاما لبناء سمعة

أيدة، خمس دقائق تكفي لهدمها.

Dec 2015




Solution:


+ ∑ 𝐹𝑦 = 0

𝐷𝑦 – 𝐹𝐷𝐸 𝑠𝑖𝑛 60° = 0

𝐹𝐷𝐸 = 1.15 𝐷 𝑦 (𝐶)

Unit load @: x = 12 m

𝐹𝐷𝐸 = 0

هناك شيئان النهائيان: الكون

وغباء اإلنسان، وبالنسبة للكون

فأنا ما زلت غير متأكد تماما.

Dec 2015




Solution:

Suppose unit load between A, C:

:بدراسة الجزء األيمن المبين بالشكل*

+ 𝐹𝑦 = 0

𝐵𝑦– (𝐹𝐶𝐽 ∗4

5) = 0

𝐹𝐶𝐽 = 1.25 𝐺𝑦

Suppose unit load between D, J:

:المبين بالشكل سربدراسة الجزء األي*

+ 𝐹𝑦 = 0

𝐴𝑦 + (𝐹𝐶𝐽 ∗ 4

5) = 0

𝐹𝐶𝐽 = − 1.25 𝐴𝑦

حيييث يمكيين اسييتخدام المعييادلتين فييي رسييم *

influence line كما بالشكل:

Maximum Force:

compressionي واضح من الشكل ان مساحة ال

. tension يأكبر من مساحة ال

(𝐴𝑟𝑒𝑎)𝑐 =1

2 ∗ (1.8 + 9) ∗ 0.625

= 3.375 𝑚2

𝐹𝑚𝑎𝑥𝑖𝑚𝑢𝑚 = 3.375 ∗ 3

= 10.125 𝑘𝑁

يتذكر الناس الذي فعلته،

وليس بالضرورة ما قلته.

Dec 2015




1.00

1.00

0.5

0.5

0.83

0.83

0.67

0.67

-1.78 -1.33

-0.89

yA

yG

JIF

يتطلب األمر أدلة كثيرة إلثبات ذكائك،

لكن يكفي دليل واحد إلثبات عكس ذلك.

6-49. Draw the influence lines for the force in member JI.

Solution:


+ ∑ 𝑀𝐸 = 0

𝐺𝑦 (8) + 𝐹𝐽𝐼 (3) = 0

𝐹𝐽𝐼 = − 𝐺𝑦 (2.67)

Unit load @: 12 ≤ x ≤ 24 m

+ ∑ 𝑀𝐸 = 0

𝐴𝑦 (16) + 𝐹𝐽𝐼 (3) = 0

𝐹𝐽𝐼 = − 𝐴𝑦 (5.33)

Dec 2015




1.00 0.83

0.67 0.5 0..33 0.17

-1.38 -1.11

-0.83

yA

ALF

6-50. Draw the influence lines for the force in member AL.

Solution:

Joint A:

+ ∑ 𝐹𝑦 = 0

𝐴𝑦 + 𝐹𝐴𝐿 3

5= 0

𝐹𝐴𝐿 = − 1.67 𝐴𝑦

يخدع المظهر األول الكثيرين.

Dec 2015




Solution:


+ ∑ 𝑀𝐶 = 0

(−𝐹𝐻𝐺 ∗ 5) + (𝐸𝑦 ∗ 10) = 0

𝐹𝐻𝐺 = 2 𝐸𝑦 (𝐶)

Unit load @: 10 ≤ x ≤ 20 m

+ ∑ 𝑀𝐶 = 0

(𝐹𝐻𝐺 ∗ 5) − (𝐴 ∗ 10) = 0

𝐹𝐻𝐺 = 2 𝐴𝑦 (𝐶)

(𝐹𝐻𝐺)𝑚𝑎𝑥(𝐶)= (

1

2∗ 20 ∗ 10) ∗ 16

= 160 𝑘𝑁 (𝐶)

(𝐹𝐻𝐺)𝑚𝑎𝑥(𝑇)= 0

يرى الحاسد زوال نعمتك نعمة عليه.

Dec 2015




Solution:


+ ∑ 𝐹𝑦 = 0

𝐴𝑦 – 𝐹𝐻𝐶 5

5√2 = 0

𝐹𝐻𝐶 = 1.414 𝐴𝑦

Unit load @: 10 ≤ x ≤ 20 m

+ ∑ 𝐹𝑦 = 0

𝐷𝑦 + 𝐹𝐻𝐶 5

5√2 = 0

𝐹𝐻𝐶 = −1.414 𝐷𝑦

a)

(𝐹𝐻𝐶)𝑚𝑎𝑥(𝑇)= (16)(20 − 6.67)(0.5) (0.77)

= 82.11 𝑘𝑁

b)

(𝐹𝐻𝐶)𝑚𝑎𝑥(𝐶)= (16)(6.67)(0.5) (−0.35)

= −18.68 𝑘𝑁

1.00 0.5

1.00 0.5

0.77

0.35-

0.35 HCF

yE

yA

يسخر من الجروح من ال يعرف األلم .

6-52. Draw the influence line for the force in member HC, then

determine the maximum live force (tension or compression) that can

be developed in this member due to a uniform live load of 16 kN/m

that acts on the bridge deck along the bottom cord of the truss.

Dec 2015




1.00 0.5 0.75

-1.06 -0.707

6-53. Draw the influence line for the force in member AH, then determine

the maximum live force (tension or compression) that can be developed in

this member due to a uniform live load of 16 kN/m that acts on the bridge

deck along the bottom cord of the truss.

yA

AHF

Solution:

Joint A:

+ ∑ 𝐹𝑦 = 0

𝐴𝑦 + 𝐹𝐴𝐻 5

5√2 = 0

𝐹𝐴𝐻 = − 1.414 𝐴𝑦

a)

(𝐹𝐴𝐻)𝑚𝑎𝑥(𝐶)

= (16)(−1.06)(0.5)(20) = 169.6 𝑘𝑁

(𝐹𝐴𝐻)𝑚𝑎𝑥(𝑇)= 0

يصبح الجيد غير أيد، إذا

كان األأود هو المتوقع.

Dec 2015




Solution:


:Gعند GFFندرس الجزء األيمن ونحلل *

+ ∑ 𝑀𝐶 = 0

(𝐸𝑦 ∗ 6) − (𝐹𝐺𝐻 𝑠𝑖𝑛 𝜃 ∗ 6) = 0

𝐹𝐺𝐹 = 2.24 𝐸𝑦 (𝐶)

Unit load @: 9 ≤ x ≤ 12m

:Gعند GFFندرس الجزء األيسر ونحلل *

+ ∑ 𝑀𝐶 = 0

(𝐹𝐺𝐹 𝑐𝑜𝑠 𝜃 ∗ 3) − (𝐴𝑦 ∗ 6) = 0

𝐹𝐺𝐹 = 2.24 𝐴𝑦 (𝐶)

GFمييين المعيييادلتين يمكييين رسيييم القيييوة فيييي *

عنييدما تكييون Compressionواسييتنتاأ أن أكبيير

60 kN عند نقطةC .

(𝐹𝐺𝐹)𝑚𝑎𝑥 = 1.12 ∗ 60

= 67.2 𝑘𝑁 (𝐶)

عدوك، سيشعر بتفاهته.عندما تحب

Dec 2015




Maximum Influence at a point due to a series of

Concentrated Loads

Once the influence line of a function has

been established for a point in a

structure, the maximum effect caused

by a live concentrated force is

determined by multiplying the peak

ordinate of the influence line by the

magnitude of the force. In some cases,

however, several concentrated forces

must be placed on the structure. An

example would be the wheel loadings of

a truck or a train. In order to determine

the maximum effect in this case, either a

trial-and-error procedure can be used or a method that is based on the change in

the function that takes place as the load is moved. Each of these methods will now

be explained specifically as it applies to shear and moment.

Shear. Consider the simply supported beam with the associated influence line for

the shear at point C in Fig. 6-27a. The maximum positive shear at point C is to be

determined due to the series of concentrated (wheel) loads which move from right

to left over the beam. The critical loading will occur when one of the loads is placed

just to the right of point C, which is coincident with the positive peak of the

influence line. By trial and error each of three possible cases can therefore be

investigated, Fig. 6-27b. We have

Case 1: (VC)1 = 4.5 (0.75) + 18 (0.625) + 18 (0.5) = 23.63 kN

Case 2: (VC)2 = 4.5 (-0.125) + 18 (0.75) + 18 (0.625) = 24.19 kN

Case 3: (VC)3 = 4.5 (0) + 18 (-0.125) + 18 (0.75) = 11.25 kN

Case 2, with the 4.5-kN force located 1.5+ m from the left support, yields the largest

value for VC and therefore represents the critical loading. Actually investigation of

Case 3 is unnecessary, since by inspection such an arrangement of loads would

yield a value of (VC)3 that would be less than (VC)2

اغفر ألعدائك فلن يغيظهم شيء أقوى من هذا.

Dec 2015




عندما تمطرالسماء

تعيش األرض

Dec 2015




When many concentrated loads act on the span, as in the case of the E-72 load of

Fig. 1-11, the trial-and-error computations used above can be tedious. Instead, the

critical position of the loads can be determined in a more direct manner by finding

the change in shear ΔV, which occurs when the loads are moved from Case 1 to

Case 2, then from Case 2 to Case 3, and so on. As long as each computed ΔV is

positive, the new position will yield a larger shear in the beam at C than the

previous position. Each movement is investigated until a negative change in shear is

computed. When this occurs, the previous position of the loads will give the critical

value. The change in shear ΔV for a load P hat moves from position x1 to x2 over a

beam can be determined by multiplying P by the change in the ordinate of the

influence line, that is, (y2 - y1). If the slope of the influence line is s, then

(y2 – y1) = s (x2 – x1), and therefore

(6-1)

If the load moves past a point where there is a discontinuity or “jump” in the

influence line, as point C in fig. 6-27a, then the change in shear is simply

(6-2)

Use of the above equations will be illustrated with reference to the beam, loading,

and influence line for VC, shown in Fig. 6-28a. Notice that the magnitude of the

slope of the influence line is s = 0.75 / (12 – 3) = 0.75 / 9 = 0.0833, and the jump C

has a magnitude of 0.75 + 0.25 = 1.

Consider the loads of Case 1 moving 1.5-m to Case 2, Fig. 6-28b. When this occurs,

the 4.5-kN load jumps down (-1) and all the loads move up the slope of the influence

line. This causes a change of shear,

ΔV1-2 = 4.5 (-1) + [4.5 +18 +18] (0.0833) (1.5) = +0.563 kN

Since ΔV1-2 is a positive, Case 2 will yield a larger value for VC than Case 1.

[Compare the answers for (VC)1 and (VC)2 previously computed, where indeed (VC)2

= (VC) 1 + 0.563] Investigating ΔV2-3, which occurs when Case 2 moves to Case 3,

Fig. 6-28b, we must account for the downward (negative) jump of the 18-kN load

and the 1.5-m horizontal movement of all the loads up the slope of the influence.

We have

ΔV2-3 = 18 (-1) + [4.5 +18 +18] (0.0833) (1.5) = +12.94 kN

Since ΔV2-3 is negative, Case 2 is the position of the critical loading, as

determined previously.

)1x – 2x(s P = VΔ Sloping line

)1y – 2y( P = VΔ Jump

تبدو األمور أفضل كثيرا إذا

نظرنا إليها من الطرف اآلخر.

Dec 2015




تكمن أعلى درأات االمتياز في

قتال.كسر مقاومة عدوك دون

Dec 2015




Moment. We can also use the foregoing methods to determine the critical

position of a series of concentrated forces so that they create the largest

internal moment at a specific point in a structure. Of course, it’s the first

necessary to draw the influence line for the moment at the point and

determine the slope x of its line segments.

For a horizontal movement (x2 – x1) of concentrated force P, the change in

moment, ΔM, is equivalent to the magnitude of the force times the change in

the influence-line ordinate under the load, that is,

As an example, consider the beam, loading, and influence line for the

moment at point C in Fig. 6-29a. If each of the three concentrated forces is

placed on the beam, coincident with the peak of the influence line, we will

obtain the greatest influence from each force. The three cases of loading are

shown in Fig. 6-29b. When the loads of Case 1 are moved 1.2m to the left to

Case 2, it’s observed that the 9-kN load decreases ΔM1-2, since the slope

(7.5) is downward, Fig. 6-29a. Likewise, the 18-kN and 13.5 kN forces cause

an increase of ΔM1-2, since the slope [2.25 / (12 – 3)] is upward. We have.

∆𝑀1−2 = −9 (2.25

3) (1.2) + (18 + 13.5) (

7.5

12 − 3) (1.2) = 22.38 𝑘𝑁. 𝑚

Since ΔM1-2 is positive, we must further investigate moving the loads 1.8m

from Case 2 to Case 3.

∆𝑀2−3 = −(9 + 18) (2.25

3) (1.8) + 13.5 (

7.5

12 − 3) (1.8) = 30.38 𝑘𝑁. 𝑚

Here the change is negative, so the greatest moment at C will occur when the

beam is loaded as shown in Case 2, Fig.6-29c. The maximum moment at C is

therefore

(𝑀𝐶)𝑚𝑎𝑥 = 9 (1.35) + 18 (2.25) + 13.5 (1.8) = 77.0 𝑘𝑁. 𝑚

)1x – 2x(s P M =Δ Sloping line

ربما تضطر لخوض المعركة

أكثر من مرة لكسبها.

Dec 2015




The following examples further illustrate this method.

كل إنسان يعتقد أنه هو الذي يمتلك

ما يعتقد، الحقيقة، وأن الحق هو

وهذا هو سبب نزاعات العالم.

Dec 2015




Example 6-18

Determine the maximum

positive shear created at

point B in the beam shown

in Fig. (a) due to the wheel

loads of the moving truck.

Solution:

The influence line for the shear at B is shown in

Fig. (b)

ال يمكن لمعاهدة أن تدوم إال إذا انتصر فيها الطرفان.

𝟎. 𝟗 − 𝒎 Movement of 18-kN Load. Imagine that the 18-Kn load acts just

to the right of point 𝐵 so that we obtain its maximum positive influence.

Since the beam segment 𝐵𝐶 is 3 𝑚 long, the 45 − 𝑘𝑛 load is not as yet on

the beam. When the truck moves 0.9 𝑚 to the left, the 18 − 𝑘𝑁 load jumps

downward on the influence line 1 unit and the 18 − 𝑘𝑁, 40.5 − 𝑘𝑁, and

67.5 − 𝑘𝑁 loads create a positive increase in ∆𝑉𝐵, since the slope is upward

to the left. Although the 45 − 𝑘𝑁 load also moves forward 0.9 m, it is still

not on the beam. Thus,

∆𝑉𝐵 = 18(−1) + (18 + 40.5 + 67.5) (0.5

3) 0.9 = +0.9

1.8-m Movement of 40-5-kN Load. When the 40.5-kN load acts just to the

right of 𝐵, and then the truck moves 1.8 𝑚 to the left, we have

∆𝑉𝐵 = 40.5(−1) + (18 + 40.5 + 67.5) (0.5

3) (1.8) + 45 (

0.5

3) (1.2)

= +6.3 𝑘𝑁

Note in the calculation that the 45-kN load only moves 1.2 m on the beam

Dec 2015




1.8-m Movement of 67.5-kN Load. If the 67.5-kN load is positioned just to

the right of B and then the truck moves 1.8 m to the left, the 18-kN load

moves only 1 ft until it’s off the beam, and likewise the 40.5-kN load moves

only 1.2 m until it’s off the beam. Hence,

Since ΔVB is now negative, the correct position of the loads occurs when the

67.5-kN is just to the right of point B, Fig. 6-30©.

Consequently,

In practices we should consider motion of the truck from left to right and then

choose the maximum value between these two situations.

لن يمكنك االنتصار إذا كان

بمقدورك قبول الهزيمة.

18-6تابع حل مثال

∆𝑉𝐵 = 67.5(−1) + 18 (0.5

3) (0.3) + 40.5 (

0.5

3) (1.2) + (67.5 + 45) (

0.5

3) (1.8)

= 24.8 𝑘𝑁

(∆𝑉𝐵)𝑚𝑎𝑥 = 18(−0.05) + 40.5(−0.2) + 67.5(0.5) + 45(0.2)

= 33.8 𝑘𝑁

Dec 2015




Example 6-19

نصف النصر أن تبقى محتفظا

بأعصابك، ال تخبر بذلك عدوك.

Determine the maximum positive moment

created at point 𝐵 in the beam shown in Fig. a

due to the wheel loads of the crane.

:Solution

The influence line for the moment at 𝐵 is shown in Fig. 6-31b.

2 − 𝑚 Movement of 3 − 𝑘𝑁 Load. If the 3-kN load is assumed to

act at 𝐵 and then moves 2 𝑚 to the right, Fig. 6-31b, the change in

moment is

∆𝑀𝐵 = −3 (1.20

3) (2) + 8 (

1.20

2) (2) = 7.20 𝑘𝑁. 𝑚

Why is the 4-kN load not included in the calculations?

3 − 𝑚 Movement of 8-kN load. If the 8-kN load is assumed to act at

𝐵 and then moves 3 𝑚 to the right, the change in moment is

∆𝑀𝐵 = −3 (1.20

3) (3) − 8 (

1.20

2) (3) + 4 (

1.20

2) (2)

= −8.40 𝑘𝑁. 𝑚

Notice here that the 4-kN load was initially 1 m off the beam, and

therefore moves only 2 𝑚 on the beam.

Since there is a sign change in ∆𝑀𝐵 , the correct position of the loads

for maximum positive moment at B occurs when the 8-kN force is at

𝐵, Fig. 6-31 b. Therefore,

(𝑀𝐵)𝑚𝑎𝑥 = 8(1.20) + 3(0.4) = 10.8 𝑘𝑁. 𝑚

Dec 2015




Example 6-20

القيادة هي أن تجعل شخصا أخر يقوم

بعمل تريده أنت ألنه يريد ذلك.

Determine the maximum compressive force developed in member 𝐵𝐺 of

the side truss in Fig. a due to the right side wheel loads of the car and

trailer. Assume the loads are applied directly to the truss and move only to

the right.

Solution:

The influence line for the force in member BG is shown in Fig. 6-32b. Here a

trial-and-error approach for the solution will be used. Since we want the greatest

negative (compressive) force in 𝐵𝐺, we begin as follows:

1.5-kN Load at Point C. In this case

𝐹𝐵𝐺 = 1.5 𝑘𝑁 (−0.625) + 4(0) + 2 𝑘𝑁 (0.3125

3𝑚) (1 𝑚)

= −0.729 𝑘𝑁

4 − 𝑘𝑁 Load at Point C. By inspection this would seem a more reasonable case

than the previous one.

𝐹𝐵𝐺 = 4 𝑘𝑁(−0.625) + 1.5 𝑘𝑁 (−0.625

6𝑚) (4𝑚) + 2 𝑘𝑁(0.3125)

= −2.50 𝑘𝑁

2-kN Load at Point C. In this case all loads will create compressive force in 𝐵𝐶.

𝐹𝐵𝐺 = 2𝑘𝑁(−0.625) + 4 𝑘𝑁 (−0.625

6 𝑚) (3𝑚) + 1.5 𝑘𝑁 (

−0.625

6𝑚) (1𝑚)

= −2.66 𝑘𝑁

Since this final case results is the largest answer, the critical loading occurs

when the 2-kN load is at C.

Dec 2015




Example 6-21


القائد الماهر هو شخص

يعلم متى يستطيع القتال

ومتى ال يستطيع ذلك.

Solution:

The magnitude and position of the resultant force of the system are

determined first, Fig. a. We have

+ ∑ 𝐹𝑅 = ∑ 𝐹 ; 𝐹𝑅 = 8 + 6 + 4 = 18 𝑘𝑁

+ 𝑀𝑅𝐶 = ∑ 𝑀𝐶 ; 18�� = 6(3) + 4(4.5)

�� = 2𝑚 Let us first assume the absolute maximum moment occurs under the 6-kN

load. The load and the resultant force are positioned equidistant from the

beam's centerline, Fig. b. Calculating 𝐴𝑦 first, Fig. b, we have

+ 𝑀𝐵 = 0; −𝐴𝑦(9) + 1.8(5) = 0 𝐴𝑦 = 10 𝑘𝑁

Now using the left section of the beam, Fig. 6-37 c, yields

+ ∑ 𝑀𝑆 = 0; −10(5) + 8(3) + 𝑀𝑆 = 0

𝑀𝑆 = 26.0 𝑘𝑁. 𝑚

Determine the absolute maximum moment in the simply supported

bridge deck shown in Fig. a.

Dec 2015




التردد والمماطلة والتأأيل أشبه

ببطاقات االئتمان: لذيذ استعمالها

إلى حين وصول الفاتورة.


There is a possibility that the absolute maximum moment may occur

under the 8-kN load, since 8 kN > 6 kN and 𝑭𝑹 is between both 8 kN and 6

kN. To investigate this case, the 2-k load and 𝑭𝑹 are positioned equidistant

from the beam's centerline, Fig. d. Show that 𝐴 = 7 𝑘𝑁 as indicated in Fig.

6-37 e and that

𝑀𝑆 = 24.5 𝑘𝑁. 𝑚

By comparison, the absolute maximum moment is

𝑀𝑆 = 26.0 𝑘𝑁. 𝑚

Which occurs under the 1.5-k load, when the loads are positioned on the

beam as shown in Fig. b.

Dec 2015




1.00 0. 5

2.5

-2.5

Solution:

+ ∑ 𝑀𝐶 = 0

− 𝑀𝐶 + 𝐵𝑦 (5) = 0

𝑀𝐶 = 𝐵𝑦 (5)

(𝑀𝐶)𝑚𝑎𝑥

= (14.71)(2.5) + (4.9)(1.5)

= 44.13 𝑘𝑁. 𝑚

19.62

4.9 14.71

1.5

CM

6-58. Determine the maximum live moment at point C on the

single girder caused by the moving dolly that has a mass of 2

Mg and a mass center at G. Assume A is a roller.

0. 5-

yB

التسويف هو المباراة بين األلم

للقيام بالمهمة ولذة الحالي

تأخيرها بألم أكبر مستقبال

Dec 2015




Solution:

:Getting influence lines of C

(A, C)بين unit loadبفرض وأود *

:(CB)ودراسة الجزء األيمن

+ ∑ 𝐹𝑦 = 0

𝑉𝐶 + 𝐵𝑦 = 0

𝑉𝐶 = −𝐵𝑦

+ ∑ 𝑀𝐶 = 0

−𝑀𝐶 + 𝐵𝑦 + 12 = 0

𝑀𝐶 = 12 𝐵𝑦

(C, B)بين unit load وأود* وبفرض

:(AC)ودراسة الجزء األيمن

+ ∑ 𝐹𝑦 = 0

𝐴𝑦 − 𝑉𝐶 = 0

𝑉𝐶 = 𝐴𝑦

+ ∑ 𝑀𝐶 = 0

𝑀𝐶 − 𝐴𝑦 ∗ 6 = 0

𝑀𝐶 = 6 𝐴𝑦

a)

(𝑉𝐶)𝑚𝑎𝑥𝑖𝑚𝑢𝑚 =(36 ∗ 0.667) + (0.417 ∗ 18)

2

= 15.76 𝑘𝑁

b)

(𝑀𝐶)𝑚𝑎𝑥𝑖𝑚𝑢𝑚 =(36 ∗ 4) + (18 ∗ 2.5)

2

= 94.5 𝑘𝑁. 𝑚

الكسل هو عادة االستراحة

قبل أن تتعب.

Dec 2015




1.00

1.00

0.5

0.5

100 25

100

5 2

4

6-61. Determine the maximum live moment at point C

on the bridge caused by the moving load.

CM

yA

yB

Solution:

+ ∑ MC = 0

− 𝑀𝐶 + 𝐵𝑦 (10) = 0

𝑀𝐶 = 𝐵𝑦 (10)

(𝑀𝐶)𝑚𝑎𝑥 = (100)(5) + (25)(4) + (100)(2)

= 800 𝑘𝑁. 𝑚

إذا كنت كسوال في سن العشرين

فستستجدي في سن األربعين.

Dec 2015




أكيد برئ

: لقد حكمت ببراءتك من القاضي للمتهم

النقود!تهمة سرقة

: وهل سأحتفظ بالنقود؟المتهم

البنت فقط

: أريد أتزوأ من ابنتك يا عمي؟الشاب

: وهل تستطيع تنفق على عائلة؟األب

أنا أريد ابنتك فقط!الشاب:

كالم صدأ!

ربما ندم من لم يتزوأ، لكن من تزوأ ال

مناص له من الندم!

احتفاال بعيد الزواأ!

عيد زواأنا، ماذا نفعل؟: اليوم يوافق الزوأة

: لنقف دقيقة حداد!الزوأ

المسامح كريم

: لهبشدة زوأته الزوأ بعد إساءة

سأمنحك ساعة كاملة لتقديم اعتذارك؟

: وماذا لو لم أعتذر؟الزوأة بعنف

ى، : ال شيء، سأمنحك ساعة أخرالزوأ

أو كأن شيئا لم يكن!

كبت

: مما يشكو زوأك!الطبيب للزوأة

ة، الحظت أنه بدأ يتكلم بعدما تزوأته بسن :الزوأة

وهو نائم، واألمر يتطور بشكل مستمر؟

: دعيه يتكلم بالنهار، وسيعود لحالته بالطبي

الطبيعية مرة أخرى!

أعرف ما بداخله

البرتقال قبل أن ال تقشر : لماذا األول

تأكله؟

: ألني أعرف ما بداخله!الثاني

عملية خطيرة

بساذأ مع مسلحين بالخطأ أمسكت المخابرات

: ما هي آخر عملية سويتها؟المحقق

ز!: اللو البرئ

أرسل حلك من إيميلك الخاص على الموقع واحصل على دينار لكل فزورة

بقى درأة واحدة، وإذا صعدنا كل ثالث درأات دفعة تواحدة إذا صعدنا سلما كل درأتين دفعة

، وإذا صعدنا كل درأات بقى ثالثتصعدنا كل أربع درأات دفعة واحدة بقى درأتان، وإذاتواحدة

بقى ت دفعة واحدة كل ست درأات ، وإذا صعدنادرأات بقى أربعت دفعة واحدة خمس درأات

السلم؟ ال يبقى شيء، فكم عدد درأات دفعة واحدة ، وإذا صعدنا كل سبع درأاتدرأات خمس

Documents

Week 6 Influence Lines-Part 2 - موقع المهندس حمادة شعبان · 2017-11-18 · Influence Lines-Part 2 Week 6 ... Influence Lines For Floor Girders ىييلإ لاوأ